solving quadratic equations homework 5

Solving Quadratic Equations: Worksheets with Answers

Whether you want a homework, some cover work, or a lovely bit of extra practise, this is the place for you. And best of all they all (well, most!) come with answers.

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Number line.

• ax^2+bx+c=0
• x^2+2x+1=3x-10
• 2x^2+4x-6=0
• How do you calculate a quadratic equation?
• To solve a quadratic equation, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
• What is the quadratic formula?
• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
• Does any quadratic equation have two solutions?
• There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

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• High School Math Solutions – Quadratic Equations Calculator, Part 3 On the last post we covered completing the square (see link). It is pretty strait forward if you follow all the...

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Quadratic Equation Worksheets (pdfs)

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• Solve Quadratic Equations by Factoring
• Solve Quadratic Equations by Completing the Square
• Quadratic Formula Worksheet (real solutions)
• Quadratic Formula Worksheet (complex solutions)
• Quadratic Formula Worksheet (both real and complex solutions)
• Discriminant Worksheet
• Sum and Product of Roots
• Radical Equations Worksheet

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2.5 Quadratic Equations

Learning objectives.

In this section, you will:

• Solve quadratic equations by factoring.
• Solve quadratic equations by the square root property.
• Solve quadratic equations by completing the square.
• Solve quadratic equations by using the quadratic formula.

The computer monitor on the left in Figure 1 is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods.

• Solving Quadratic Equations by Factoring

An equation containing a second-degree polynomial is called a quadratic equation . For example, equations such as 2 x 2 + 3 x − 1 = 0 2 x 2 + 3 x − 1 = 0 and x 2 − 4 = 0 x 2 − 4 = 0 are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.

Often the easiest method of solving a quadratic equation is factoring . Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.

If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if a ⋅ b = 0 , a ⋅ b = 0 , then a = 0 a = 0 or b = 0 , b = 0 , where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.

Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression ( x − 2 ) ( x + 3 ) ( x − 2 ) ( x + 3 ) by multiplying the two factors together.

The product is a quadratic expression. Set equal to zero, x 2 + x − 6 = 0 x 2 + x − 6 = 0 is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.

The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , where a , b , and c are real numbers, and a ≠ 0. a ≠ 0. The equation x 2 + x − 6 = 0 x 2 + x − 6 = 0 is in standard form.

We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.

The Zero-Product Property and Quadratic Equations

The zero-product property states

where a and b are real numbers or algebraic expressions.

A quadratic equation is an equation containing a second-degree polynomial; for example

where a , b , and c are real numbers, and if a ≠ 0 , a ≠ 0 , it is in standard form.

Solving Quadratics with a Leading Coefficient of 1

In the quadratic equation x 2 + x − 6 = 0 , x 2 + x − 6 = 0 , the leading coefficient, or the coefficient of x 2 , x 2 , is 1. We have one method of factoring quadratic equations in this form.

Given a quadratic equation with the leading coefficient of 1, factor it.

• Find two numbers whose product equals c and whose sum equals b .
• Use those numbers to write two factors of the form ( x + k ) or  ( x − k ) , ( x + k ) or  ( x − k ) , where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and −2 , −2 , the factors are ( x + 1 ) ( x − 2 ) . ( x + 1 ) ( x − 2 ) .
• Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

Factoring and Solving a Quadratic with Leading Coefficient of 1

Factor and solve the equation: x 2 + x − 6 = 0. x 2 + x − 6 = 0.

To factor x 2 + x − 6 = 0 , x 2 + x − 6 = 0 , we look for two numbers whose product equals −6 −6 and whose sum equals 1. Begin by looking at the possible factors of −6. −6.

The last pair, 3 ⋅ ( −2 ) 3 ⋅ ( −2 ) sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

The two solutions are 2 2 and −3. −3. We can see how the solutions relate to the graph in Figure 2 . The solutions are the x- intercepts of y = x 2 + x − 6 = 0. y = x 2 + x − 6 = 0.

Factor and solve the quadratic equation: x 2 − 5 x − 6 = 0. x 2 − 5 x − 6 = 0.

Solve the Quadratic Equation by Factoring

Solve the quadratic equation by factoring: x 2 + 8 x + 15 = 0. x 2 + 8 x + 15 = 0.

Find two numbers whose product equals 15 15 and whose sum equals 8. 8. List the factors of 15. 15.

The numbers that add to 8 are 3 and 5. Then, write the factors, set each factor equal to zero, and solve.

The solutions are −3 −3 and −5. −5.

Solve the quadratic equation by factoring: x 2 − 4 x − 21 = 0. x 2 − 4 x − 21 = 0.

Using the Zero-Product Property to Solve a Quadratic Equation Written as the Difference of Squares

Solve the difference of squares equation using the zero-product property: x 2 − 9 = 0. x 2 − 9 = 0.

Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.

The solutions are 3 3 and −3. −3.

Solve by factoring: x 2 − 25 = 0. x 2 − 25 = 0.

Solving a Quadratic Equation by Factoring when the Leading Coefficient is not 1

When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:

• With the quadratic in standard form, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , multiply a ⋅ c . a ⋅ c .
• Find two numbers whose product equals a c a c and whose sum equals b . b .
• Rewrite the equation replacing the b x b x term with two terms using the numbers found in step 2 as coefficients of x.
• Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
• Factor out the expression in parentheses.
• Set the expressions equal to zero and solve for the variable.

