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Mathematics LibreTexts

9.6: Solve Applications of Quadratic Equations

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Learning Objectives

By the end of this section, you will be able to:

  • Solve applications modeled by quadratic equations

Before you get started, take this readiness quiz.

  • The sum of two consecutive odd numbers is \(−100\). Find the numbers. If you missed this problem, review Example 2.18.
  • Solve: \(\frac{2}{x+1}+\frac{1}{x-1}=\frac{1}{x^{2}-1}\). If you missed this problem, review Example 7.35.
  • Find the length of the hypotenuse of a right triangle with legs \(5\) inches and \(12\) inches. If you missed this problem, review Example 2.34.

Solve Applications Modeled by Quadratic Equations

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.

Let’s first summarize the methods we now have to solve quadratic equations.

Methods to Solve Quadratic Equations

  • Square Root Property
  • Completing the Square
  • Quadratic Formula

As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.

Use a Problem-Solving Strategy

  • Read the problem. Make sure all the words and ideas are understood.
  • Identify what we are looking for.
  • Name what we are looking for. Choose a variable to represent that quantity.
  • Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
  • Solve the equation using algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is \(2\) more than the number preceding it. If we call the first one \(n\), then the next one is \(n+2\). The next one would be \(n+2+2\) or \(n+4\). This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.

\(\begin{array}{cl}{}&{\text{Consecutive even integers}}\\{}& {64,66,68}\\ {n} & {1^{\text { st }} \text { even integer }} \\ {n+2} & {2^{\text { nd }} \text { consecutive even integer }} \\ {n+4} & {3^{\text { rd }} \text { consecutive even integer }}\end{array}\)

\(\begin{array}{cl}{}&{\text{Consecutive odd integers}}\\{}& {77,79,81}\\ {n} & {1^{\text { st }} \text { odd integer }} \\ {n+2} & {2^{\text { nd }} \text { consecutive odd integer }} \\ {n+4} & {3^{\text { rd }} \text { consecutive odd integer }}\end{array}\)

Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

Example \(\PageIndex{1}\)

The product of two consecutive odd integers is \(195\). Find the integers.

Step 1 : Read the problem

Step 2 : Identify what we are looking for.

We are looking for two consecutive odd integers.

Step 3 : Name what we are looking for.

Let \(n=\) the first odd integer.

\(n+2=\) the next odd integer.

Step 4 : Translate into an equation. State the problem in one sentence.

“The product of two consecutive odd integers is \(195\).” The product of the first odd integer and the second odd integer is \(195\).

Translate into an equation.

\(n(n+2)=195\)

Step 5 : Solve the equation. Distribute.

\(n^{2}+2 n=195\)

Write the equation in standard form.

\(n^{2}+2 n-195=0\)

\((n+15)(n-13)=0\)

Use the Zero Product Property.

\(n+15=0 \quad n-13=0\)

Solve each equation.

\(n=-15, \quad n=13\)

There are two values of \(n\) that are solutions. This will give us two pairs of consecutive odd integers for our solution.

\(\begin{array}{cc}{\text { First odd integer } n=13} & {\text { First odd integer } n=-15} \\ {\text { next odd integer } n+2} & {\text { next odd integer } n+2} \\ {13+2} & {-15+2} \\ {15} & {-13}\end{array}\)

Step 6 : Check the answer.

Do these pairs work? Are they consecutive odd integers?

\(\begin{aligned} 13,15 & \text { yes } \\-13,-15 & \text { yes } \end{aligned}\)

Is their product \(195\)?

\(\begin{aligned} 13 \cdot 15 &=195 &\text{yes} \\-13(-15) &=195 & \text { yes } \end{aligned}\)

Step 7 : Answer the question.

Two consecutive odd integers whose product is \(195\) are \(13,15\) and \(-13,-15\).

Exercise \(\PageIndex{1}\)

The product of two consecutive odd integers is \(99\). Find the integers.

The two consecutive odd integers whose product is \(99\) are \(9, 11\), and \(−9, −11\).

Exercise \(\PageIndex{2}\)

The product of two consecutive even integers is \(168\). Find the integers.

The two consecutive even integers whose product is \(128\) are \(12, 14\) and \(−12, −14\).

We will use the formula for the area of a triangle to solve the next example.

Definition \(\PageIndex{1}\)

Area of a Triangle

For a triangle with base, \(b\), and height, \(h\), the area, \(A\), is given by the formula \(A=\frac{1}{2} b h\).

Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.

Recall that when we solve geometric applications, it is helpful to draw the figure.

Example \(\PageIndex{2}\)

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of \(120\) square feet and the architect wants the base to be \(4\) feet more than twice the height. Find the base and height of the window.

Exercise \(\PageIndex{3}\)

Find the base and height of a triangle whose base is four inches more than six times its height and has an area of \(456\) square inches.

The height of the triangle is \(12\) inches and the base is \(76\) inches.

Exercise \(\PageIndex{4}\)

If a triangle that has an area of \(110\) square feet has a base that is two feet less than twice the height, what is the length of its base and height?

The height of the triangle is \(11\) feet and the base is \(20\) feet.

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

We will use the formula for the area of a rectangle to solve the next example.

Definition \(\PageIndex{2}\)

Area of a Rectangle

For a rectangle with length, \(L\), and width, \(W\), the area, \(A\), is given by the formula \(A=LW\).

Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.

Example \(\PageIndex{3}\)

Mike wants to put \(150\) square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than \(3\) times the width. Find the length and width. Round to the nearest tenth of a foot.

Exercise \(\PageIndex{5}\)

The length of a \(200\) square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.

The length of the garden is approximately \(18\) feet and the width \(11\) feet.

Exercise \(\PageIndex{6}\)

A rectangular tablecloth has an area of \(80\) square feet. The width is \(5\) feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot?

The length of the tablecloth is approximately \(11.8\) feet and the width \(6.8\) feet.

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

Definition \(\PageIndex{3}\)

Pythagorean Theorem

  • In any right triangle, where \(a\) and \(b\) are the lengths of the legs, and \(c\) is the length of the hypotenuse, \(a^{2}+b^{2}=c^{2}\).

Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.

Example \(\PageIndex{4}\)

Rene is setting up a holiday light display. He wants to make a ‘tree’ in the shape of two right triangles, as shown below, and has two \(10\)-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

Exercise \(\PageIndex{7}\)

The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is \(20\) feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.

The length of the flag pole’s shadow is approximately \(6.3\) feet and the height of the flag pole is \(18.9\) feet.

Exercise \(\PageIndex{8}\)

The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.

The distance between the opposite corners is approximately \(7.2\) feet.

The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, \(v_{0}\), propels the object up until gravity causes the object to fall back down.

Definition \(\PageIndex{4}\)

The height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds is given by the formula

\(h=-16 t^{2}+v_{0} t\)

We can use this formula to find how many seconds it will take for a firework to reach a specific height.

Example \(\PageIndex{5}\)

A firework is shot upwards with initial velocity \(130\) feet per second. How many seconds will it take to reach a height of \(260\) feet? Round to the nearest tenth of a second.

Exercise \(\PageIndex{9}\)

An arrow is shot from the ground into the air at an initial speed of \(108\) ft/s. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the arrow will be \(180\) feet from the ground. Round the nearest tenth.

The arrow will reach \(180\) feet on its way up after \(3\) seconds and again on its way down after approximately \(3.8\) seconds.

Exercise \(\PageIndex{10}\)

A man throws a ball into the air with a velocity of \(96\) ft/s. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the height of the ball will be \(48\) feet. Round to the nearest tenth.

The ball will reach \(48\) feet on its way up after approximately \(.6\) second and again on its way down after approximately \(5.4\) seconds.

We have solved uniform motion problems using the formula \(D=rt\) in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

Image shows the template for a table with three rows and four columns. The first column is empty. The second column is labeled “Rate.” The third column is labeled “Time.” The fourth column is labeled “Distance.” The labels are written in the equation Rate times Time equals Distance. There is one extra cell at the bottom of the fourth column.

The formula \(D=rt\) assumes we know \(r\) and \(t\) and use them to find \(D\). If we know \(D\) and \(r\) and need to find \(t\), we would solve the equation for \(t\) and get the formula \(t=\frac{D}{r}\).

Some uniform motion problems are also modeled by quadratic equations.

Example \(\PageIndex{6}\)

Professor Smith just returned from a conference that was \(2,000\) miles east of his home. His total time in the airplane for the round trip was \(9\) hours. If the plane was flying at a rate of \(450\) miles per hour, what was the speed of the jet stream?

This is a uniform motion situation. A diagram will help us visualize the situation.

Diagram first shows motion of the plane at 450 miles per hour with an arrow to the right. The plane is traveling 2000 miles with the wind, represented by the expression 450 plus r. The jet stream motion is to the right. The round trip takes 9 hours. At the bottom of the diagram, an arrow to the left models the return motion of the plane. The plane’s velocity is 450 miles per hour, and the motion is 2000 miles against the wind modeled by the expression 450 – r.

We fill in the chart to organize the information.

We are looking for the speed of the jet stream. Let \(r=\) the speed of the jet stream.

When the plane flies with the wind, the wind increases its speed and so the rate is \(450 + r\).

When the plane flies against the wind, the wind decreases its speed and the rate is \(450 − r\).

The speed of the jet stream was \(50\) mph.

Exercise \(\PageIndex{11}\)

MaryAnne just returned from a visit with her grandchildren back east. The trip was \(2400\) miles from her home and her total time in the airplane for the round trip was \(10\) hours. If the plane was flying at a rate of \(500\) miles per hour, what was the speed of the jet stream?

The speed of the jet stream was \(100\) mph.

Exercise \(\PageIndex{12}\)

Gerry just returned from a cross country trip. The trip was \(3000\) miles from his home and his total time in the airplane for the round trip was \(11\) hours. If the plane was flying at a rate of \(550\) miles per hour, what was the speed of the jet stream?

Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We’ll use a similar scenario now.

Example \(\PageIndex{7}\)

The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes \(12\) hours more than Press #2 to do the job and when both presses are running they can print the job in \(8\) hours. How long does it take for each press to print the job alone?

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

Exercise \(\PageIndex{13}\)

The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes \(6\) hours more than Press #2 to do the job and when both presses are running they can print the job in \(4\) hours. How long does it take for each press to print the job alone?

Press #1 would take \(12\) hours, and Press #2 would take \(6\) hours to do the job alone.

Exercise \(\PageIndex{14}\)

Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes \(3\) hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in \(2\) hours. How long does it take for each hose to fill the hot tub?

The red hose take \(6\) hours and the green hose take \(3\) hours alone.

Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.

  • Word Problems Involving Quadratic Equations
  • Quadratic Equation Word Problems
  • Applying the Quadratic Formula

Key Concepts

  • Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
  • Solve the equation using good algebra techniques.
  • For a triangle with base, \(b\), and height, \(h\), the area, \(A\), is given by the formula \(A=\frac{1}{2}bh\).

Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.

  • For a rectangle with length,\(L\), and width, \(W\), the area, \(A\), is given by the formula \(A=LW\).

Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.

