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Solving Quadratic Equations

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra</link> by Denny Burzynski and Wade Ellis, Jr. Methods of solving quadratic equations as well as the logic underlying each method are discussed. Factoring, extraction of roots, completing the square, and the quadratic formula are carefully developed. The zero-factor property of real numbers is reintroduced. The chapter also includes graphs of quadratic equations based on the standard parabola, y = x^2, and applied problems from the areas of manufacturing, population, physics, geometry, mathematics (numbers and volumes), and astronomy, which are solved using the five-step method. Objectives of this module: be able to place a quadratic equation into standard form, be familiar with the zero-factor property of real numbers.

Overview

  • Standard Form of A Quadratic Equation
  • Zero-Factor Property of Real Numbers

Standard Form of A Quadratic Equation

In Chapter ((Reference)) we studied linear equations in one and two variables and methods for solving them. We observed that a linear equation in one variable was any equation that could be written in the form ax+b=0, a≠0, ax+b=0, a≠0, and a linear equation in two variables was any equation that could be written in the form ax+by=c, ax+by=c, where a a and b b are not both 0. We now wish to study quadratic equations in one variable.

Quadratic Equation

A quadratic equation is an equation of the form a x 2 +bx+c=0, a≠0. a x 2 +bx+c=0, a≠0.

The standard form of the quadratic equation is a x 2 +bx+c=0, a≠0. a x 2 +bx+c=0, a≠0.

For a quadratic equation in standard form a x 2 +bx+c=0, a x 2 +bx+c=0,

a a is the coefficient of x 2 . x 2 .
b b is the coefficient of x. x.
c c is the constant term.

Sample Set A

The following are quadratic equations.

Example 1

3 x 2 +2x−1=0. a=3,b=2,c=−1. 3 x 2 +2x−1=0. a=3,b=2,c=−1.

Example 2

5 x 2 +8x=0. a=5,b=8,c=0. 5 x 2 +8x=0. a=5,b=8,c=0.
Notice that this equation could be written 5 x 2 +8x+0=0. 5 x 2 +8x+0=0. Now it is clear that c=0. c=0.

Example 3

x 2 +7=0. a=1,b=0,c=7. x 2 +7=0. a=1,b=0,c=7.
Notice that this equation could be written x 2 +0x+7=0. x 2 +0x+7=0. Now it is clear that b=0. b=0.

The following are not quadratic equations.

Example 4

3x+2=0. a=0. This equation is linear. 3x+2=0. a=0. This equation is linear.

Example 5

8 x 2 + 3 x −5=0. 8 x 2 + 3 x −5=0.
The expression on the left side of the equal sign has a variable in the denominator and, therefore, is not a quadratic.

Practice Set A

Which of the following equations are quadratic equations? Answer “yes” or “no” to each equation.

Exercise 1

6 x 2 −4x+9=0 6 x 2 −4x+9=0

Solution

yes

Exercise 2

5x+8=0 5x+8=0

Solution

no

Exercise 3

4 x 3 −5 x 2 +x+6=8 4 x 3 −5 x 2 +x+6=8

Solution

no

Exercise 4

4 x 2 −2x+4=1 4 x 2 −2x+4=1

Solution

yes

Exercise 5

2 x −5 x 2 =6x+4 2 x −5 x 2 =6x+4

Solution

no

Exercise 6

9 x 2 −2x+6=4 x 2 +8 9 x 2 −2x+6=4 x 2 +8

Solution

yes

Zero-Factor Property

Our goal is to solve quadratic equations. The method for solving quadratic equations is based on the zero-factor property of real numbers. We were introduced to the zero-factor property in Section (Reference). We state it again.

Zero-Factor Property

If two numbers a a and b b are multiplied together and the resulting product is 0, then at least one of the numbers must be 0. Algebraically, if a · b=0, a · b=0, then a=0 a=0 or b=0, b=0, or both a=0 a=0 and b=0. b=0.

Sample Set B

Use the zero-factor property to solve each equation.

Example 6

If 9x=0,then x must be 0. 9x=0,then x must be 0.

Example 7

If −2 x 2 =0, −2 x 2 =0, then x 2 =0, x=0. x 2 =0, x=0.