Solving a Quadratic Equation Using Grouping

Use grouping to factor and solve the quadratic equation: 4 x 2 + 15 x + 9 = 0. 4 x 2 + 15 x + 9 = 0.

First, multiply a c : 4 ( 9 ) = 36. a c : 4 ( 9 ) = 36. Then list the factors of 36. 36.

The only pair of factors that sums to 15 15 is 3 + 12. 3 + 12. Rewrite the equation replacing the b term, 15 x , 15 x , with two terms using 3 and 12 as coefficients of x . Factor the first two terms, and then factor the last two terms.

Solve using the zero-product property.

The solutions are − 3 4 , − 3 4 , and −3. −3. See Figure 3 .

Solve using factoring by grouping: 12 x 2 + 11 x + 2 = 0. 12 x 2 + 11 x + 2 = 0.

Solving a Polynomial of Higher Degree by Factoring

Solve the equation by factoring: −3 x 3 − 5 x 2 − 2 x = 0. −3 x 3 − 5 x 2 − 2 x = 0.

This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out − x − x from all of the terms and then proceed with grouping.

Use grouping on the expression in parentheses.

Now, we use the zero-product property. Notice that we have three factors.

The solutions are 0 , 0 , − 2 3 , − 2 3 , and −1. −1.

Solve by factoring: x 3 + 11 x 2 + 10 x = 0. x 3 + 11 x 2 + 10 x = 0.

Using the Square Root Property

When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property , in which we isolate the x 2 x 2 term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the x 2 x 2 term so that the square root property can be used.

The Square Root Property

With the x 2 x 2 term isolated, the square root property states that:

where k is a nonzero real number.

Given a quadratic equation with an x 2 x 2 term but no x x term, use the square root property to solve it.

• Isolate the x 2 x 2 term on one side of the equal sign.
• Take the square root of both sides of the equation, putting a ± ± sign before the expression on the side opposite the squared term.
• Simplify the numbers on the side with the ± ± sign.

Solving a Simple Quadratic Equation Using the Square Root Property

Solve the quadratic using the square root property: x 2 = 8. x 2 = 8.

Take the square root of both sides, and then simplify the radical. Remember to use a ± ± sign before the radical symbol.

The solutions are 2 2 , 2 2 , −2 2 . −2 2 .

Solving a Quadratic Equation Using the Square Root Property

Solve the quadratic equation: 4 x 2 + 1 = 7. 4 x 2 + 1 = 7.

First, isolate the x 2 x 2 term. Then take the square root of both sides.

The solutions are 6 2 , 6 2 , and − 6 2 . − 6 2 .

Solve the quadratic equation using the square root property: 3 ( x − 4 ) 2 = 15. 3 ( x − 4 ) 2 = 15.

• Completing the Square

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square . Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a , must equal 1. If it does not, then divide the entire equation by a . Then, we can use the following procedures to solve a quadratic equation by completing the square.

We will use the example x 2 + 4 x + 1 = 0 x 2 + 4 x + 1 = 0 to illustrate each step.

Given a quadratic equation that cannot be factored, and with a = 1 , a = 1 , first add or subtract the constant term to the right side of the equal sign.

Multiply the b term by 1 2 1 2 and square it.

Add ( 1 2 b ) 2 ( 1 2 b ) 2 to both sides of the equal sign and simplify the right side. We have

The left side of the equation can now be factored as a perfect square.

Use the square root property and solve.

The solutions are −2 + 3 , −2 + 3 , and −2 − 3 . −2 − 3 .

Solving a Quadratic by Completing the Square

Solve the quadratic equation by completing the square: x 2 − 3 x − 5 = 0. x 2 − 3 x − 5 = 0.

First, move the constant term to the right side of the equal sign.

Then, take 1 2 1 2 of the b term and square it.

Add the result to both sides of the equal sign.

Factor the left side as a perfect square and simplify the right side.

The solutions are 3 + 29 2 3 + 29 2 and 3 - 29 2 3 - 29 2 .

Solve by completing the square: x 2 − 6 x = 13. x 2 − 6 x = 13.

Using the Quadratic Formula

The fourth method of solving a quadratic equation is by using the quadratic formula , a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.

We can derive the quadratic formula by completing the square . We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by −1 −1 and obtain a positive a . Given a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , a ≠ 0 , a ≠ 0 , we will complete the square as follows:

First, move the constant term to the right side of the equal sign:

As we want the leading coefficient to equal 1, divide through by a :

Then, find 1 2 1 2 of the middle term, and add ( 1 2 b a ) 2 = b 2 4 a 2 ( 1 2 b a ) 2 = b 2 4 a 2 to both sides of the equal sign:

Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:

Now, use the square root property, which gives

Finally, add − b 2 a − b 2 a to both sides of the equation and combine the terms on the right side. Thus,

The Quadratic Formula

Written in standard form, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , any quadratic equation can be solved using the quadratic formula :

where a , b , and c are real numbers and a ≠ 0. a ≠ 0.

Given a quadratic equation, solve it using the quadratic formula

• Make sure the equation is in standard form: a x 2 + b x + c = 0. a x 2 + b x + c = 0.
• Make note of the values of the coefficients and constant term, a , b , a , b , and c . c .
• Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
• Calculate and solve.