  • The height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds is given by the formula \(h=-16 t^{2}+v_{0} t\).
  • 5.6 Quadratic Equations with Two Variables with Applications
  • Introduction
  • 1.1 Basic Set Concepts
  • 1.2 Subsets
  • 1.3 Understanding Venn Diagrams
  • 1.4 Set Operations with Two Sets
  • 1.5 Set Operations with Three Sets
  • Key Concepts
  • Formula Review
  • Chapter Review
  • Chapter Test
  • 2.1 Statements and Quantifiers
  • 2.2 Compound Statements
  • 2.3 Constructing Truth Tables
  • 2.4 Truth Tables for the Conditional and Biconditional
  • 2.5 Equivalent Statements
  • 2.6 De Morgan’s Laws
  • 2.7 Logical Arguments
  • 3.1 Prime and Composite Numbers
  • 3.2 The Integers
  • 3.3 Order of Operations
  • 3.4 Rational Numbers
  • 3.5 Irrational Numbers
  • 3.6 Real Numbers
  • 3.7 Clock Arithmetic
  • 3.8 Exponents
  • 3.9 Scientific Notation
  • 3.10 Arithmetic Sequences
  • 3.11 Geometric Sequences
  • 4.1 Hindu-Arabic Positional System
  • 4.2 Early Numeration Systems
  • 4.3 Converting with Base Systems
  • 4.4 Addition and Subtraction in Base Systems
  • 4.5 Multiplication and Division in Base Systems
  • 5.1 Algebraic Expressions
  • 5.2 Linear Equations in One Variable with Applications
  • 5.3 Linear Inequalities in One Variable with Applications
  • 5.4 Ratios and Proportions
  • 5.5 Graphing Linear Equations and Inequalities
  • 5.7 Functions
  • 5.8 Graphing Functions
  • 5.9 Systems of Linear Equations in Two Variables
  • 5.10 Systems of Linear Inequalities in Two Variables
  • 5.11 Linear Programming
  • 6.1 Understanding Percent
  • 6.2 Discounts, Markups, and Sales Tax
  • 6.3 Simple Interest
  • 6.4 Compound Interest
  • 6.5 Making a Personal Budget
  • 6.6 Methods of Savings
  • 6.7 Investments
  • 6.8 The Basics of Loans
  • 6.9 Understanding Student Loans
  • 6.10 Credit Cards
  • 6.11 Buying or Leasing a Car
  • 6.12 Renting and Homeownership
  • 6.13 Income Tax
  • 7.1 The Multiplication Rule for Counting
  • 7.2 Permutations
  • 7.3 Combinations
  • 7.4 Tree Diagrams, Tables, and Outcomes
  • 7.5 Basic Concepts of Probability
  • 7.6 Probability with Permutations and Combinations
  • 7.7 What Are the Odds?
  • 7.8 The Addition Rule for Probability
  • 7.9 Conditional Probability and the Multiplication Rule
  • 7.10 The Binomial Distribution
  • 7.11 Expected Value
  • 8.1 Gathering and Organizing Data
  • 8.2 Visualizing Data
  • 8.3 Mean, Median and Mode
  • 8.4 Range and Standard Deviation
  • 8.5 Percentiles
  • 8.6 The Normal Distribution
  • 8.7 Applications of the Normal Distribution
  • 8.8 Scatter Plots, Correlation, and Regression Lines
  • 9.1 The Metric System
  • 9.2 Measuring Area
  • 9.3 Measuring Volume
  • 9.4 Measuring Weight
  • 9.5 Measuring Temperature
  • 10.1 Points, Lines, and Planes
  • 10.2 Angles
  • 10.3 Triangles
  • 10.4 Polygons, Perimeter, and Circumference
  • 10.5 Tessellations
  • 10.7 Volume and Surface Area
  • 10.8 Right Triangle Trigonometry
  • 11.1 Voting Methods
  • 11.2 Fairness in Voting Methods
  • 11.3 Standard Divisors, Standard Quotas, and the Apportionment Problem
  • 11.4 Apportionment Methods
  • 11.5 Fairness in Apportionment Methods
  • 12.1 Graph Basics
  • 12.2 Graph Structures
  • 12.3 Comparing Graphs
  • 12.4 Navigating Graphs
  • 12.5 Euler Circuits
  • 12.6 Euler Trails
  • 12.7 Hamilton Cycles
  • 12.8 Hamilton Paths
  • 12.9 Traveling Salesperson Problem
  • 12.10 Trees
  • 13.1 Math and Art
  • 13.2 Math and the Environment
  • 13.3 Math and Medicine
  • 13.4 Math and Music
  • 13.5 Math and Sports
  • A | Co-Req Appendix: Integer Powers of 10

Learning Objectives

After completing this section, you should be able to:

  • Multiply binomials.
  • Factor trinomials.
  • Solve quadratic equations by graphing.
  • Solve quadratic equations by factoring.
  • Solve quadratic equations using square root method.
  • Solve quadratic equations using the quadratic formula.
  • Solve real world applications modeled by quadratic equations.

In this section, we will discuss quadratic equations. There are several real-world scenarios that can be represented by the graph of a quadratic equation. Think of the Gateway Arch in St. Louis, Missouri. Both ends of the arch are 630 feet apart and the arch is 630 feet tall. You can plot these points on a coordinate system and create a parabola to graph the quadratic equation.

Identify Polynomials, Monomials, Binomials and Trinomials

You have learned that a term is a constant, or the product of a constant and one or more variables. When it is of the form a x m a x m , where a a is a constant and x m x m is a positive whole number, it is called a monomial . Some examples of monomial are 8, - 2 x 2 - 2 x 2 , 4 y 3 4 y 3 , and 11 z 11 z .

A monomial or two or more monomials combined by addition or subtraction is a polynomial . Some examples include: b + 11 b + 11 , 4 y 2 − 7 y + 2 4 y 2 − 7 y + 2 , and 4 x 4 + x 3 + 8 x 2 − 9 x + 1 4 x 4 + x 3 + 8 x 2 − 9 x + 1 . Some polynomials have special names, based on the number of terms. A monomial is a polynomial with exactly one term (examples: 14, 8 y 2 8 y 2 , - 9 x 3 y 5 - 9 x 3 y 5 , and − 13 − 13 ). A binomial has exactly two terms (examples: a + 7 a + 7 , 4 b − 5 4 b − 5 , y 2 − 16 y 2 − 16 , and 3 x 3 − 9 x 2 3 x 3 − 9 x 2 ), and a trinomial has exactly three terms (examples: x 2 − 7 x + 12 x 2 − 7 x + 12 , 9 y 2 + 2 y − 8 9 y 2 + 2 y − 8 , 6 m 4 − m 3 + 8 m 6 m 4 − m 3 + 8 m , and x 4 + 3 x 2 − 1 x 4 + 3 x 2 − 1 ).

Notice that every monomial, binomial, and trinomial is also a polynomial. They are just special members of the “family” of polynomials and so they have special names. We use the words monomial, binomial, and trinomial when referring to these special polynomials and just call all the rest polynomials.

Multiply Binomials

Recall multiplying algebraic expressions from Algebraic Expressions . In this section, we will continue that work and multiply binomials as well. We can use an area model to do multiplication.

Example 5.47

Multiply ( x + 2 ) ( x + 3 ) ( x + 2 ) ( x + 3 ) .

Step 1: Use the distributive property:

In the area model ( Figure 5.40 ) multiply each term on the side by each term on the top (think of it as a multiplication table).

Step 2: After we multiply, we get the following equation:

Step 3: Combine the like terms to arrive at:

Your Turn 5.47

Example 5.48, multiplying more complex binomials.

Multiply ( 2 x + 7 ) ( 3 x − 5 ) ( 2 x + 7 ) ( 3 x − 5 ) .

Step 1: Use the Distributive Property:

Step 2: After multiplying, get the following equation:

Your Turn 5.48

They are teaching multiplication of binomials in elementary school.

Manipulatives are often used in elementary school for students to experience a hands-on way to experience the mathematics they are learning. Base Ten Blocks, or Dienes Blocks, are often used to introduce place value and the operation of whole numbers. When multiplying two-digit numbers, students can make an array to visualize the Distributive Property. Figure 5.42 shows the value of each Base Ten Block and Figure 5.43 shows how to multiply 17 and 23 using an area model and Base Ten Blocks. You can see how this helps students visualize the multiplication using the Distributive Property. Consider how ( 10 + 7 ) ( 20 + 3 ) ( 10 + 7 ) ( 20 + 3 ) can easily extend to ( 10 + x ) ( 20 + y ) ( 10 + x ) ( 20 + y ) in algebra!

Factoring Trinomials

We’ve just covered how to multiply binomials. Now you will need to “undo” this multiplication—to start with the product and end up with the factors. Let us review an example of multiplying binomials to refresh your memory.

To factor the trinomial means to start with the product, x 2 + 5 x + 6 x 2 + 5 x + 6 , and end with the factors, ( x + 2 ) ( x + 3 ) ( x + 2 ) ( x + 3 ) . You need to think about where each of the terms in the trinomial came from. The first term came from multiplying the first term in each binomial. So, to get x 2 x 2 in the product, each binomial must start with an x x .

The last term in the trinomial came from multiplying the last term in each binomial. So, the last terms must multiply to 6. What two numbers multiply to 6? The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use? Consider the middle term. It came from adding the outer and inner terms. So the numbers that must have a product of 6 will need a sum of 5.

We’ll test both possibilities and summarize the results in the following table, which will be very helpful when you work with numbers that can be factored in many different ways.

We see that 2 and 3 are the numbers that multiply to 6 and add to 5. We have the factors of x 2 + 5 x + 6 x 2 + 5 x + 6 . They are ( x + 2 ) ( x + 3 ) ( x + 2 ) ( x + 3 ) .

You can check if the factors are correct by multiplying. Looking back, we started with x 2 + 5 x + 6 x 2 + 5 x + 6 , which is of the form x 2 + b x + c x 2 + b x + c , where b = 5 b = 5 and c = 6 c = 6 . We factored it into two binomials of the form ( x + m ) ( x + m ) and ( x + n ) ( x + n ) .

To get the correct factors, we found two number m m and n n whose product is c c and sum is b b . With the area model ( Figure 5.44 ), start with an empty box and then put in the x 2 x 2 term and c c .

Continue by putting in two terms that add up to 5 x : 2 x 5 x : 2 x and 3 x 3 x ( Figure 5.45 ):

Then you find the terms of the binomials on the top and side ( Figure 5.46 ):

Example 5.49

Factor x 2 + 7 x + 12 x 2 + 7 x + 12 .

The numbers that must have a product of 12 will need a sum of 7. We will summarize the results in a table below.

We see that 3 and 4 are the numbers that multiply to 12 and add to 7. The factors of x 2 + 7 x + 12 x 2 + 7 x + 12 are ( x + 3 ) ( x + 4 ) ( x + 3 ) ( x + 4 ) .

Your Turn 5.49

Example 5.50, factoring more complex trinomials.

Factor x 2 - 11 x + 28 x 2 - 11 x + 28 .

The numbers that must have a product of 28 will need a sum of − 11 − 11 . We will summarize the results in a table.

We see that 4 and 7 are the numbers that multiply to 28 and add to 11. But we needed − 11 − 11 , so we will need to use − 4 − 4 and − 7 − 7 because ( − 4 ) ( − 7 ) = 28 ( − 4 ) ( − 7 ) = 28 and ( − 4 ) + ( − 7 ) = − 11 ( − 4 ) + ( − 7 ) = − 11 . The factors of x 2 − 11 x + 28 x 2 − 11 x + 28 are ( x − 4 ) ( x − 7 ) ( x − 4 ) ( x − 7 ) .

