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Solving Quadratic Equations Using the Quadratic Formula

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: recognize the standard form of a quadratic equation, understand the derivation of the quadratic formula, solve quadratic equations using the quadratic formula.

Overview

  • Standard Form Of A Quadratic Equation
  • The Quadratic Formula
  • Derivation Of The Quadratic Formula

Standard Form Of A Quadratic Equation

We have observed that 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

where

a a is the coefficient of the quadratic term,
 b b is the coefficient of the linear term, and
c c is the constant term.

Standard Form

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

Sample Set A

Determine the values of a,b, a,b, and c. c.

Example 1

In the equation 3 x 2 +5x+2=0, 3 x 2 +5x+2=0,

a = 3 b = 5 c = 2 a = 3 b = 5 c = 2

Example 2

In the equation 12 x 2 −2x−1=0, 12 x 2 −2x−1=0,

a = 12 b = −2 c = −1 a = 12 b = −2 c = −1

Example 3

In the equation 2 y 2 +3=0, 2 y 2 +3=0,

a = 2 b = 0 Because the equation could be written 2 y 2 +0y+3=0 c = 3 a = 2 b = 0 Because the equation could be written 2 y 2 +0y+3=0 c = 3

Example 4

In the equation −8 y 2 +11y=0, −8 y 2 +11y=0,

a = −8 b = 11 c = 0 Since − 8y 2 +11y+0=0. a = −8 b = 11 c = 0 Since − 8y 2 +11y+0=0.

Example 5

In the equation z 2 =z+8, z 2 =z+8,

a = 1 b = −1 c = −8 When we write the equation in standard form, we get  z 2 −z−8=0 a = 1 b = −1 c = −8 When we write the equation in standard form, we get  z 2 −z−8=0

Practice Set A

Determine the values of a, b, a, b, and c c in the following quadratic equations.

Exercise 1

4 x 2 −3x+5=0 4 x 2 −3x+5=0

Solution

a = 4 b = −3 c = 5 a = 4 b = −3 c = 5

Exercise 2

3 y 2 −2y+9=0 3 y 2 −2y+9=0

Solution

a = 3 b = −2 c = 9 a = 3 b = −2 c = 9

Exercise 3

x 2 −5x−1=0 x 2 −5x−1=0

Solution

a = 1 b = −5 c = −1 a = 1 b = −5 c = −1

Exercise 4

z 2 −4=0 z 2 −4=0

Solution

a = 1 b = 0 c = −4 a = 1 b = 0 c = −4

Exercise 5

x 2 −2x=0 x 2 −2x=0

Solution

a = 1 b = −2 c = 0 a = 1 b = −2 c = 0

Exercise 6

y 2 =5y−6 y 2 =5y−6

Solution

a = 1 b = −5 c = 6 a = 1 b = −5 c = 6

Exercise 7

2 x 2 −4x=−1 2 x 2 −4x=−1

Solution

a = 2 b = −4 c = 1 a = 2 b = −4 c = 1

Exercise 8

5x−3=−3 x 2 5x−3=−3 x 2

Solution

a = 3 b = 5 c = −3 a = 3 b = 5 c = −3

Exercise 9

2x−11−3 x 2 =0 2x−11−3 x 2 =0

Solution

a = −3 b = 2 c = −11 a = −3 b = 2 c = −11

Exercise 10

y 2 =0 y 2 =0

Solution

a = 1 b = 0 c = 0 a = 1 b = 0 c = 0

The solutions to all quadratic equations depend only and completely on the values a, b, a, b, and c. c.

The Quadratic Formula

When a quadratic equation is written in standard form so that the values a, b, a, b, and c c are readily determined, the equation can be solved using the quadratic formula. The values that satisfy the equation are found by substituting the values a, b, a, b, and c c into the formula

Quadratic Formula

x= −b± b 2 −4ac 2a x= −b± b 2 −4ac 2a

Keep in mind that the plus or minus symbol, ±, ±, is just a shorthand way of denoting the two possibilities:

 x= −b+ b 2 −4ac 2a and x= −b− b 2 −4ac 2a x= −b+ b 2 −4ac 2a and x= −b− b 2 −4ac 2a

The quadratic formula can be derived by using the method of completing the square.

Derivation Of The Quadratic Formula

Solve a x 2 +bx+c=0 a x 2 +bx+c=0 for x x by completing the square.

Example 6

Subtract c c from both sides.

 a x 2 +bx=−c a x 2 +bx=−c

Example 7

Divide both sides by a, a, the coefficient of x 2 . x 2 .

x 2 + b a x= −c a x 2 + b a x= −c a

Example 8

Now we have the proper form to complete the square. Take one half the coefficient of x, x, square it, and add the result to both sides of the equation found in step 2.

(a) 1 2  ·  b a = b 2a 1 2  ·  b a = b 2a is one half the coefficient of x. x.

(b) ( b 2a ) 2 ( b 2a ) 2 is the square of one half the coefficient of x. x.