Solve the Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation: x 2 + 5 x + 1 = 0. x 2 + 5 x + 1 = 0.

Identify the coefficients: a = 1 , b = 5 , c = 1. a = 1 , b = 5 , c = 1. Then use the quadratic formula.

Solving a Quadratic Equation with the Quadratic Formula

Use the quadratic formula to solve x 2 + x + 2 = 0. x 2 + x + 2 = 0.

First, we identify the coefficients: a = 1 , b = 1 , a = 1 , b = 1 , and c = 2. c = 2.

Substitute these values into the quadratic formula.

The solutions to the equation are − 1 + i 7 2 − 1 + i 7 2 and − 1 − i 7 2 − 1 − i 7 2

Solve the quadratic equation using the quadratic formula: 9 x 2 + 3 x − 2 = 0. 9 x 2 + 3 x − 2 = 0.

The Discriminant

The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant , or the expression under the radical, b 2 − 4 a c . b 2 − 4 a c . The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table 1 relates the value of the discriminant to the solutions of a quadratic equation.

For a x 2 + b x + c = 0 a x 2 + b x + c = 0 , where a a , b b , and c c are real numbers, the discriminant is the expression under the radical in the quadratic formula: b 2 − 4 a c . b 2 − 4 a c . It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.

Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation

Use the discriminant to find the nature of the solutions to the following quadratic equations:

• ⓐ x 2 + 4 x + 4 = 0 x 2 + 4 x + 4 = 0
• ⓑ 8 x 2 + 14 x + 3 = 0 8 x 2 + 14 x + 3 = 0
• ⓒ 3 x 2 − 5 x − 2 = 0 3 x 2 − 5 x − 2 = 0
• ⓓ 3 x 2 − 10 x + 15 = 0 3 x 2 − 10 x + 15 = 0

Calculate the discriminant b 2 − 4 a c b 2 − 4 a c for each equation and state the expected type of solutions.

x 2 + 4 x + 4 = 0 x 2 + 4 x + 4 = 0

b 2 − 4 a c = ( 4 ) 2 − 4 ( 1 ) ( 4 ) = 0. b 2 − 4 a c = ( 4 ) 2 − 4 ( 1 ) ( 4 ) = 0. There will be one rational double solution.

8 x 2 + 14 x + 3 = 0 8 x 2 + 14 x + 3 = 0

b 2 − 4 a c = ( 14 ) 2 − 4 ( 8 ) ( 3 ) = 100. b 2 − 4 a c = ( 14 ) 2 − 4 ( 8 ) ( 3 ) = 100. As 100 100 is a perfect square, there will be two rational solutions.

3 x 2 − 5 x − 2 = 0 3 x 2 − 5 x − 2 = 0

b 2 − 4 a c = ( −5 ) 2 − 4 ( 3 ) ( −2 ) = 49. b 2 − 4 a c = ( −5 ) 2 − 4 ( 3 ) ( −2 ) = 49. As 49 49 is a perfect square, there will be two rational solutions.

3 x 2 −10 x + 15 = 0 3 x 2 −10 x + 15 = 0

b 2 − 4 a c = ( −10 ) 2 − 4 ( 3 ) ( 15 ) = −80. b 2 − 4 a c = ( −10 ) 2 − 4 ( 3 ) ( 15 ) = −80. There will be two complex solutions.

Using the Pythagorean Theorem

One of the most famous formulas in mathematics is the Pythagorean Theorem . It is based on a right triangle, and states the relationship among the lengths of the sides as a 2 + b 2 = c 2 , a 2 + b 2 = c 2 , where a a and b b refer to the legs of a right triangle adjacent to the 90° 90° angle, and c c refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.

We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.

The Pythagorean Theorem is given as

where a a and b b refer to the legs of a right triangle adjacent to the 90 ∘ 90 ∘ angle, and c c refers to the hypotenuse, as shown in Figure 4 .

Finding the Length of the Missing Side of a Right Triangle

Find the length of the missing side of the right triangle in Figure 5 .

As we have measurements for side b and the hypotenuse, the missing side is a.

Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.

Access these online resources for additional instruction and practice with quadratic equations.

• The Zero-Product Property
• Quadratic Formula with Two Rational Solutions
• Length of a leg of a right triangle

2.5 Section Exercises

How do we recognize when an equation is quadratic?

When we solve a quadratic equation, how many solutions should we always start out seeking? Explain why when solving a quadratic equation in the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 we may graph the equation y = a x 2 + b x + c y = a x 2 + b x + c and have no zeroes ( x -intercepts).

When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?

In the quadratic formula, what is the name of the expression under the radical sign b 2 − 4 a c , b 2 − 4 a c , and how does it determine the number of and nature of our solutions?

Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.