Your Turn 5.50

Factoring with the Box Method (Area Model)

Solving Quadratic Equations by Graphing

We have already solved and graphed linear equations in Graphing Linear Equations and Inequalities , equations of the form A x + B y = C A x + B y = C . In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. The following are some examples of quadratic equations:

The last equation does not appear to have the variable squared, but when we simplify the expression on the left, we will get n 2 + n n 2 + n . The general form of a quadratic equation is a x 2 + b x + c = 0 a x 2 + b x + c = 0 , where a , b , a , b , and c c are real numbers, with a ≠ 0 a ≠ 0 . Remember that a solution of an equation is a value of a variable that makes a true statement when substituted into the equation. The solutions of quadratic equations are the values of the variables that make the quadratic equation a x 2 + b x + c = 0 a x 2 + b x + c = 0 true.

To solve quadratic equations, we need methods different than the ones we used in solving linear equations. We will start by solving a quadratic equation from its graph. Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations. Let us look first at graphing the quadratic equation y = x 2 y = x 2 . We will choose integer values of x x between − 2 − 2 and 2 and find their y y values, as shown in the table below.

Notice when we let x = 1 x = 1 and x = − 1 x = − 1 , we got the same value for y y .

The same thing happened when we let x = 2 x = 2 and x = − 2 x = − 2 . Now, we will plot the points to show the graph of y = x 2 y = x 2 . See Figure 5.47 .

The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this. When y = 0 y = 0 the solution to the quadratic y = x 2 y = x 2 is 0 because x 2 = 0 x 2 = 0 at x = 0 x = 0 .

Example 5.51

Graphing a quadratic equation.

Graph y = x 2 - 1 y = x 2 - 1 and list the solutions to the quadratic equation.

We will graph the equation by plotting points.

Step 1: Choose integer values for x x , substitute them into the equation, and solve for y y .

Step 2: Record the values of the ordered pairs in the chart.

Step 3: Plot the points and then connect them with a smooth curve. The result will be the graph of the equation y = x 2 - 1 y = x 2 - 1 ( Figure 5.48 ). The solutions are x = 1 x = 1 and x = - 1 x = - 1 .

Your Turn 5.51

Example 5.52, solving a quadratic equation from its graph.

Find the solutions of y = x 2 + 5 x + 4 y = x 2 + 5 x + 4 from its graph ( Figure 5.49 ).

The solutions of a quadratic equations are the values of x x that make the equation a true statement when set equal to zero (i.e. when y = 0 y = 0 ). x 2 + 5 x + 4 = 0 x 2 + 5 x + 4 = 0 at x = - 4 x = - 4 and x = - 1 x = - 1 .

Your Turn 5.52

Solving quadratic equations by factoring.

Another way of solving quadratic equations is by factoring. We will use the Zero Product Property that says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

Example 5.53

Solving a quadratic equation by factoring.

Solve ( x + 1 ) ( x - 4 ) = 0 ( x + 1 ) ( x - 4 ) = 0 .

Your Turn 5.53

Example 5.54, solve another quadratic equation by factoring.

Solve x 2 + 2 x − 8 = 0 x 2 + 2 x − 8 = 0 .

Your Turn 5.54

Solving Quadratics with the Zero Property

Be careful to write the quadratic equation in standard form first. The equation must be set equal to zero in order for you to use the Zero Product Property! Often students start in Step 2 resulting in an incorrect solution. For example, x 2 + 2 x − 15 = - 7 x 2 + 2 x − 15 = - 7 cannot be factored to ( x − 3 ) ( x + 5 ) = − 7 ( x − 3 ) ( x + 5 ) = − 7 and then solved by setting each factor equal to − 7 − 7 .

The correct way to solve this quadratic equation is to set it equal to zero FIRST: x 2 + 2 x − 15 + 7 = − 7 + 7 x 2 + 2 x − 15 + 7 = − 7 + 7 which becomes x 2 + 2 x − 8 = 0 x 2 + 2 x − 8 = 0 , then continue to factor. See the table below for the correct way to apply the Zero Product Property.

Solving Quadratic Equations Using the Square Root Property

We just solved some quadratic equations by factoring. Let us use factoring to solve the quadratic equation x 2 = 9 x 2 = 9 .

The solution is read as “ x x is equal to positive or negative 3.”

What happens when we have an equation like x 2 = 7 x 2 = 7 ? Since 7 is not a perfect square, we cannot solve the equation by factoring. These equations are all of the form x 2 = k x 2 = k . We define the square root of a number in this way: If n 2 = m n 2 = m , then n n is a square root of m m . This leads to the Square Root Property .

Example 5.55

Using the square root property to solve a quadratic equation.

Solve using the square Root Property: x 2 = 169 x 2 = 169 .

Your Turn 5.55

Example 5.56, using the square root property to solve another quadratic equation.

Solve using the Square Root Property: 144 q 2 = 25 144 q 2 = 25 .

Step 1: Solve for q q .

q 2 = 25 144 q 2 = 25 144

Step 2: Use the Square Root Property.

q = ± 25 144 q = ± 25 144

Step 3: Simplify the radical.

Step 4: Rewrite to show the two solutions. q = 5 12 q = 5 12 , q = - 5 12 q = - 5 12

Your Turn 5.56

Solving quadratic equations using the quadratic formula.

This last method we will look at for solving quadratic equations is the quadratic formula . This method works for all quadratic equations, even the quadratic equations we could not factor! To use the quadratic formula, we substitute the values of a a , b b , and c c into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation.

Example 5.57

Solving a quadratic equation using the quadratic formula.

Solve using the quadratic formula: x 2 - 6 x + 5 = 0 x 2 - 6 x + 5 = 0 .

x 2 - 6 x + 5 = 0 x 2 - 6 x + 5 = 0

This equation is in standard form.

a x 2 + b x + c = 0 x 2 - 6 x + 5 = 0 a x 2 + b x + c = 0 x 2 - 6 x + 5 = 0

Step 1: Identify the a a , b b , and c c values.

a = 1 , b = - 6 , c = 5 a = 1 , b = - 6 , c = 5

Step 2: Write the quadratic formula.

x = - b ± b 2 - 4 a c 2 a x = - b ± b 2 - 4 a c 2 a

Step 3: Substitute in the values of a a , b b , c c .

x = ( − 6 ) ± ( − 6 ) 2 − 4 ( 1 ) ( 5 ) 2.1 x = ( − 6 ) ± ( − 6 ) 2 − 4 ( 1 ) ( 5 ) 2.1

Step 4: Simplify.

x = 6 ± 36 - 20 2 x = 6 ± 36 - 20 2

x = 6 ± 16 2 x = 6 ± 16 2

x = 6 ± 4 2 x = 6 ± 4 2

Step 5: Rewrite to show two solutions.

x = 6 + 4 2 , x = 6 - 4 2 x = 6 + 4 2 , x = 6 - 4 2

Step 6: Simplify.

x = 10 2 , x = 2 2 x = 10 2 , x = 2 2

x = 5 , x = 1 x = 5 , x = 1

Step 7: Check.

x 2 − 6 x + 5 = 0 x 2 − 6 x + 5 = 0 5 2 − 6 ⋅ 5 + 5 = ? 0 1 2 − 6 ⋅ 1 + 5 = ? 0 25 − 30 + 5 = ? 0 1 − 6 + 5 = ? 0 0 = 0 ✓ 0 = 0 ✓ x 2 − 6 x + 5 = 0 x 2 − 6 x + 5 = 0 5 2 − 6 ⋅ 5 + 5 = ? 0 1 2 − 6 ⋅ 1 + 5 = ? 0 25 − 30 + 5 = ? 0 1 − 6 + 5 = ? 0 0 = 0 ✓ 0 = 0 ✓

Your Turn 5.57

Example 5.58, solving another quadratic equation using the quadratic formula.

Solve using the quadratic formula: 2 x 2 + 9 x − 5 = 0 2 x 2 + 9 x − 5 = 0 .

Your Turn 5.58

Solving Quadratics with the Quadratic Formula

Solving Real-World Applications Modeled by Quadratic Equations

There are problem solving strategies that will work well for applications that translate to quadratic equations. Here’s a problem-solving strategy to solve word problems:

Step 1: Read the problem. Make sure all the words and ideas are understood.

Step 2: Identify what we are looking for.

Step 3: Name what we are looking for. Choose a variable to represent that quantity.

Step 4: Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.

Step 5: Solve the equation using good algebra techniques.

Step 6: Check the answer in the problem and make sure it makes sense.

Step 7: Answer the question with a complete sentence.

Example 5.59

Finding consecutive integers.

The product of two consecutive integers is 132. Find the integers.

Step 1: Read the problem.

We are looking for two consecutive integers.

Step 3: Name what we are looking for.

Let n = n = the first integer

Let n + 1 = n + 1 = the next consecutive integer.

Step 4: Translate into an equation. Restate the problem in a sentence.

n ( n + 1 ) = 132 n ( n + 1 ) = 132

The product of the two consecutive integers is 132. The first integer times the next integer is 132.

Step 5: Solve the equation.

n 2 + n = 132 n 2 + n = 132

Bring all the terms to one side.

n 2 + n - 132 = 0 n 2 + n - 132 = 0

Factor the trinomial.

( n - 11 ) ( n + 12 ) = 0 ( n - 11 ) ( n + 12 ) = 0

Use the zero product property.

n - 11 = 0 or n + 12 = 0 n - 11 = 0 or n + 12 = 0

Solve the equations.

n = 11 , n = - 12 n = 11 , n = - 12

There are two values for n n that are solutions to this problem. So, there are two sets of consecutive integers that will work.

If the first integer is n = 11 n = 11 , then the next integer is 12. If the first integer is n = - 12 n = - 12 , then the next integer is − 11 − 11 .

Step 6: Check the answer.

The consecutive integers are 11, 12 and − 11 − 11 , − 12 − 12 . The product of 11 and 12 = 132 12 = 132 and the product of - 11 ( - 12 ) = 132 - 11 ( - 12 ) = 132 . Both pairs of consecutive integers are solutions.

Step 7: Answer the question.

The consecutive integers are 11, 12, and − 11 − 11 , − 12 − 12 .

Your Turn 5.59

Were you surprised by the pair of negative integers that is one of the solutions? In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

Example 5.60

Finding length and width of a garden.

A rectangular garden has an area 15 square feet. The length of the garden is 2 feet more than the width. Find the length and width of the garden.

Step 1: Read the problem. In problems involving geometric figures, a sketch can help you visualize the situation ( Figure 5.50 ).

Step 2: Identify what you are looking for.

We are looking for the length and width.

Step 3: Name what you are looking for. The length is 2 feet more than width.

Let W = W = the width of the garden.

W + 2 = W + 2 = the length of the garden.

Step 4: Translate into an equation.

Restate the important information in a sentence.

The area of the rectangular garden is 15 square feet.

Use the formula for the area of a rectangle.

A = L × W A = L × W

Substitute in the variables.

15 = ( W + 2 ) W 15 = ( W + 2 ) W

Step 5: Solve the equation. Distribute first.

15 = W 2 + 2 W 15 = W 2 + 2 W

Get zero on one side.

0 = W 2 + 2 W - 15 0 = W 2 + 2 W - 15

0 = ( W + 5 ) ( W - 3 ) 0 = ( W + 5 ) ( W - 3 )

Use the Zero Product Property.

0 = W + 5 0 = W + 5

0 = W - 3 0 = W - 3

Solve each equation.

- 5 = W - 5 = W

3 = W 3 = W

Since W W is the width of the garden, it does not make sense for it to be negative. We eliminate that value for W W .

W = 3 W = 3

Width is 3 feet.

Find the value of the length.

W + 2 = W + 2 = length.

3 + 2 3 + 2

Length is 5 feet.

Step 6: Check the answer ( Figure 5.51 ).