Example 8

If 5 5( x−1 )=0, then x−1 x−1 must be 0, since 5 is not zero.

 x−1 = 0 x = 1 x−1 = 0 x = 1

Example 9

If x( x+6 )=0, x( x+6 )=0, then

 x = 0 or x+6 = 0 x = −6 x = 0, −6. x = 0 or x+6 = 0 x = −6 x = 0, −6.

Example 10

If ( x+2 )( x+3 )=0, ( x+2 )( x+3 )=0, then

 x+2 = 0 or x+3 = 0 x = −2 x = −3 x = −2, −3. x+2 = 0 or x+3 = 0 x = −2 x = −3 x = −2, −3.

Example 11

If ( x+10 )( 4x−5 )=0, ( x+10 )( 4x−5 )=0, then

 x+10 = 0 or 4x−5 = 0 x = −10 4x = 5 x = 5 4 x = −10, 5 4 . x+10 = 0 or 4x−5 = 0 x = −10 4x = 5 x = 5 4 x = −10, 5 4 .

Practice Set B

Use the zero-factor property to solve each equation.

Exercise 7

6( a−4 )=0 6( a−4 )=0

Solution

a=4 a=4

Exercise 8

( y+6 )( y−7 )=0 ( y+6 )( y−7 )=0

Solution

y=−6, 7 y=−6, 7

Exercise 9

( x+5 )( 3x−4 )=0 ( x+5 )( 3x−4 )=0

Solution

x=−5, 4 3 x=−5, 4 3

Exercises

For the following problems, write the values of a, b, a, b, and c c in quadratic equations.

Exercise 10

3 x 2 +4x−7=0 3 x 2 +4x−7=0

Solution

3,4,−7 3,4,−7

Exercise 11

7 x 2 +2x+8=0 7 x 2 +2x+8=0

Exercise 12

2 y 2 −5y+5=0 2 y 2 −5y+5=0

Solution

2,−5,5 2,−5,5

Exercise 13

7 a 2 +a−8=0 7 a 2 +a−8=0

Exercise 14

−3 a 2 +4a−1=0 −3 a 2 +4a−1=0

Solution

−3,4,−1 −3,4,−1

Exercise 15

7 b 2 +3b=0 7 b 2 +3b=0

Exercise 16

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

Solution

2, 5, 0

Exercise 17

4 y 2 +9=0 4 y 2 +9=0

Exercise 18

8 a 2 −2a=0 8 a 2 −2a=0

Solution

8,−2,0 8,−2,0

Exercise 19

6 x 2 =0 6 x 2 =0

Exercise 20

4 y 2 =0 4 y 2 =0

Solution

4, 0, 0

Exercise 21

5 x 2 −3x+9=4 x 2 5 x 2 −3x+9=4 x 2

Exercise 22

7 x 2 +2x+1=6 x 2 +x−9 7 x 2 +2x+1=6 x 2 +x−9

Solution

1, 1, 10

Exercise 23

−3 x 2 +4x−1=−4 x 2 −4x+12 −3 x 2 +4x−1=−4 x 2 −4x+12

Exercise 24

5x−7=−3 x 2 5x−7=−3 x 2

Solution

3,5,−7 3,5,−7

Exercise 25

3x−7=−2 x 2 +5x 3x−7=−2 x 2 +5x

Exercise 26

0= x 2 +6x−1 0= x 2 +6x−1

Solution

1,6,−1 1,6,−1

Exercise 27

9= x 2 9= x 2

Exercise 28

x 2 =9 x 2 =9

Solution

1,0,−9 1,0,−9

Exercise 29

0=− x 2 0=− x 2

For the following problems, use the zero-factor property to solve the equations.