For the following exercises, solve the quadratic equation by factoring.

x 2 + 4 x − 21 = 0 x 2 + 4 x − 21 = 0

x 2 − 9 x + 18 = 0 x 2 − 9 x + 18 = 0

2 x 2 + 9 x − 5 = 0 2 x 2 + 9 x − 5 = 0

6 x 2 + 17 x + 5 = 0 6 x 2 + 17 x + 5 = 0

4 x 2 − 12 x + 8 = 0 4 x 2 − 12 x + 8 = 0

3 x 2 − 75 = 0 3 x 2 − 75 = 0

8 x 2 + 6 x − 9 = 0 8 x 2 + 6 x − 9 = 0

4 x 2 = 9 4 x 2 = 9

2 x 2 + 14 x = 36 2 x 2 + 14 x = 36

5 x 2 = 5 x + 30 5 x 2 = 5 x + 30

4 x 2 = 5 x 4 x 2 = 5 x

7 x 2 + 3 x = 0 7 x 2 + 3 x = 0

x 3 − 9 x = 2 x 3 − 9 x = 2

For the following exercises, solve the quadratic equation by using the square root property.

x 2 = 36 x 2 = 36

x 2 = 49 x 2 = 49

( x − 1 ) 2 = 25 ( x − 1 ) 2 = 25

( x − 3 ) 2 = 7 ( x − 3 ) 2 = 7

( 2 x + 1 ) 2 = 9 ( 2 x + 1 ) 2 = 9

( x − 5 ) 2 = 4 ( x − 5 ) 2 = 4

For the following exercises, solve the quadratic equation by completing the square. Show each step.

x 2 − 9 x − 22 = 0 x 2 − 9 x − 22 = 0

2 x 2 − 8 x − 5 = 0 2 x 2 − 8 x − 5 = 0

x 2 − 6 x = 13 x 2 − 6 x = 13

x 2 + 2 3 x − 1 3 = 0 x 2 + 2 3 x − 1 3 = 0

2 + z = 6 z 2 2 + z = 6 z 2

6 p 2 + 7 p − 20 = 0 6 p 2 + 7 p − 20 = 0

2 x 2 − 3 x − 1 = 0 2 x 2 − 3 x − 1 = 0

For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.

2 x 2 − 6 x + 7 = 0 2 x 2 − 6 x + 7 = 0

x 2 + 4 x + 7 = 0 x 2 + 4 x + 7 = 0

3 x 2 + 5 x − 8 = 0 3 x 2 + 5 x − 8 = 0

9 x 2 − 30 x + 25 = 0 9 x 2 − 30 x + 25 = 0

2 x 2 − 3 x − 7 = 0 2 x 2 − 3 x − 7 = 0

6 x 2 − x − 2 = 0 6 x 2 − x − 2 = 0

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution .

2 x 2 + 5 x + 3 = 0 2 x 2 + 5 x + 3 = 0

x 2 + x = 4 x 2 + x = 4

3 x 2 − 5 x + 1 = 0 3 x 2 − 5 x + 1 = 0

x 2 + 4 x + 2 = 0 x 2 + 4 x + 2 = 0

4 + 1 x − 1 x 2 = 0 4 + 1 x − 1 x 2 = 0

For the following exercises, enter the expressions into your graphing utility and find the zeroes to the equation (the x -intercepts) by using 2 nd CALC 2:zero . Recall finding zeroes will ask left bound (move your cursor to the left of the zero,enter), then right bound (move your cursor to the right of the zero,enter), then guess (move your cursor between the bounds near the zero, enter). Round your answers to the nearest thousandth.

Y 1 = 4 x 2 + 3 x − 2 Y 1 = 4 x 2 + 3 x − 2

Y 1 = −3 x 2 + 8 x − 1 Y 1 = −3 x 2 + 8 x − 1

Y 1 = 0.5 x 2 + x − 7 Y 1 = 0.5 x 2 + x − 7

To solve the quadratic equation x 2 + 5 x − 7 = 4 , x 2 + 5 x − 7 = 4 , we can graph these two equations

Y 1 = x 2 + 5 x − 7 Y 2 = 4 Y 1 = x 2 + 5 x − 7 Y 2 = 4

and find the points of intersection. Recall 2 nd CALC 5:intersection. Do this and find the solutions to the nearest tenth.

To solve the quadratic equation 0.3 x 2 + 2 x − 4 = 2 , 0.3 x 2 + 2 x − 4 = 2 , we can graph these two equations

Y 1 = 0.3 x 2 + 2 x − 4 Y 2 = 2 Y 1 = 0.3 x 2 + 2 x − 4 Y 2 = 2

Beginning with the general form of a quadratic equation, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , solve for x by using the completing the square method, thus deriving the quadratic formula.

Show that the sum of the two solutions to the quadratic equation is − b a − b a .

A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is 119 ft. 2 . Solve the quadratic equation to find the length and width.

Abercrombie and Fitch stock had a price given as P = 0.2 t 2 − 5.6 t + 50.2 , P = 0.2 t 2 − 5.6 t + 50.2 , where t t is the time in months from 1999 to 2001. ( t = 1 t = 1 is January 1999). Find the two months in which the price of the stock was $30.

Suppose that an equation is given p = −2 x 2 + 280 x − 1000 , p = −2 x 2 + 280 x − 1000 , where x x represents the number of items sold at an auction and p p is the profit made by the business that ran the auction. How many items sold would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using 2 nd CALC maximum. To obtain a good window for the curve, set x x [0,200] and y y [0,10000].

Real-World Applications

A formula for the normal systolic blood pressure for a man age A , A , measured in mmHg, is given as P = 0.006 A 2 − 0.02 A + 120. P = 0.006 A 2 − 0.02 A + 120. Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.