Does the answer make sense?

Yes, this makes sense.

The width of the garden is 3 feet and the length is 5 feet.

Your Turn 5.60

Work it out, completing the square.

Recall the two methods used to solve quadratic equations of the form a x 2 + b x + c : a x 2 + b x + c : by factoring and by using the quadratic formula. There are, however, many different methods for solving quadratic equations that were developed throughout history. Egyptian, Mesopotamian, Chinese, Indian, and Greek mathematicians all solved various types of quadratic equations, as did Arab mathematicians of the ninth through the twelfth centuries. It is one of these Arab mathematicians' methods that we wish to investigate with this activity.

Muhammad ibn Musa al-Khwarizmi was employed as a scholar at the House of Wisdom in Baghdad, located in present day Iraq. One of the many accomplishments of Al-Khwarizmi was his book on the topic of algebra. In that book, he asks, “What must be the square which, when increased by ten of its own roots, amounts to 39?” Al-Khwarizmi, like many Arab mathematicians of his time, was well versed in Euclid's Elements. Like Euclid, he viewed algebra very geometrically, and thus had a geometric approach to solving a problem like the one above. In his approach, he used a method which today we refer to as completing the square .

His description of the solution method for the above problem is: halve the number of roots, which in the present instance yields 5. This you multiply by itself; the product is 25. Add this to 39; the sum is 64. Now take the root of this which is 8, and subtract from it half the number of the roots, which is 5; the remainder is 3. This is the root of the square which you sought for. Thus the square is 9.

x 2 + 10 x + 25 = 39 + 25 x 2 + 10 x + 25 = 39 + 25 , or x 2 + 10 x + 25 = 64 x 2 + 10 x + 25 = 64 .

Notice that the completed square has side length x + 5 x + 5 , so the large square has area ( x + 5 ) 2 ( x + 5 ) 2 . (Notice algebraically that the left half of the equation x 2 + 10 x + 25 = 64 x 2 + 10 x + 25 = 64 factors to ( x + 5 ) 2 = 64. ) ( x + 5 ) 2 = 64. ) This means the area of large square equals 64. If ( x + 5 ) 2 = 64 ( x + 5 ) 2 = 64 , then x + 5 = 8 x + 5 = 8 ; so x x must be equal to 3 or − 13 − 13 to make this true. Note that Al-Kwarimi would not have considered the possibility of a negative solution, since he approached the solution geometrically, and negative distances do not exist.

Check Your Understanding

  • x 2 − 3 x + 15
  • x 2 − 2 x − 15
  • x 2 + 2 x − 15
  • ( x − 3 ) ( x + 5 )
  • ( x − 3 ) ( x − 5 )
  • ( x + 3 ) ( x + 5 )
  • ( x + 3 ) ( x − 5 )
  • x 2 − x + 5 = 0
  • x 2 − 5 x + 1 = 0
  • x 2 + 6 x + 5 = 0
  • x 2 − 4 x + 5 = 0

Section 5.6 Exercises

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Quadratic Equations

An example of a Quadratic Equation :

The function makes nice curves like this one:

The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x 2 ).

It is also called an "Equation of Degree 2" (because of the "2" on the x )

Standard Form

The Standard Form of a Quadratic Equation looks like this:

  • a , b and c are known values. a can't be 0.
  • " x " is the variable or unknown (we don't know it yet).

Here are some examples:

Have a Play With It

Play with the " Quadratic Equation Explorer " so you can see:

  • the function's graph, and
  • the solutions (called "roots").

Hidden Quadratic Equations!

As we saw before, the Standard Form of a Quadratic Equation is

ax 2 + bx + c = 0

But sometimes a quadratic equation does not look like that!

For example:

How To Solve Them?

The " solutions " to the Quadratic Equation are where it is equal to zero .

They are also called " roots ", or sometimes " zeros "

There are usually 2 solutions (as shown in this graph).

And there are a few different ways to find the solutions:

Just plug in the values of a, b and c, and do the calculations.

We will look at this method in more detail now.

About the Quadratic Formula

First of all what is that plus/minus thing that looks like ± ?

The ± means there are TWO answers:

x = −b + √(b 2 − 4ac) 2a

x = −b − √(b 2 − 4ac) 2a

Here is an example with two answers:

But it does not always work out like that!

  • Imagine if the curve "just touches" the x-axis.
  • Or imagine the curve is so high it doesn't even cross the x-axis!

This is where the "Discriminant" helps us ...

Discriminant

Do you see b 2 − 4ac in the formula above? It is called the Discriminant , because it can "discriminate" between the possible types of answer:

  • when b 2 − 4ac is positive, we get two Real solutions
  • when it is zero we get just ONE real solution (both answers are the same)
  • when it is negative we get a pair of Complex solutions

Complex solutions? Let's talk about them after we see how to use the formula.

Using the Quadratic Formula

Just put the values of a, b and c into the Quadratic Formula, and do the calculations.

Example: Solve 5x 2 + 6x + 1 = 0

Answer: x = −0.2 or x = −1

And we see them on this graph.

Remembering The Formula

A kind reader suggested singing it to "Pop Goes the Weasel":

Try singing it a few times and it will get stuck in your head!

Or you can remember this story:

x = −b ± √(b 2 − 4ac) 2a

"A negative boy was thinking yes or no about going to a party, at the party he talked to a square boy but not to the 4 awesome chicks. It was all over at 2 am. "

Complex Solutions?

When the Discriminant (the value b 2 − 4ac ) is negative we get a pair of Complex solutions ... what does that mean?

It means our answer will include Imaginary Numbers . Wow!

Example: Solve 5x 2 + 2x + 1 = 0

√(−16) = 4 i (where i is the imaginary number √−1)

Answer: x = −0.2 ± 0.4 i

The graph does not cross the x-axis. That is why we ended up with complex numbers.

In a way it is easier: we don't need more calculation, we leave it as −0.2 ± 0.4 i .

Example: Solve x 2 − 4x + 6.25 = 0

√(−9) = 3 i (where i is the imaginary number √−1)

Answer: x = 2 ± 1.5 i

BUT an upside-down mirror image of our equation does cross the x-axis at 2 ± 1.5 (note: missing the i ).

Just an interesting fact for you!

  • Quadratic Equation in Standard Form: ax 2 + bx + c = 0
  • Quadratic Equations can be factored
  • Quadratic Formula: x = −b ± √(b 2 − 4ac) 2a
  • positive, there are 2 real solutions
  • zero, there is one real solution
  • negative, there are 2 complex solutions
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Need Help Solving Those Dreaded Word Problems Involving Quadratic Equations?

Yes, I know it's tough. You've finally mastered factoring and using the quadratic formula and now you are asked to solve more problems!

Except these are even more tough. Now you have to figure out what the problem even means before trying to solve it. I completely understand and here's where I am going to try to help!

There are many types of problems that can easily be solved using your knowledge of quadratic equations. You may come across problems that deal with money and predicted incomes (financial) or problems that deal with physics such as projectiles. You may also come across construction type problems that deal with area or geometry problems that deal with right triangles.

Lucky for you, you can solve the quadratic equations, now you just have to learn how to apply this useful skill.

On this particular page, we are going to take a look at a physics "projectile problem".

Projectiles - Example 1

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. Find the maximum height attained by the ball.

Let's first take a minute to understand this problem and what it means. We know that a ball is being shot from a cannon. So, in your mind, imagine a cannon firing a ball. We know that the ball is going to shoot from the cannon, go into the air, and then fall to the ground.

So, here's a mathematical picture that I see in my head.

Now let's talk about what each part of this problem means. In our equation that we are given we must be given the value for the force of gravity (coefficient of t 2 ). We must also use our upward velocity (coefficient of t) and our original height of the cannon/ball (the constant or 1.5). Take a look...

Now that you have a mental picture of what's happening and you understand the formula given, we can go ahead and solve the problem.

  • First, ask yourself, "What am I solving for?" "What do I need to find?" You are asked to find the maximum height (go back and take a look at the diagram). What part of the parabola is this? Yes, it's the vertex! We will need to use the vertex formula and I will need to know the y coordinate of the vertex because it's asking for the height.
  • Next Step: Solve! Now that I know that I need to use the vertex formula, I can get to work.

Just as simple as that, this problem is solved.

Let's not stop here. Let's take this same problem and put a twist on it. There are many other things that we could find out about this ball!

Projectiles - Example 2

Same problem - different question. Take a look...

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long did it take for the ball to reach the ground?

Now, we've changed the question and we want to know how long did it take the ball to reach the ground.

What ground, you may ask. The problem didn't mention anything about a ground. Let's take a look at the picture "in our mind" again.

Do you see where the ball must fall to the ground. The x-axis is our "ground" in this problem. What do we know about points on the x-axis when we are dealing with quadratic equations and parabolas?

Yes, the points on the x-axis are our "zeros" or x-intercepts. This means that we must solve the quadratic equation in order to find the x-intercept.

Let's do it! Let's solve this equation. I'm thinking that this may not be a factorable equation. Do you agree? So, what's our solution?

Hopefully, you agree that we can use the quadratic formula to solve this equation.

The first time doesn't make sense because it's negative. This is the calculation for when the ball was on the ground initially before it was shot.

This actually never really occurred because the ball was shot from the cannon and was never shot from the ground. Therefore, we will disregard this answer.

The other answer was 2.54 seconds which is when the ball reached the ground (x-axis) after it was shot. Therefore, this is the only correct answer to this problem.

Ok, one more spin on this problem. What would you do in this case?

Projectiles - Example 3

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long does it take the ball to reach a height of 20 feet?

Yes, this problem is a little trickier because the question is not asking for the maximum height (vertex) or the time it takes to reach the ground (zeros), instead it it asking for the time it takes to reach a height of 20 feet.

Since the ball reaches a maximum height of 26.5 ft, we know that it will reach a height of 20 feet on the way up and on the way down.

Let's just estimate on our graph and also make sure that we get this visual in our head.

From looking at this graph, I would estimate the times to be about 0.7 sec and 1.9 sec. Do you see how the ball will reach 20 feet on the way up and on the way down?

Now, let's find the actual values. Where will we substitute 20 feet?

Yes, we must substitute 20 feet for h(t) because this is the given height. We will now be solving for t using the quadratic formula. Take a look.

Our actual times were pretty close to our estimates. Just don't forget that when you solve a quadratic equation, you must have the equation set equal to 0. Therefore, we had to subtract 20 from both sides in order to have the equation set to 0.

You've now seen it all when it comes to projectiles!

Great Job! Hopefully you've been able to understand how to solve problems involving quadratic equations. I also hope that you better understand these common velocity equations and how to think about what this problem looks like graphically in order to help you to understand which process or formula to use in order to solve the problem.

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Quadratic Equation

Quadratic equations are second-degree algebraic expressions and are of the form ax 2 + bx + c = 0. The term "quadratic" comes from the Latin word "quadratus" meaning square, which refers to the fact that the variable x is squared in the equation. In other words, a quadratic equation is an “equation of degree 2.” There are many scenarios where a quadratic equation is used. Did you know that when a rocket is launched, its path is described by a quadratic equation? Further, a quadratic equation has numerous applications in physics, engineering, astronomy, etc.

Quadratic equations have maximum of two solutions, which can be real or complex numbers. These two solutions (values of x) are also called the roots of the quadratic equations and are designated as (α, β). We shall learn more about the roots of a quadratic equation in the below content.

What is Quadratic Equation?