Exercise 30

4x=0 4x=0

Solution

x=0 x=0

Exercise 31

16y=0 16y=0

Exercise 32

9a=0 9a=0

Solution

a=0 a=0

Exercise 33

4m=0 4m=0

Exercise 34

3( k+7 )=0 3( k+7 )=0

Solution

k=−7 k=−7

Exercise 35

8( y−6 )=0 8( y−6 )=0

Exercise 36

−5( x+4 )=0 −5( x+4 )=0

Solution

x=−4 x=−4

Exercise 37

−6( n+15 )=0 −6( n+15 )=0

Exercise 38

y( y−1 )=0 y( y−1 )=0

Solution

y=0,1 y=0,1

Exercise 39

a( a−6 )=0 a( a−6 )=0

Exercise 40

n( n+4 )=0 n( n+4 )=0

Solution

n=0,−4 n=0,−4

Exercise 41

x( x+8 )=0 x( x+8 )=0

Exercise 42

9( a−4 )=0 9( a−4 )=0

Solution

a=4 a=4

Exercise 43

−2( m+11 )=0 −2( m+11 )=0

Exercise 44

x( x+7 )=0 x( x+7 )=0

Solution

x=−7 or x=0 x=−7 or x=0

Exercise 45

n( n−10 )=0 n( n−10 )=0

Exercise 46

( y−4 )( y−8 )=0 ( y−4 )( y−8 )=0

Solution

y=4 or y=8 y=4 or y=8

Exercise 47

( k−1 )( k−6 )=0 ( k−1 )( k−6 )=0

Exercise 48

( x+5 )( x+4 )=0 ( x+5 )( x+4 )=0

Solution

x=−4 or x=−5 x=−4 or x=−5

Exercise 49

( y+6 )( 2y+1 )=0 ( y+6 )( 2y+1 )=0

Exercise 50

( x−3 )( 5x−6 )=0 ( x−3 )( 5x−6 )=0

Solution

x= 6 5  or x=3 x= 6 5  or x=3

Exercise 51

( 5a+1 )( 2a−3 )=0 ( 5a+1 )( 2a−3 )=0

Exercise 52

( 6m+5 )( 11m−6 )=0 ( 6m+5 )( 11m−6 )=0

Solution

m=− 5 6  or m= 6 11 m=− 5 6  or m= 6 11

Exercise 53

( 2m−1 )( 3m+8 )=0 ( 2m−1 )( 3m+8 )=0

Exercise 54

( 4x+5 )( 2x−7 )=0 ( 4x+5 )( 2x−7 )=0

Solution

x= −5 4 ,  7 2 x= −5 4 ,  7 2

Exercise 55

( 3y+1 )( 2y+1 )=0 ( 3y+1 )( 2y+1 )=0

Exercise 56

( 7a+6 )( 7a−6 )=0 ( 7a+6 )( 7a−6 )=0

Solution

a= −6 7 ,  6 7 a= −6 7 ,  6 7

Exercise 57

( 8x+11 )( 2x−7 )=0 ( 8x+11 )( 2x−7 )=0

Exercise 58

( 5x−14 )( 3x+10 )=0 ( 5x−14 )( 3x+10 )=0

Solution

x= 14 5 ,  −10 3 x= 14 5 ,  −10 3

Exercise 59

( 3x−1 )( 3x−1 )=0 ( 3x−1 )( 3x−1 )=0

Exercise 60

( 2y+5 )( 2y+5 )=0 ( 2y+5 )( 2y+5 )=0

Solution

y= −5 2 y= −5 2

Exercise 61

( 7a−2 ) 2 =0 ( 7a−2 ) 2 =0

Exercise 62

( 5m−6 ) 2 =0 ( 5m−6 ) 2 =0

Solution

m= 6 5 m= 6 5

Exercises For Review

Exercise 63

((Reference)) Factor 12ax−3x+8a−2 12ax−3x+8a−2 by grouping.

Exercise 64

((Reference)) Construct the graph of 6x+10y−60=0. 6x+10y−60=0.
An xy coordinate plane with gridlines, labeled negative five and five with increments of one units on both axes.

Solution

A graph of a line passing through two points coordinates zero, six and five, three.

Exercise 65

((Reference)) Find the difference: 1 x 2 +2x+1 − 1 x 2 −1 . 1 x 2 +2x+1 − 1 x 2 −1 .

Exercise 66

((Reference)) Simplify 7 ( 2 +2 ). 7 ( 2 +2 ).

Solution

14 +2 7 14 +2 7

Exercise 67

((Reference)) Solve the radical equation 3x+10 =x+4. 3x+10 =x+4.

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