The cost function for a certain company is C = 60 x + 300 C = 60 x + 300 and the revenue is given by R = 100 x − 0.5 x 2 . R = 100 x − 0.5 x 2 . Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $300.

A falling object travels a distance given by the formula d = 5 t + 16 t 2 d = 5 t + 16 t 2 ft, where t t is measured in seconds. How long will it take for the object to travel 74 ft?

A vacant lot is being converted into a community garden. The garden and the walkway around its perimeter have an area of 378 ft 2 . Find the width of the walkway if the garden is 12 ft. wide by 15 ft. long.

An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, P P , who contracted the flu t t days after it broke out is given by the model P = − t 2 + 13 t + 130 , P = − t 2 + 13 t + 130 , where 1 ≤ t ≤ 6. 1 ≤ t ≤ 6. Find the day that 160 students had the flu. Recall that the restriction on t t is at most 6.

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• Book title: College Algebra 2e
• Publication date: Dec 21, 2021
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• Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra-2e/pages/2-5-quadratic-equations

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2.5: Quadratic Equations

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Learning Objectives

• Solve quadratic equations by factoring.
• Solve quadratic equations by the square root property.
• Solve quadratic equations by completing the square.
• Solve quadratic equations by using the quadratic formula.

The computer monitor on the left in Figure \(\PageIndex{1}\) is a \(23.6\)-inch model and the one on the right is a \(27\)-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods.

• Solving Quadratic Equations by Factoring

An equation containing a second-degree polynomial is called a quadratic equation . For example, equations such as \(2x^2 +3x−1=0\) and \(x^2−4= 0\) are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.

Often the easiest method of solving a quadratic equation is factoring . Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.

If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if \(a⋅b=0\), then \(a = 0\) or \(b =0\), where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.

Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression \((x−2)(x+3)\) by multiplying the two factors together.

\[\begin{align*} (x-2)(x+3)&= x^2+3x-2x-6\\ &= x^2+x-6\\ \end{align*}\]

The product is a quadratic expression. Set equal to zero, \(x^2+x−6= 0\) is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.

The process of factoring a quadratic equation depends on the leading coefficient, whether it is \(1\) or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are real numbers, and \(a≠0\). The equation \(x^2 +x−6= 0\) is in standard form.

We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor(GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.

ZERO-PRODUCT PROPERTY AND QUADRATIC EQUATIONS

The zero-product property states

If \(a⋅b=0\), then \(a=0\) or \(b=0\),

where \(a\) and \(b\) are real numbers or algebraic expressions.

A quadratic equation is an equation containing a second-degree polynomial; for example

\[ax^2+bx+c=0\]

where \(a\), \(b\), and \(c\) are real numbers, and if \(a≠0\), it is in standard form.

Solving Quadratics with a Leading Coefficient of \(1\)

In the quadratic equation \(x^2 +x−6=0\), the leading coefficient, or the coefficient of \(x^2\), is \(1\). We have one method of factoring quadratic equations in this form.

How to: Factor a quadratic equation with the leading coefficient of 1

• Find two numbers whose product equals \(c\) and whose sum equals \(b\).
• Use those numbers to write two factors of the form \((x+k)\) or \((x−k)\), where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are \(1\) and \(−2\), the factors are \((x+1)(x−2)\).
• Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

Example \(\PageIndex{1}\): Solving a Quadratic with Leading Coefficient of \(1\)

Factor and solve the equation: \(x^2+x−6=0\).

To factor \(x^2 +x−6=0\), we look for two numbers whose product equals \(−6\) and whose sum equals \(1\). Begin by looking at the possible factors of \(−6\).

\[1⋅(−6) \nonumber \]

\[(−6)⋅1 \nonumber \]

\[2⋅(−3) \nonumber \]

\[3⋅(−2) \nonumber \]

The last pair, \(3⋅(−2)\) sums to \(1\), so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

\[(x−2)(x+3)=0 \nonumber \]

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

\[\begin{align*} (x-2)(x+3)&= 0\\ (x-2)&= 0\\ x&= 2\\ (x+3)&= 0\\ x&= -3 \end{align*}\]

The two solutions are \(2\) and \(−3\). We can see how the solutions relate to the graph in Figure \(\PageIndex{2}\). The solutions are the x-intercepts of \(x^2 +x−6=0\).

Exercise \(\PageIndex{1}\)

Factor and solve the quadratic equation: \(x^2−5x−6=0\).

\((x−6)(x+1)=0\), \(x=6\), \(x=−1\)

Example \(\PageIndex{2}\): Solve the Quadratic Equation by Factoring

Solve the quadratic equation by factoring: \(x^2+8x+15=0\).

Find two numbers whose product equals \(15\) and whose sum equals \(8\). List the factors of \(15\).

\[1⋅15 \nonumber \]

\[3⋅5 \nonumber \]

\[(−1)⋅(−15) \nonumber \]

\[(−3)⋅(−5) \nonumber \]

The numbers that add to \(8\) are \(3\) and \(5\). Then, write the factors, set each factor equal to zero, and solve.

\[\begin{align*} (x+3)(x+5)&= 0\\ (x+3)&= 0\\ x&= -3\\ (x+5)&= 0\\ x&= -5 \end{align*}\]

The solutions are \(−3\) and \(−5\).