A quadratic equation is an algebraic equation of the second degree in x. The quadratic equation in its standard form is ax 2 + bx + c = 0, where a and b are the coefficients, x is the variable, and c is the constant term. The important condition for an equation to be a quadratic equation is the coefficient of x 2 is a non-zero term (a ≠ 0). For writing a quadratic equation in standard form, the x 2 term is written first, followed by the x term, and finally, the constant term is written.

Quadratic equation is of the form a x squared plus b x plus c equals 0

Further, in real math problems the quadratic equations are presented in different forms: (x - 1)(x + 2) = 0, -x 2 = -3x + 1, 5x(x + 3) = 12x, x 3 = x(x 2 + x - 3). All of these equations need to be transformed into standard form of the quadratic equation before performing further operations.

Roots of a Quadratic Equation

The roots of a quadratic equation are the two values of x, which are obtained by solving the quadratic equation. These roots of the quadratic equation are also called the zeros of the equation. For example, the roots of the equation x 2 - 3x - 4 = 0 are x = -1 and x = 4 because each of them satisfies the equation. i.e.,

  • At x = -1, (-1) 2 - 3(-1) - 4 = 1 + 3 - 4 = 0
  • At x = 4, (4) 2 - 3(4) - 4 = 16 - 12 - 4 = 0

There are various methods to find the roots of a quadratic equation. The usage of the quadratic formula is one of them.

Quadratic Formula

Quadratic formula is the simplest way to find the roots of a quadratic equation . There are certain quadratic equations that cannot be easily factorized, and here we can conveniently use this quadratic formula to find the roots in the quickest possible way. The two roots in the quadratic formula are presented as a single expression. The positive sign and the negative sign can be alternatively used to obtain the two distinct roots of the equation.

Quadratic Formula: The roots of a quadratic equation ax 2 + bx + c = 0 are given by x = [-b ± √(b 2 - 4ac)]/2a.

Quadratic formula is used to find the roots of quadratic equation

This formula is also known as the Sridharacharya formula .

Example: Let us find the roots of the same equation that was mentioned in the earlier section x 2 - 3x - 4 = 0 using the quadratic formula.

a = 1, b = -3, and c = -4.

x = [-b ± √(b 2 - 4ac)]/2a = [-(-3) ± √((-3) 2 - 4(1)(-4))]/2(1) = [3 ± √25] / 2 = [3 ± 5] / 2 = (3 + 5)/2 or (3 - 5)/2 = 8/2 or -2/2 = 4 or -1 are the roots.

Proof of Quadratic Formula

Consider an arbitrary quadratic equation: ax 2 + bx + c = 0, a ≠ 0

To determine the roots of this equation, we proceed as follows:

ax 2 + bx = -c ⇒ x 2 + bx/a = -c/a

Now, we express the left-hand side as a perfect square , by introducing a new term (b/2a) 2 on both sides:

  • x 2 + bx/a + (b/2a) 2 = -c/a + (b/2a) 2

The left-hand side is now a perfect square:

(x + b/2a) 2 = -c/a + b 2 /4a 2 ⇒ (x + b/2a) 2 = (b 2 - 4ac)/4a 2

This is good for us, because now we can take square roots to obtain:

x + b/2a = ±√(b 2 - 4ac)/2a

x = (-b ± √(b 2 - 4ac))/2a

Thus, by completing the squares, we were able to isolate x and obtain the two roots of the equation.

Nature of Roots of the Quadratic Equation

The roots of a quadratic equation are usually represented to by the symbols alpha (α), and beta (β). Here we shall learn more about how to find the nature of roots of a quadratic equation without actually finding the roots of the equation.

The nature of roots of a quadratic equation can be found without actually finding the roots (α, β) of the equation. This is possible by taking the discriminant value, which is part of the formula to solve the quadratic equation. The value b 2 - 4ac is called the discriminant of a quadratic equation and is designated as 'D'. Based on the discriminant value the nature of the roots of the quadratic equation can be predicted.

Discriminant: D = b 2 - 4ac

  • D > 0, the roots are real and distinct
  • D = 0, the roots are real and equal.
  • D < 0, the roots do not exist or the roots are imaginary .

Nature of roots of a quadratic equation is determined by discriminant.

Now, check out the formulas to find the sum and the product of the roots of the equation.

Sum and Product of Roots of Quadratic Equation

The coefficient of x 2 , x term, and the constant term of the quadratic equation ax 2 + bx + c = 0 are useful in determining the sum and product of the roots of the quadratic equation. The sum and product of the roots of a quadratic equation can be directly calculated from the equation, without actually finding the roots of the quadratic equation. For a quadratic equation ax 2 + bx + c = 0, the sum and product of the roots are as follows.

  • Sum of the Roots: α + β = -b/a = - Coefficient of x/ Coefficient of x 2
  • Product of the Roots: αβ = c/a = Constant term/ Coefficient of x 2

Writing Quadratic Equations Using Roots

The quadratic equation can also be formed for the given roots of the equation. If α, β, are the roots of the quadratic equation, then the quadratic equation is as follows.

x 2 - (α + β)x + αβ = 0

Example: What is the quadratic equation whose roots are 4 and -1?

Solution: It is given that α = 4 and β = -1. The corresponding quadratic equation is found by:

x 2 - (α + β)x + αβ = 0 x 2 - (α + β)x + αβ = 0 x 2 - (4 - 1)x + (4)(-1) = 0 x 2 - 3x - 4 = 0

Formulas Related to Quadratic Equations

The following list of important formulas is helpful to solve quadratic equations.

  • The quadratic equation in its standard form is ax 2 + bx + c = 0
  • For D > 0 the roots are real and distinct.
  • For D = 0 the roots are real and equal.
  • For D < 0 the real roots do not exist, or the roots are imaginary.
  • The formula to find the roots of the quadratic equation is x = [-b ± √(b 2 - 4ac)]/2a.
  • The sum of the roots of a quadratic equation is α + β = -b/a.
  • The product of the Root of the quadratic equation is αβ = c/a.
  • The quadratic equation whose roots are α, β, is x 2 - (α + β)x + αβ = 0.
  • The condition for the quadratic equations a 1 x 2 + b 1 x + c 1 = 0, and a 2 x 2 + b 2 x + c 2 = 0 having the same roots is (a 1 b 2 - a 2 b 1 ) (b 1 c 2 - b 2 c 1 ) = (a 2 c 1 - a 1 c 2 ) 2 .
  • When a > 0, the quadratic expression f(x) = ax 2 + bx + c has a minimum value at x = -b/2a.
  • When a < 0, the quadratic expression f(x) = ax 2 + bx + c has a maximum value at x = -b/2a.
  • The domain of any quadratic function is the set of all real numbers.

Methods to Solve Quadratic Equations

A quadratic equation can be solved to obtain two values of x or the two roots of the equation. There are four different methods to find the roots of the quadratic equation. The four methods of solving the quadratic equations are as follows.

  • Factorizing of Quadratic Equation
  • Using quadratic formula (which we have seen already)

Method of Completing the Square

  • Graphing Method to Find the Roots

Let us look in detail at each of the above methods to understand how to use these methods, their applications, and their uses.

Solving Quadratic Equations by Factorization

Factorization of quadratic equation follows a sequence of steps. For a general form of the quadratic equation ax 2 + bx + c = 0, we need to first split the middle term into two terms, such that the product of the terms is equal to the constant term. Further, we can take the common terms from the available term, to finally obtain the required factors as follows:

  • x 2 + (a + b)x + ab = 0
  • x 2 + ax + bx + ab = 0
  • x(x + a) + b(x + a)
  • (x + a)(x + b) = 0

Here is an example to understand the factorization process.

  • x 2 + 5x + 6 = 0
  • x 2 + 2x + 3x + 6 = 0
  • x(x + 2) + 3(x + 2) = 0
  • (x + 2)(x + 3) = 0

Thus the two obtained factors of the quadratic equation are (x + 2) and (x + 3). To find its roots, just set each factor to zero and solve for x. i.e., x + 2 = 0 and x + 3 = 0 which gives x = -2 and x = -3. Thus, x = -2 and x = -3 are the roots of x 2 + 5x + 6 = 0.

Further, there is another important method of solving a quadratic equation. The method of completing the square for a quadratic equation is also useful to find the roots of the equation.

The method of completing the square in a quadratic equation is to algebraically square and simplify, to obtain the required roots of the equation. Consider a quadratic equation ax 2 + bx + c = 0, a ≠ 0. To determine the roots of this equation, we simplify it as follows:

  • ax 2 + bx + c = 0
  • ax 2 + bx = -c
  • x 2 + bx/a = -c/a

Now, we express the left-hand side as a perfect square, by introducing a new term (b/2a) 2 on both sides:

  • (x + b/2a) 2 = -c/a + b 2 /4a 2
  • (x + b/2a) 2 = (b 2 - 4ac)/4a 2
  • x + b/2a = + √(b 2 - 4ac)/2a
  • x = - b/2a + √(b 2 - 4ac)/2a
  • x = [-b ± √(b 2 - 4ac)]/2a

Here the '+' sign gives one root and the '-' sign gives another root of the quadratic equation. Generally, this detailed method is avoided, and only the quadratic formula is used to obtain the required roots.

Graphing a Quadratic Equation

Quadratic equation graph intersecting the x-axis at two points

The point(s) where the graph cuts the horizontal x-axis (typically the x-intercepts ) is the solution of the quadratic equation. These points can also be algebraically obtained by equalizing the y value to 0 in the function y = ax 2 + bx + c and solving for x.

Quadratic Equations Having Common Roots

Consider two quadratic equations having common roots a 1 x 2 + b 1 x + c 1 = 0, and a 2 x 2 + b 2 x + c 2 = 0. Let us solve these two equations to find the conditions for which these equations have a common root. The two equations are solved for x 2 and x respectively.

(x 2 )(b 1 c 2 - b 2 c 1 ) = (-x)/(a 1 c 2 - a 2 c 1 ) = 1/(a 1 b 2 - a 2 b 1 )

x 2 = (b 1 c 2 - b 2 c 1 ) / (a 1 b 2 - a 2 b 1 )

x = (a 2 c 1 - a 1 c 2 ) / (a 1 b 2 - a 2 b 1 )

Hence, by simplifying the above two expressions we have the following condition for the two equations having the common root.

(a 1 b 2 - a 2 b 1 ) (b 1 c 2 - b 2 c 1 ) = (a 2 c 1 - a 1 c 2 ) 2

Maximum and Minimum Value of Quadratic Expression

Maximum and minimum value of quadratic equation on graphs.

The maximum and minimum values of the quadratic expressions are of further help to find the range of the quadratic expression: The range of the quadratic expressions also depends on the value of a. For positive values of a( a > 0), the range is [ F(-b/2a), ∞), and for negative values of a ( a < 0), the range is (-∞, F(-b/2a)].

  • For a > 0, Range: [ f(-b/2a), ∞)
  • For a < 0, Range: (-∞, f(-b/2a)]

Note that the domain of a quadratic function is the set of all real numbers, i.e., (-∞, ∞).

Tips and Tricks on Quadratic Equation:

Some of the below-given tips and tricks on quadratic equations are helpful to more easily solve quadratic equations.

  • The quadratic equations are generally solved through factorization. But in instances when it cannot be solved by factorization, the quadratic formula is used.
  • The roots of a quadratic equation are also called the zeroes of the equation.
  • For quadratic equations having negative discriminant values, the roots are represented by complex numbers.
  • The sum and product of the roots of a quadratic equation can be used to find higher algebraic expressions involving these roots.