Exercise \(\PageIndex{2}\)

Solve the quadratic equation by factoring: \(x^2−4x−21=0\).

\((x−7)(x+3)=0\), \(x=7\), \(x=−3\)

Example \(\PageIndex{3}\): Using Zero-Product Property to Solve a Quadratic Equation

Solve the difference of squares equation using the zero-product property: \(x^2−9=0\).

Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.

\[\begin{align*} x^2-9&= 0\\ (x-3)(x+3)&= 0\\ x-3&= 0\\ x&= 3\\ (x+3)&= 0\\ x&= -3 \end{align*}\]

The solutions are \(3\) and \(−3\).

Exercise \(\PageIndex{3}\)

Solve by factoring: \(x^2−25=0\).

\((x+5)(x−5)=0, x=−5, x=5\)

Factoring and Solving a Quadratic Equation of Higher Order

When the leading coefficient is not \(1\), we factor a quadratic equation using the method called grouping , which requires four terms.

Grouping: Steps for factoring quadratic equations

With the equation in standard form, let’s review the grouping procedures

• With the quadratic in standard form, \(ax^2+bx+c=0\), multiply \(a⋅c\).
• Find two numbers whose product equals ac and whose sum equals \(b\).
• Rewrite the equation replacing the \(bx\) term with two terms using the numbers found in step \(1\) as coefficients of \(x\).
• Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
• Factor out the expression in parentheses.
• Set the expressions equal to zero and solve for the variable.

Example \(\PageIndex{4}\): Solving a Quadratic Equation Using Grouping

Use grouping to factor and solve the quadratic equation: \(4x^2+15x+9=0\).

First, multiply \(ac:4(9)=36\). Then list the factors of \(36\).

\[1⋅36 \nonumber\]

\[2⋅18 \nonumber\]

\[3⋅12 \nonumber\]

\[4⋅9 \nonumber\]

\[6⋅6 \nonumber\]

The only pair of factors that sums to \(15\) is \(3+12\). Rewrite the equation replacing the b term, \(15x\), with two terms using \(3\) and \(12\) as coefficients of \(x\). Factor the first two terms, and then factor the last two terms.

\[\begin{align*} 4x^2+3x+12x+9&= 0\\ x(4x+3)+3(4x+3)&= 0\\ (4x+3)(x+3)&= 0 \qquad \text{Solve using the zero-product property}\\ (4x+3)&= 3\\ x&= -\dfrac{3}{4}\\ (x+3)&= 0\\ x&= -3 \end{align*}\]

The solutions are \(−\dfrac{3}{4}\), and \(−3\). See Figure \(\PageIndex{3}\).

Exercise \(\PageIndex{4}\)

Solve using factoring by grouping: \(12x^2+11x+2=0\).

\((3x+2)(4x+1)=0\), \(x=−\dfrac{2}{3}\), \(x=−\dfrac{1}{4}\)

Example \(\PageIndex{5}\): Solving a Higher Degree Quadratic Equation by Factoring

Solve the equation by factoring: \(−3x^3−5x^2−2x=0\).

This equation does not look like a quadratic, as the highest power is \(3\), not \(2\). Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out \(−x\) from all of the terms and then proceed with grouping.

\[\begin{align*} -3x^3-5x^2-2x&= 0\\ -x(3x^2+5x+2)&= 0\\ -x(3x^2+3x+2x+2)&= 0 \qquad \text{Use grouping on the expression in parentheses}\\ -x[3x(x+1)+2(x+1)]&= 0\\ -x(3x+2)(x+1)&= 0\\ \text{Now, we use the zero-product property. Notice that we have three factors.}\\ -x&= 0\\ x&= 0\\ 3x+2&= 0\\ x&= -\dfrac{2}{3}\\ x+1&= 0\\ x&= -1 \end{align*}\]

The solutions are \(0\), \(−\dfrac{2}{3}\), and \(−1\).

Exercise \(\PageIndex{5}\)

Solve by factoring: \(x^3+11x^2+10x=0\).

\(x=0, x=−10, x=−1\)

Using the Square Root Property

When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property , in which we isolate the \(x^2\) term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the \(x^2\) term so that the square root property can be used.

THE SQUARE ROOT PROPERTY

With the \(x^2\) term isolated, the square root property states that:

where \(k\) is a nonzero real number.

Howto: Given a quadratic equation with an \(x^2\) term but no \(x\) term, use the square root property to solve it

• Isolate the \(x^2\) term on one side of the equal sign.
• Take the square root of both sides of the equation, putting a \(±\) sign before the expression on the side opposite the squared term.
• Simplify the numbers on the side with the \(±\) sign.

Example \(\PageIndex{6}\): Solving a Simple Quadratic Equation Using the Square Root Property

Solve the quadratic using the square root property: \(x^2=8\).

Take the square root of both sides, and then simplify the radical. Remember to use a \(±\) sign before the radical symbol.

\[\begin{align*} x^2&= 8\\ x&= \pm \sqrt{8}\\ &= \pm 2\sqrt{2} \end{align*}\]

The solutions are \(2\sqrt{2}\),\(-2\sqrt{2}\)

Example \(\PageIndex{7}\): Solving a Quadratic Equation Using the Square Root Property

Solve the quadratic equation: \(4x^2+1=7\).