☛Related Topics:

  • Roots Calculator
  • Quadratic Factoring Calculator
  • Roots of Quadratic Equation Calculator

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Quadratic Equations Examples

Example 1: James is a fitness enthusiast and goes for a jog every morning. The park where he jogs is rectangular in shape and measures 12m × 8m. An environmentalist group plans to revamp the park and decides to build a pathway surrounding the park. This would increase the total area to 140 sq m. What will be the width of the pathway?

Let’s denote the width of the pathway by x.

Then, the length and breadth of the outer rectangle is (12+2x) m and (8+2x) m.

Given that, area = 140

(12 + 2x)(8 + 2x) = 140

2(6 + x) 2(4 + x) = 140

(6 + x)(4 + x) = 35

24 + 6x + 4x + x 2 = 35

x 2 + 10x -11 = 0

x 2 + 11x - x - 11 = 0

x(x + 11) - 1(x + 11) = 0

(x + 11)(x - 1) = 0

(x + 11) =0 and (x - 1) = 0

x = -11 and x = 1

Since length can’t be negative, we take x = 1.

Answer: Therefore the width of the pathway is 1 m.

Example 2: Rita throws a ball upwards from a platform that is 20m above the ground. The height of the ball from the ground at a time 't', is denoted by 'h'. Suppose h = -4t 2 + 16t + 20. Find the maximum height attained by the ball.

We can rearrange the terms of the quadratic equation

h = -4t 2 + 16t + 20

in such a way that it is easy to find the maximum value of this equation.

= -4t 2 + 16t + 20

= -4(t 2 - 4t - 5)

= -4((t - 2) 2 - 9)

= -4(t - 2) 2 + 36

We should keep the value of (t - 2) 2 minimum in order to find the maximum value of h.

So, the minimum value (t - 2) 2 can take is 0.

Answer: Therefore the maximum height attained is 36m.

Example 3: Find the quadratic equation having the roots 5 and 8 respectively.

The quadratic equation having the roots α, β, is x 2 - (α + β)x + αβ = 0.

Given α = 5, and β = 8.

Therefore the quadratic equation is:

x 2 - (5 + 8)x + 5×8 = 0

x 2 - 13x + 40 = 0

Answer: Hence the required quadratic equation is x 2 - 13x + 40 = 0

Example 4: The quad equation 2x 2 + 9x + 7 = 0 has roots α, β. Find the quadratic equation having the roots 1/α, and 1/β.

The quadratic equation having roots that are reciprocal to the roots of the equation ax 2 + bx + c = 0, is cx 2 + bx + a = 0.

The given quadratic equation is 2x 2 + 9x + 7 = 0.

Hence the required equation having reciprocal roots is 7x 2 + 9x + 2 = 0.

Method 2: From the given equation,

α + β = -9/2 and α β = 2/7.

The new equation should have its roots to be 1/α and 1/β.

Their sum = 1/α + 1/β = (α + β) / α β = -9/7 Their product = 1/α β = 2/7 Thus, the required equation is, x 2 - (1/α + 1/β)x + 1/α β = 0 x 2 - (-9/7)x + 2/7 = 0 Multiplying both sides by 7, 7x 2 + 9x + 2 = 0

Answer: Therefore the equation is 7x 2 + 9x + 2 = 0.

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Practice Questions on Quadratic Equation

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FAQs on Quadratic Equation

What is the definition of a quadratic equation.

A quadratic equation in math is a second-degree equation of the form ax 2 + bx + c = 0. Here a and b are the coefficients, c is the constant term, and x is the variable. Since the variable x is of the second degree, there are two roots or answers for this quadratic equation. The roots of the quadratic equation can be found by either solving by factorizing or through the use of the quadratic formula.

What is the Quadratic Formula?

The quadratic equation formula to solve the equation ax 2 + bx + c = 0 is x = [-b ± √(b 2 - 4ac)]/2a. Here we obtain the two values of x, by applying the plus and minus symbols in this formula. Hence the two possible values of x are [-b + √(b 2 - 4ac)]/2a, and [-b - √(b 2 - 4ac)]/2a.

How do You Solve a Quadratic Equation?

There are several methods to solve quadratic equations, but the most common ones are factoring, using the quadratic formula, and completing the square.

  • Factoring involves finding two numbers that multiply to equal the constant term, c, and add up to the coefficient of x, b.
  • The quadratic formula is used when factoring is not possible, and it is given by x = [-b ± √(b 2 - 4ac)]/2a.
  • Completing the square involves rewriting the quadratic equation in a different form that allows you to easily solve for x.

What is Determinant in Quadratic Formula?

The value b 2 - 4ac is called the discriminant and is designated as D. The discriminant is part of the quadratic formula. The discriminants help us to find the nature of the roots of the quadratic equation, without actually finding the roots of the quadratic equation.

What are Some Real-Life Applications of Quadratic Equations?

Quadratic equations are used to find the zeroes of the parabola and its axis of symmetry . There are many real-world applications of quadratic equations.

  • They can be used in running time problems to evaluate the speed, distance or time while traveling by car, train or plane.
  • Quadratic equations describe the relationship between quantity and the price of a commodity.
  • Similarly, demand and cost calculations are also considered quadratic equation problems.
  • It can also be noted that a satellite dish or a reflecting telescope has a shape that is defined by a quadratic equation.

How are Quadratic Equations Different From Linear Equations?

A linear degree is an equation of a single degree and one variable, and a quadratic equation is an equation in two degrees and a single variable. A linear equation is of the the form ax + b = 0 and a quadratic equation is of the form ax 2 + bx + c = 0. A linear equation has a single root and a quadratic equation has two roots or two answers. Also, a quadratic equation is a product of two linear equations.

What Are the 4 Ways To Solve A Quadratic Equation?

The four ways of solving a quadratic equation are as follows.

  • Factorizing method
  • Roots of Quadratic Equation Formula Method
  • Method of Completing Squares
  • Graphing Method

How to Solve a Quadratic Equation by Completing the Square?

The quadratic equation is solved by the method of completing the square and it uses the formula (a + b)^2 = a 2 + 2ab + b 2 (or) (a - b)^2 = a 2 - 2ab + b 2 .

How to Find the Value of the Discriminant?

The value of the discriminant in a quadratic equation can be found from the variables and constant terms of the standard form of the quadratic equation ax 2 + bx + c = 0. The value of the discriminant is D = b 2 - 4ac, and it helps to predict the nature of roots of the quadratic equation, without actually finding the roots of the equation.

How Do You Solve Quadratic Equations With Graphing?

The quadratic equation can be solved similarly to a linear equal by graphing. Let us take the quadratic equation ax 2 + bx + c = 0 as y = ax 2 + bx + c . Here we take the set of values of x and y and plot the graph. The two points where this graph meets the x-axis, are the solutions of this quadratic equation.

How Important Is the Discriminant of a Quadratic Equation?

The discriminant is very much needed to easily find the nature of the roots of the quadratic equation. Without the discriminant, finding the nature of the roots of the equation is a long process, as we first need to solve the equation to find both the roots. Hence the discriminant is an important and needed quantity, which helps to easily find the nature of the roots of the quadratic equation.

Where Can I Find Quadratic Equation Solver?

To get the quadratic equation solver, click here . Here, we can enter the values of a, b, and c for the quadratic equation ax 2 + bx + c = 0, then it will give you the roots along with a step-by-step procedure.

What is the Use of Discriminants in Quadratic Formula?

The discriminant (D = b 2 - 4ac) is useful to predict the nature of the roots of the quadratic equation. For D > 0, the roots are real and distinct, for D = 0 the roots are real and equal, and for D < 0, the roots do not exist or the roots are imaginary complex numbers . With the help of this discriminant and with the least calculations, we can find the nature of the roots of the quadratic equation.

How do you Solve a Quadratic Equation without Using the Quadratic Formula?

There are two alternative methods to the quadratic formula. One method is to solve the quadratic equation through factorization, and another method is by completing the squares. In total there are three methods to find the roots of a quadratic equation.

How to Derive Quadratic Formula?

The algebra formula (a + b) 2 = a 2 + 2ab + b 2 is used to solve the quadratic equation and derive the quadratic formula. This algebraic formula is used to manipulate the quadratic equation and derive the quadratic formula to find the roots of the equation.

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Quadratic Functions Problems with Solutions

quadratic functions problems with detailed solutions are presented along with graphical interpretations of the solutions.

Review Vertex and Discriminant of Quadratic Functions

f(x) = a x 2 + b x + c

If a > 0 , the vertex is a minimum point and the minimum value of the quadratic function f is equal to k. This minimum value occurs at x = h. If a < 0 , the vertex is a maximum point and the maximum value of the quadratic function f is equal to k. This maximum value occurs at x = h. The quadratic function f(x) = a x 2 + b x + c can be written in vertex form as follows:

f(x) = a (x - h) 2 + k

Problems with Solutions

Problem 1 The profit (in thousands of dollars) of a company is given by.

P(x) = 5000 + 1000 x - 5 x 2

  • Function P that gives the profit is a quadratic function with the leading coefficient a = - 5. This function (profit) has a maximum value at x = h = - b / (2a) x = h = -1000 / (2(-5)) = 100
  • The maximum profit Pmax, when x = 100 thousands is spent on advertising, is given by the maximum value of function P k = c - b 2 / (4 a) b)
  • The maximum profit Pmax, when x = 100 thousands is spent on advertising, is also given by P(h = 100) P(100) = 5000 + 1000 (100) - 5 (100) 2 = 55000.
  • When the company spends 100 thousands dollars on advertising, the profit is maximum and equals 55000 dollars.

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Problem 2 An object is thrown vertically upward with an initial velocity of V o feet/sec. Its distance S(t), in feet, above ground is given by

  • S(t) is a quadratic function and the maximum value of S(t)is given by k = c - b 2 /(4 a) = 0 - (v o ) 2 / (4(-16))
  • This maximum value of S(t) has to be 300 feet in order for the object to reach a maximum distance above ground of 300 feet. - (v o ) 2 / (4(-16)) = 300
  • we now solve - (v o ) 2 / (4(-16)) = 300 for v o v o = 64*300 = 80 √3 feet/sec.

Problem 3 Find the equation of the quadratic function f whose graph passes through the point (2 , -8) and has x intercepts at (1 , 0) and (-2 , 0). Solution to Problem 3

  • Since the graph has x intercepts at (1 , 0) and (-2 , 0), the function has zeros at x = 1 and x = - 2 and may be written as follows. f(x) = a (x - 1)(x + 2)
  • The graph of f passes through the point (2 , -8), it follows that f(2) = - 8
  • which leads to - 8 = a (2 - 1)(2 + 2)
  • expand the right side of the above equation and group like terms -8 = 4 a
  • Solve the above equation for a to obtain a = - 2
  • The equation of f is given by f(x) = - 2 (x - 1)(x + 2)
  • Check answer f(1) = 0 f(-2) = 0 f(2) = - 2 (2 - 1)(2 + 2) = -8

Problem 4 Find values of the parameter m so that the graph of the quadratic function f given by

  • To find the points of intersection, you need to solve the system of equations y = x 2 + x + 1 y = m x
  • Substitute m x for y in the first equation to obtain mx = x 2 + x + 1
  • Write the above quadratic equation in standard form. x 2 + x (1 - m) + 1 = 0
  • Find the discriminant D of the above equation. D = (1 - m) 2 - 4(1)(1) D = (1 - m) 2 - 4 a)
  • For the graph of f and that of the line to have 2 points of intersection, D must be positive, which leads to (1 - m) 2 - 4 > 0
  • Solve the above inequality to obtain solution set for m in the intervals (- ∞ , -1) U (3 , + ∞) b)
  • For the graph of f and that of the line to have 1 point of intersection, D must be zero, which leads to (1 - m) 2 - 4 = 0
  • Solve the above equation to obtain 2 solutions for m. m = -1 m = 3 c)
  • For the graph of f and that of the line to have no points of intersection, D must be negative, which leads to (1 - m) 2 - 4 < 0

Problem 5 The quadratic function C(x) = a x 2 + b x + c represents the cost, in thousands of Dollars, of producing x items. C(x) has a minimum value of 120 thousands for x = 2000 and the fixed cost is equal to 200 thousands. Find the coefficients a,b and c. Solution to Problem 5

  • Function C is a quadratic function. Its minimum point, which is given as (2000,120) is the vertex of the graph of C. Hence we can write C(x) in vertex form as follows C(x) = a (x - 2000) 2 + 120
  • The fixed cost is the value of C(x) when x = 0. Hence C(0) = a (0 - 2000) 2 + 120 = 200
  • Solve for a a = 80 / 2000 2 = 0.00002
  • We expand C(x) and identify the coefficients a, b and c. C(x) = 0.00002 (x - 2000) 2 + 120 = 0.00002 x 2 - 0.08 x + 200 a = 0.00002 , b = -0.08 and c = 200.

Problem 6 Find the equation of the tangent line to the the graph of f(x) = - x 2 + x - 2 at x = 1. Solution to Problem 6

  • There are at least two methods to solve the above question. Method 1
  • Let the equation of the tangent line be of the form y = m x + b
  • and we therefore need to find m and b. The tangent line passes through the point (1 , f(1)) = (1 , -2)
  • Hence the equation in m and b - 2 = m (1) + b or m + b = -2
  • To find the point of tangency of the line and the graph of the quadratic function, we need to solve the system y = m x + b and y = - x 2 + x - 2
  • Substitute y by m x + b in the second equation of the system to obtain m x + b = - x 2 + x - 2
  • Write the above equation in standard form - x 2 + x (1 - m) - 2 - b = 0
  • For the line to be tangent to the graph of the quadratic function, the discriminant D of the above equation must be equal to zero. Hence D = b 2 - 4 a c = (1 - m) 2 - 4 (-1) (- 2 - b) = 0
  • which gives (1 - (- 2 - b) ) 2 + 4 (- 2 - b) = 0
  • Expand, simplify and write the above equation in standard form b 2 2 b + 1 = 0 (b + 1) 2 = 0
  • Solve for b b = - 1
  • Find m m = - 2 - b = -1

Questions with Solutions

Find the equation of the quadratic function f whose graph has x intercepts at (-1 , 0) and (3 , 0) and a y intercept at (0 , -4).

Question 2 Find values of the parameter c so that the graphs of the quadratic function f given by f(x) = x 2 + x + c and the graph of the line whose equation is given by y = 2 x have: a) 2 points of intersection, b) 1 point of intersection, c) no points of intersection.

Solutions to the Above Questions

Solution to Question 1

  • The x intercepts of the graph of f are the zero of f(x). Hence f(x) is of the form f(x) = a (x + 1)(x - 3)
  • We now need to find coefficient a using the y intercept f(0) = a(0 + 1)(0 - 3) = - 4
  • Solve for a a = 4 / 3
  • Hence f(x) = (4 / 3) (x + 1)(x - 3)

Solution to Question 2

  • To find the coordinates of the point of intersections of the graphs of f(x) = x 2 + x + c and y = 2 x, we need to solve the system y = x 2 + x + c and y = 2 x
  • which by substitution , gives the equation x 2 + x + c = 2x
  • Rewrite the above equation in standard form x 2 - x + c = 0
  • Find the discriminant D D = 1 - 4 c
  • Conclusion If D is positive or c < 1 / 4 , the two graphs intersect at two points. If D is equal to 0 or c = 1 / 4 , the two graphs intersect (touch) at 1 point. If D is negative or c > 1 / 4 , the two graphs have no point of intersection.

More References and Links to Quadratic Functions

  • Math Questions With Answers (13): Quadratic Functions.
  • Vertex and Intercepts Parabola Problems.
  • Find Vertex and Intercepts of Quadratic Functions - Calculator: An applet to solve calculate the vertex and x and y intercepts of the graph of a quadratic function.
  • Quadratic functions (general form).
  • Quadratic functions (standard form).
  • Graphing quadratic functions.
  • Solver to Analyze and Graph a Quadratic Function

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

Use this quiz to check your grade 6 to 8 students’ understanding of math equations. 10+ questions with answers covering a range of 6th, 7th and 8th grade math equations topics to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

  • Math equations
  • Rearranging equations
  • How to find the equation of a line
  • Solve equations with fractions
  • Linear equations
  • Writing linear equations
  • Substitution
  • Identity math
  • One step equation

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

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Quadratics or Quadratic Equations

Quadratics   can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations . The general form of the quadratic equation is: 

ax² + bx + c = 0

where x is an unknown variable and a, b, c are numerical coefficients. For example, x 2 + 2x +1 is a quadratic or quadratic equation. Here, a ≠ 0 because if it equals zero then the equation will not remain quadratic anymore and it will become a linear equation, such as: 

Thus, this equation cannot be called a quadratic equation.

The terms a, b and c are also called quadratic coefficients. 

The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations . The roots of any polynomial are the solutions for the given equation.

What is Quadratic Equation?

The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of:

where x is the unknown variable and a, b and c are the constant terms.

Standard Form of Quadratic Equation

Standard Form of Quadratic Equation

Since the quadratic includes only one unknown term or variable, thus it is called univariate. The power of variable x is always non-negative integers. Hence the equation is a polynomial equation with the highest power as 2.

The solution for this equation is the values of x, which are also called zeros. Zeros of the polynomial are the solution for which the equation is satisfied. In the case of quadratics, there are two roots or zeros of the equation. And if we put the values of roots or x on the left-hand side of the equation, it will equal to zero. Therefore, they are called zeros.

Quadratics Formula

The formula for a quadratic equation is used to find the roots of the equation. Since quadratics have a degree equal to two, therefore there will be two solutions for the equation. Suppose ax² + bx + c = 0 is the quadratic equation, then the formula to find the roots of this equation will be:

x = [-b±√(b 2 -4ac)]/2a

The sign of plus/minus indicates there will be two solutions for x. Learn in detail the quadratic formula here.

Examples of Quadratics

Beneath are the illustrations of quadratic equations of the form (ax² + bx + c = 0)

  • x² –x – 9 = 0
  • 5x² – 2x – 6 = 0
  • 3x² + 4x + 8 = 0
  • -x² +6x + 12 = 0

Examples of a quadratic equation with the absence of a ‘ C ‘- a constant term.

  • -x² – 9x = 0
  • x² + 2x = 0
  • -6x² – 3x = 0
  • -5x² + x = 0
  • -12x² + 13x = 0
  • 11x² – 27x = 0

Following are the examples of a quadratic equation in factored form

  • (x – 6)(x + 1) = 0  [ result obtained after solving is x² – 5x – 6 = 0]
  • –3(x – 4)(2x + 3) = 0  [result obtained after solving is -6x² + 15x + 36 = 0]
  • (x − 5)(x + 3) = 0  [result obtained after solving is x² − 2x − 15 = 0]
  • (x – 5)(x + 2) = 0  [ result obtained after solving is x² – 3x – 10 = 0]
  • (x – 4)(x + 2) = 0  [result obtained after solving is x² – 2x – 8 = 0]
  • (2x+3)(3x – 2) = 0  [result obtained after solving is 6x² + 5x – 6]

Below are the examples of a quadratic equation with an absence of linear co – efficient ‘ bx’

  • 2x² – 64 = 0
  • x² – 16 = 0
  • 9x² + 49 = 0
  • -2x² – 4 = 0
  • 4x² + 81 = 0
  • -x² – 9 = 0

How to Solve Quadratic Equations?

There are basically four methods of solving quadratic equations. They are:

  • Completing the square

Using Quadratic Formula

  • Taking the square root

Factoring of Quadratics

  • Begin with a equation of the form ax² + bx + c = 0
  • Ensure that it is set to adequate zero.
  • Factor the left-hand side of the equation by assuming zero on the right-hand side of the equation.
  • Assign each factor equal to zero.
  • Now solve the equation in order to determine the values of x.

Suppose if the main coefficient is not equal to one then deliberately, you have to follow a methodology in the arrangement of the factors.

(2x+3)(x-2)=0

Learn more about the factorization of quadratic equations here.

Completing the Square Method

Let us learn this method with example.

Example: Solve 2x 2 – x – 1 = 0.

First, move the constant term to the other side of the equation.

2x 2 – x = 1

Dividing both sides by 2.

x 2 – x/2 = ½

Add the square of half of the coefficient of x, (b/2a) 2 , on both the sides, i.e., 1/16

x 2 – x/2 + 1/16 = ½ + 1/16

Now we can factor the right side,

(x-¼) 2 = 9/16 = (¾) 2

Taking root on both sides;

X – ¼ = ±3/4

Add ¼ on both sides

X = ¼ + ¾ = 4/4 = 1

X = ¼ – ¾ = -2/4 = -½ 

To learn more about completing the square method, click here .

For the given Quadratic equation of the form, ax² + bx + c = 0

Therefore the roots of the given equation can be found by:

\(\begin{array}{l}x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\end{array} \)

where ± (one plus and one minus) represent two distinct roots of the given equation.

Taking the Square Root

We can use this method for the equations such as:

x 2 + a 2 = 0

Example: Solve x 2 – 50 = 0.

x 2 – 50 = 0

Taking the roots both sides

√x 2 = ±√50

x = ±√(2 x 5 x 5)

Thus, we got the required solution.

Related Articles

Video lesson on quadratic equations, range of quadratic equations.

solving problems involving quadratic equation examples

Solved Problems on Quadratic Equations

Applications of quadratic equations.

Many real-life word problems can be solved using quadratic equations. While solving word problems, some common quadratic equation applications include speed problems and Geometry area problems.

  • Solving the problems related to finding the area of quadrilateral such as rectangle, parallelogram and so on
  • Solving Word Problems involving Distance, speed, and time, etc.,

Example: Find the width of a rectangle of area 336 cm2 if its length is equal to the 4 more than twice its width. Solution: Let x cm be the width of the rectangle. Length = (2x + 4) cm We know that Area of rectangle = Length x Width x(2x + 4) = 336 2x 2 + 4x – 336 = 0 x 2 + 2x – 168 = 0 x 2 + 14x – 12x – 168 = 0 x(x + 14) – 12(x + 14) = 0 (x + 14)(x – 12) = 0 x = -14, x = 12 Measurement cannot be negative. Therefore, Width of the rectangle = x = 12 cm

Practice Questions

  • Solve x 2 + 2 x + 1 = 0.
  • Solve 5x 2 + 6x + 1 = 0
  • Solve 2x 2 + 3 x + 2 = 0.
  • Solve x 2 − 4x + 6.25 = 0

Frequently Asked Questions on Quadratics

What is a quadratic equation, what are the methods to solve a quadratic equation, is x 2 – 1 a quadratic equation, what is the solution of x 2 + 4 = 0, write the quadratic equation in the form of sum and product of roots., leave a comment cancel reply.

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solving problems involving quadratic equation examples

Thanks a lot ,This was very useful for me

x=√9 Squaring both the sides, x^2 = 9 x^2 – 9 = 0 It is a quadratic equation.

solving problems involving quadratic equation examples

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Procedure Steps, Example Solved Problem - Solving Problems Involving Quadratic Equations | 10th Mathematics : UNIT 3 : Algebra

Chapter: 10th mathematics : unit 3 : algebra.