First, isolate the \(x^2\) term. Then take the square root of both sides.

\[\begin{align*} 4x^2+1&= 7\\ 4x^2&= 6\\ x^2&= \dfrac{6}{4}\\ x&= \pm \dfrac{\sqrt{6}}{2} \end{align*}\]

The solutions are \(\dfrac{\sqrt{6}}{2}\), and \(-\dfrac{\sqrt{6}}{2}\).

Exercise \(\PageIndex{6}\)

Solve the quadratic equation using the square root property: \(3{(x−4)}^2=15\).

\(x=4±\sqrt{5}\)

• Completing the Square

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square . Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, \(a\), must equal \(1\). If it does not, then divide the entire equation by \(a\). Then, we can use the following procedures to solve a quadratic equation by completing the square.

We will use the example \(x^2+4x+1=0\) to illustrate each step.

Given a quadratic equation that cannot be factored, and with \(a=1\), first add or subtract the constant term to the right sign of the equal sign.

\[\begin{align*} x^2+4x+1&= 0\\ x^2+4x&= -1 \qquad \text{Multiply the b} \text{ term by } \dfrac{1}{2} \text{ and square it.}\\ \dfrac{1}{2}(4)&= 2 \\ 2^2&= 4 \qquad \text{Add } \left ({\dfrac{1}{2}} \right )^2 \text{ to both sides of the equal sign and simplify the right side. We have}\\ x^2+4x+4&= -1+4\\ x^2+4x+4&= 3 \qquad \text{The left side of the equation can now be factored as a perfect square.}\\ {(x+2)}^2&=3\\ \sqrt{{(x+2)}^2}&= \pm \sqrt{3} \qquad \text{Use the square root property and solve.}\\ \sqrt{{(x+2)}^2}&= \pm \sqrt{3}\\ x+2&= \pm \sqrt{3}\\ x&= -2 \pm \sqrt{3} \end{align*}\]

The solutions are \(−2+\sqrt{3}\), and \(−2−\sqrt{3}\).

Example \(\PageIndex{8}\): Solving a Quadratic by Completing the Square

Solve the quadratic equation by completing the square: \(x^2−3x−5=0\).

First, move the constant term to the right side of the equal sign.

\[\begin{align*} x^2-3x&= 5 \qquad \text{Then, take } \dfrac{1}{2} \text{ of the b term and square it.} \\ \dfrac{1}{2}(-3)&= -\dfrac{3}{2}\\ {\left (-\dfrac{3}{2} \right )}^2=\dfrac{9}{4}\\ x^2-3x+{\left (-\dfrac{3}{2} \right )}^2&= 5+{\left (-\dfrac{3}{2} \right )}^2 \qquad \text{Add the result to both sides of the equal sign.}\\ x^2-3x+\dfrac{9}{4}&= 5+\dfrac{9}{4}\\ \text{Factor the left side as a perfect square and simplify the right side.}\\ {\left (x-\dfrac{3}{2} \right )}^2&= \dfrac{29}{4}\\ (x-\dfrac{3}{2})&= \pm \dfrac{\sqrt{29}}{2} \qquad \text{Use the square root property and solve.}\\ x&= \dfrac{3}{2} \pm \dfrac{\sqrt{29}}{2}\\ \end{align*}\]

The solutions are \(\dfrac{3}{2}+\dfrac{\sqrt{29}}{2}\), and \(\dfrac{3}{2}-\dfrac{\sqrt{29}}{2}\)

Exercise \(\PageIndex{7}\)

Solve by completing the square: \(x^2−6x=13\).

\(x=3±\sqrt{22}\)

Using the Quadratic Formula

The fourth method of solving a quadratic equation is by using the quadratic formula , a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.

We can derive the quadratic formula by completing the square . We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by \(−1\) and obtain a positive a. Given \(ax^2+bx+c=0, a≠0\), we will complete the square as follows:

First, move the constant term to the right side of the equal sign:

\[ax^2+bx=−c \nonumber \]

As we want the leading coefficient to equal \(1\), divide through by \(a\):

\[x^2+\dfrac{b}{a}x=−\dfrac{c}{a} \nonumber \]

Then, find \(\dfrac{1}{2}\) of the middle term, and add \({(\dfrac{1}{2}\dfrac{b}{a})}^2=\dfrac{b^2}{4a^2}\) to both sides of the equal sign:

\[x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}=\dfrac{b^2}{4a^2}-\dfrac{c}{a} \nonumber \]

Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:

\[{(x+\dfrac{b}{2a})}^2=\dfrac{b^2-4ac}{4a^2} \nonumber \]

Now, use the square root property, which gives

\[x+\dfrac{b}{2a}=±\sqrt{\dfrac{b^2-4ac}{4a^2}} \nonumber \]

\[x+\dfrac{b}{2a}=\dfrac{±\sqrt{b^2-4ac}}{2a} \nonumber \]

Finally, add \(-\dfrac{b}{2a}\) to both sides of the equation and combine the terms on the right side. Thus,

\[x=\dfrac{-b±\sqrt{b^2-4ac}}{2a} \nonumber \]

THE QUADRATIC FORMULA

Written in standard form, \(ax^2+bx+c=0\), any quadratic equation can be solved using the quadratic formula :

where \(a\), \(b\), and \(c\) are real numbers and \(a≠0\).