Solving Problems Involving Quadratic Equations

Steps to solve a problem

Step 1 Convert the word problem to a quadratic equation form

Step 2 Solve the quadratic equation obtained in any one of the above three methods.

Step 3 Relate the mathematical solution obtained to the statement asked in the question.

Example 3.37

The product of Kumaran’s age (in years) two years ago and his age four years from now is one more than twice his present age. What is his present age?

Let the present age of Kumaran be x years. 

Two years ago, his age = ( x − 2)  years.

Four years from now, his age = ( x + 4)  years.

( x − 2)( x + 4)  = 1 +2 x

x 2 + 2 x − 8 = 1 +2 x gives ( x − 3)( x + 3) = 0 then, x = ± 3

Therefore, x = 3 (Rejecting −3 as age cannot be negative)

Kumaran’s present age is 3 years.

Example 3.38

A ladder 17 feet long is leaning against a wall. If the ladder, vertical wall and the floor from the bottom of the wall to the ladder form a right triangle, find the height of the wall where the top of the ladder meets if the distance between bottom of the wall to bottom of the ladder is 7 feet less than the height of the wall?

Let the height of the wall AB = x feet

As per the given data BC = ( x –7) feet

In the right triangle ABC , AC =17 ft, BC = ( x –7) feet

By Pythagoras theorem, AC 2 = AB 2 + BC 2

solving problems involving quadratic equation examples

(17) 2   = x 2   + ( x − 7) 2 ; 289 = x 2 + x 2 − 14 x + 49

x 2 − 7 x −120 = 0 hence,   ( x − 15)( x + 8) = 0 then, x = 15 (or) −8

Therefore, height of the wall AB = 15 ft (Rejecting −8 as height cannot be negative)

Example 3.39

A flock of swans contained x 2 members. As the clouds gathered, 10 x went to a lake and one-eighth of the members flew away to a garden. The remaining three pairs played about in the water. How many swans were there in total?

As given there are x 2 swans.

As per the given data x 2 − 10 x – (1/8) x 2 = 6 we get, 7 x 2 − 680 x − 48 = 0

solving problems involving quadratic equation examples

Therefore, x = 12, -4/7

Here x = 4/7 is not possible as the number of swans cannot be negative.

Hence, x = 12. Therefore total number of swans is x 2 = 144.

Example 3.40

A passenger train takes 1 hr more than an express train to travel a distance of 240 km from Chennai to Virudhachalam. The speed of passenger train is less than that of an express train by 20 km per hour. Find the average speed of both the trains.

Let the average speed of passenger train be x km/hr.

Then the average speed of express train will be ( x + 20) km/hr

Time taken by the passenger train to cover distance of 240 km = 240/ x hr

Time taken by express train to cover distance of 240 km = 240 / ( x +20) hr

solving problems involving quadratic equation examples

x 2 + 20 x – 4800 = 0 gives, ( x + 80)( x − 60) = 0 we get, x = –80 or 60.

Therefore x = 60 (Rejecting -80 as speed cannot be negative)

Average speed of the passenger train is 60 km/hr

Average speed of the express train is 80 km/hr.

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SOLVING WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS

Problem 1 :

If the difference between a number and its reciprocal is ²⁴⁄₅ , find the number.

Let x be the required number. Then its reciprocal is  ¹⁄ₓ be its reciprocal.

5 (x 2  - 1) = 24x

5 x 2  - 5 = 24x

5x 2  - 24x - 5 = 0

5x 2  - 25x + 1x - 5 = 0

5x(x - 5) + 1(x - 5) = 0

(5x + 1)(x - 5) = 0

5x + 1 = 0  or  x - 5 = 0

x = - ⅕   or  x = 5

So, the required number is - ⅕  or -5.

Problem 2 :

A garden measuring 12m by 16m is to have a pedestrian pathway that is w meters wide installed all the way around so that it increases the total area to 285 m 2 . What is the width of the pathway?

10thnewsylabusex3.12q2

From the picture above, length of the garden including pathway is (12 + 2w) and width is (16 + 2w).

Total Area = 285 m 2

Length  ⋅ Width = 285

(12 + 2w) (16 + 2w) = 285

  192 + 24w + 32w + 4w 2  = 285

 4w 2 + 56w + 192 - 285 = 0

4w 2  + 56w - 93 = 0

Solve the above quadratic equation using quadratic formula.

w = -15.5  or  w = 1.5 

Since w is the width of the pathway, it can not be negative. So, w = 1.5.

Therefore , the width of the pathway is 1.5 m .

Problem 3 :

A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the bus.

Distance covered = 90 km

Let x be the original speed  of the bus.

Increased speed = x + 15

Time taken by the bus in original speed :

Time taken by the bus in increased speed :

= ⁹⁰⁄₍ₓ ₊ ₁₅₎

From the given information, the difference between and  ⁹⁰⁄ₓ  and ⁹⁰⁄₍ₓ ₊ ₁₅₎  is 30 minutes or ½ hour .

2700 = x 2  + 15x 

x 2  + 15x - 2700 = 0

x 2  + 60x - 45x - 2700 = 0

x(x + 60) - 45(x + 60) = 0

(x - 45)(x + 60) = 0

x = 45  or  x = -60

Since x represents the original speed of the bus, it can never be negative. So, x = 45.

Therefore, the original speed of the bus is 45 km per hour.

Problem 4 :

John and Jivanti together had 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they have now is 124. Find the number marbles each one them had initially.

Let x be the number of marbles that john has.

Then, the number of marbles that Jivanthi has

After losing 5 marbles, 

number of marbles with John = x - 5

number of marbles with Jivanti = 45 - x - 5 = 40 - x

The product of number of marbles = 124

(x - 5)(40 - x) = 124

40x - x 2  - 200 + 5x = 124

45x - x 2  - 200 = 124

x 2  - 45x + 124 + 200 = 0

x 2  - 45x + 324 = 0

x 2 - 36x - 9x + 324 = 0

x(x - 36) - 9(x - 36) = 0

(x - 9)(x - 36) = 0

x - 9 = 0  or  x - 36 = 0

x = 9  or  x = 36

If x = 9, 

If x = 36, 

So, John had 36 marbles, when Jivanti had 9 marbles or Jivanti had 36 marbles, when John had 9 marbles. 

Problem 5 :

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in dollars) was found to be 55 minus the number of toys produced in a day, the total cost of production was $750. Find the number of toys produced on that day.

Let x be the number of toys produced in a particular day.

Cost of production  of one toy (in dollars) :

Total cost =  Number of toys  ⋅  Cost of one toy

750 = x (55 - x)

750 = 55x - x 2

x 2  - 55x + 750 = 0

x 2  - 30x - 25x + 750 = 0

x(x - 30) - 25(x - 30) = 0

(x - 30)(x - 25) = 0

  x - 30 = 0  or  x - 25 = 0

   x = 30  or  x = 25

So, the number of toys produced on that particular day is 30 or 25.

Problem 6 :

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Let x be the length of shorter side of the rectangle

length of diagonal = x + 60

length of longer side = x + 30

6thquestion

Using Pythagorean Theorem in the right triangle ABC,

(x + 60) 2  = x 2  + (x + 30 ) 2

x 2  + 60 2  + 2(x)(60) = x 2 + x 2  + 2(x)(30) + 30 2

x 2  + 3600 + 120x = 2x 2  + 60x + 900

2 x 2  - x 2  + 60x - 120x + 900 - 3600 = 0

x 2  - 60x - 2700 = 0

x 2  - 90x + 30x - 2700 = 0

x(x - 90) + 30(x - 90) = 0

 (x + 30)(x - 90) = 0

  x + 30 = 0  or  x - 90 = 0

   x = -30  or  x = 90

Because x represents breadth of the rectangle, it can never be negative. So,  x = 90.

Therefore, 

length of the shorter side = 90 m

length of the longer side = 90 + 30 = 120 m

Problem 7 :

The difference of squares of two positive numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Let x be the larger number and  y be the smaller number.

The square of the smaller number is 8 times the larger number.

y 2  = 8x -----(1)

The difference of squares of two numbers is 180.

 x 2  - y 2  = 180

Substitute  y 2  = 8x and solve for x.

    x 2  - 8x = 180

x 2  - 8x - 180 = 0

    x 2  - 18x + 10x - 180 = 0

x(x - 18) + 10(x - 18) = 0

 (x - 18)(x + 10) = 0 

 x - 18 = 0  or  x + 10 = 0

 x = 18  or  x = -10

Because the numbers are positive, x = -10 can not be accepted. So,  x = 18.

Substitute x = 18 into (1).

y 2  = 8(18)

y 2  = 144

 y =  √144

 y = 12

Therefore, the larger number is 18  and the smaller number is 12.

Problem 8 :

A train travels 360 miles at a uniform speed. If the speed had been 5 miles/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Distance covered = 360 miles

Let x be the original speed of the train.

If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey.

Increased speed = x + 5

Time taken by the train in original speed :

=  ³⁶⁰⁄ₓ

Time taken by the train in increased speed :

=  ³⁶⁰⁄₍ₓ ₊ ₅₎

From the given information, the difference between and  ³⁶⁰⁄ₓ   and ³⁶⁰⁄₍ₓ ₊ ₅₎  is 1  hour .

1800 = x 2  + 5x

x 2  + 5x - 1800 = 0

x 2  - 40x + 45x - 1800 = 0

x(x - 40) + 45(x - 40) = 0

(x - 40)(x + 45) = 0

 x - 40 = 0  or  x + 45 = 0

 x = 40  or  x = -45

Since x represents the original speed of the train, it can never be negative. So, x = 40.

Therefore, the original speed of the train is 40 miles/hr.

Problem 9 :

Find two consecutive positive even integers whose squares have the sum 340.

Let x and ( x + 2) be the two positive even integers.

Sum of their squares = 340

x 2 + (x + 2) 2  = 340

x 2 + (x + 2)(x + 2) = 340

x 2 + x 2  + 2x + 2x + 4 = 340

2x 2 + 4x + 4 = 340

2x 2 + 4x - 336 = 0

Divide both sides by 2.

x 2 + 2x - 168 = 0

Solve by factoring.

x 2  - 12x + 14x - 168 = 0

x(x - 12) + 14(x - 12) = 0

(x - 12)(x + 14) = 0

x - 12 = 0  or  x + 14 = 0

x = 12  or  x = -14

Since the integers are positive, x can not be negative.

Therefore, the two positive even integers are 12 and 14.

Problem 10 :

The sum of squares of three consecutive natural numbers is 194. Determine the numbers.

Let x , x + 1 and x + 2 be three consecutive natural numbers.

Sum of their squares = 194

x 2  + (x + 1) 2  + (x + 2) 2  = 194

x 2  + (x + 1)(x + 1) + (x + 2)(x + 2) = 194

x 2  + x 2  + x + x + 1 + x 2  + 2x + 2x + 4 = 194

3x 2  + 6x + 5 = 194

 3x 2  + 6x - 189 = 0

Divide both sides by 3.

x 2  + 2x - 63 = 0

x 2  - 7x + 9x - 63 = 0

x(x - 7) + 9(x - 7) = 0

(x - 7)(x + 9) = 0

x - 7 = 0  or  x + 9 = 0

x = 7  or  x = -9

Since the numbers are natural numbers, x  can not be negative.

Therefore, the three  consecutive natural numbers are 7, 8 and 9.

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