Given a quadratic equation, solve it using the quadratic formula

• Make sure the equation is in standard form: \(ax^2+bx+c=0\).
• Make note of the values of the coefficients and constant term, \(a\), \(b\), and \(c\).
• Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
• Calculate and solve.

Example \(\PageIndex{9}\): Solve the Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation: \(x^2+5x+1=0\).

Identify the coefficients: \(a=1,b=5,c=1\). Then use the quadratic formula.

\[\begin{align*} x&= \dfrac{-(5) \pm \sqrt{(5)^2-4(1)(1)}}{2(1)}\\ &= \dfrac{-5 \pm \sqrt{25-4}}{2}\\ &= \dfrac{-5 \pm \sqrt{21}}{2} \end{align*}\]

Example \(\PageIndex{10}\): Solving a Quadratic Equation with the Quadratic Formula

Use the quadratic formula to solve \(x^2+x+2=0\).

First, we identify the coefficients: \(a=1\),\(b=1\), and \(c=2\).

Substitute these values into the quadratic formula.

\[\begin{align*} x&= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\ &= \dfrac{-(1) \pm \sqrt{(1)^2-4(1)(2)}}{2(1)}\\ &= \dfrac{-1 \pm \sqrt{1-8}}{2}\\ &= \dfrac{-1 \pm \sqrt{-7}}{2}\\ &= \dfrac{-1 \pm i\sqrt{7}}{2} \end{align*}\]

Exercise \(\PageIndex{8}\)

Solve the quadratic equation using the quadratic formula: \(9x^2+3x−2=0\).

\(x=-\dfrac{2}{3},x=\dfrac{1}{3}\)

The Discriminant

The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant , or the expression under the radical, \(b^2−4ac\). The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table \(\PageIndex{1}\) relates the value of the discriminant to the solutions of a quadratic equation.

THE DISCRIMINANT

For \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are real numbers, the discriminant is the expression under the radical in the quadratic formula: \(b^2−4ac\). It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.

Example \(\PageIndex{11}\): Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation

Use the discriminant to find the nature of the solutions to the following quadratic equations:

\(x^2+4x+4=0\)

\(8x^2+14x+3=0\)

\(3x^2−5x−2=0\)

\(3x^2−10x+15=0\)

Calculate the discriminant \(b^2−4ac\) for each equation and state the expected type of solutions.

\(b^2-4ac={(4)}^2-4(1)(4)=0\) There will be one rational double solution.

\(b^2-4ac={(14)}^2-4(8)(3)=100\) As \(100\) is a perfect square, there will be two rational solutions.

\(b^2-4ac={(-5)}^2-4(3)(-2)=49\) As \(49\) is a perfect square, there will be two rational solutions.

\(b^2-4ac={(-10)}^2-4(3)(15)=-80\) There will be two complex solutions.

Using the Pythagorean Theorem

One of the most famous formulas in mathematics is the Pythagorean Theorem . It is based on a right triangle, and states the relationship among the lengths of the sides as \(a^2+b^2=c^2\), where \(a\) and \(b\) refer to the legs of a right triangle adjacent to the \(90°\) angle, and \(c\) refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.

We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.

The Pythagorean Theorem is given as

\[a^2+b^2=c^2\]

where \(a\) and \(b\) refer to the legs of a right triangle adjacent to the \(90°\) angle, and \(c\) refers to the hypotenuse, as shown in .

Example \(\PageIndex{12}\): Finding the Length of the Missing Side of a Right Triangle

Find the length of the missing side of the right triangle in Figure \(\PageIndex{5}\).

As we have measurements for side \(b\) and the hypotenuse, the missing side is \(a\).

\[\begin{align*} a^2+b^2&= c^2\\ a^2+{(4)}^2&= {(12)}^2\\ a^2+16&= 144\\ a^2&= 128\\ a&= \sqrt{128}\\ &= 8\sqrt{2} \end{align*}\]

Exercise \(\PageIndex{9}\)

Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.

\(5\) units

Access these online resources for additional instruction and practice with quadratic equations.

• The Zero-Product Property
• Quadratic Formula with Two Rational Solutions
• Length of a leg of a right triangle

Key Equations

Key concepts.

• Many quadratic equations can be solved by factoring when the equation has a leading coefficient of \(1\) or if the equation is a difference of squares. The zero-factor property is then used to find solutions. See Example , Example , and Example .
• Many quadratic equations with a leading coefficient other than \(1\) can be solved by factoring using the grouping method. See Example and Example .
• Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution. See Example and Example .
• Completing the square is a method of solving quadratic equations when the equation cannot be factored. See Example .
• A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. See Example .
• The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: real or complex, rational or irrational, and how many of each. See Example .
• The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation. See Example .

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• Mathway currently does not support Ask an Expert Live in Chemistry. If this is what you were looking for, please contact support.
• Mathway currently only computes linear regressions.
• We are here to assist you with your math questions. You will need to get assistance from your school if you are having problems entering the answers into your online assignment.
• Have a great day!
• Hope that helps!
• You're welcome!
• Per our terms of use, Mathway's live experts will not knowingly provide solutions to students while they are taking a test or quiz.

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