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

Example 9

The left side of the equation is now a perfect square trinomial and can be factored. This gives us

 ( x+ b 2a ) 2 = −c a + b 2 4 a 2 ( x+ b 2a ) 2 = −c a + b 2 4 a 2

Example 10

Add the two fractions on the right side of the equation. The LCD =4 a 2 . =4 a 2 .

( x+ b 2a ) 2 = −4ac 4 a 2 + b 2 4 a 2 ( x+ b 2a ) 2 = −4ac+ b 2 4 a 2 ( x+ b 2a ) 2 = b 2 −4ac 4 a 2 ( x+ b 2a ) 2 = −4ac 4 a 2 + b 2 4 a 2 ( x+ b 2a ) 2 = −4ac+ b 2 4 a 2 ( x+ b 2a ) 2 = b 2 −4ac 4 a 2

Example 11

Solve for x x using the method of extraction of roots.

 x+ b 2a = ± b 2 −4ac 4 a 2 x+ b 2a = ± b 2 −4ac 4 a 2 4 a 2 =| 2a |=2| a |=±2a x+ b 2a = ± b 2 −4ac 2a x = − b 2a ± b 2 −4ac 2a Add these two fractions. x = −b± b 2 −4ac 2a x+ b 2a = ± b 2 −4ac 4 a 2 x+ b 2a = ± b 2 −4ac 4 a 2 4 a 2 =| 2a |=2| a |=±2a x+ b 2a = ± b 2 −4ac 2a x = − b 2a ± b 2 −4ac 2a Add these two fractions. x = −b± b 2 −4ac 2a

Sample Set B

Solve each of the following quadratic equations using the quadratic formula.

Example 12

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

  1. Identify a, b, a, b, and c. c.

    a=3, b=5, and c=2 a=3, b=5, and c=2
  2. Write the quadratic formula.

     x= −b± b 2 −4ac 2a x= −b± b 2 −4ac 2a
  3. Substitute.

     x = −5± ( 5 ) 2 −4( 3 )( 2 ) 2( 3 ) = −5± 25−24 6 = −5± 1 6 = −5±1 6 −5+1=−4 and −5−1=−6 = −4 6 , −6 6 x = −2 3 ,−1 x = −5± ( 5 ) 2 −4( 3 )( 2 ) 2( 3 ) = −5± 25−24 6 = −5± 1 6 = −5±1 6 −5+1=−4 and −5−1=−6 = −4 6 , −6 6 x = −2 3 ,−1
    Note: Since these roots are rational numbers, this equation could have been solved by factoring. Note: Since these roots are rational numbers, this equation could have been solved by factoring.

Example 13

12 x 2 −2x−1=0. 12 x 2 −2x−1=0.

  1. Identify a, b, a, b, and c. c.

    a=12, b=−2, and c=−1 a=12, b=−2, and c=−1
  2. Write the quadratic formula.

     x= −b± b 2 −4ac 2a x= −b± b 2 −4ac 2a
  3. Substitute.

     x = −( −2 )± ( −2 ) 2 −4( 12 )( −1 ) 2( 12 ) = 2± 4+48 24 Simplify. = 2± 52 24 Simplify. = 2± 4 · 13 24 Simplify. = 2±2 13 24 Reduce. Factor 2 from the terms of the numerator. = 2( 1± 13 ) 24 x = 1± 13 12 x = −( −2 )± ( −2 ) 2 −4( 12 )( −1 ) 2( 12 ) = 2± 4+48 24 Simplify. = 2± 52 24 Simplify. = 2± 4 · 13 24 Simplify. = 2±2 13 24 Reduce. Factor 2 from the terms of the numerator. = 2( 1± 13 ) 24 x = 1± 13 12

Example 14

2 y 2 +3=0 2 y 2 +3=0

  1. Identify a, b, a, b, and c. c.

    a=2, b=0, and c=3 a=2, b=0, and c=3
  2. Write the quadratic formula.

     x= −b± b 2 −4ac 2a x= −b± b 2 −4ac 2a
  3. Substitute.

     x = −0± 0 2 −4( 2 )( 3 ) 2( 2 ) x = 0± −24 4 x = −0± 0 2 −4( 2 )( 3 ) 2( 2 ) x = 0± −24 4

    This equation has no real number solution since we have obtained a negative number under the radical sign.

Example 15

−8 x 2 +11x=0 −8 x 2 +11x=0

  1. Identify a, b, a, b, and c. c.

    a=−8, b=11, and c=0 a=−8, b=11, and c=0
  2. Write the quadratic formula.

     x= −b± b 2 −4ac 2a x= −b± b 2 −4ac 2a
  3. Substitute.

     x = −11± 11 2 −4( −8 )( 0 ) 2( −8 ) = −11± 121−0 −16 Simplify. = −11± 121 −16 Simplify. = −11±11 −16 x = 0,  11 8 x = −11± 11 2 −4( −8 )( 0 ) 2( −8 ) = −11± 121−0 −16 Simplify. = −11± 121 −16 Simplify. = −11±11 −16 x = 0,  11 8

Example 16

( 3x+1 )( x−4 )= x 2 +x−2 ( 3x+1 )( x−4 )= x 2 +x−2

  1. Write the equation in standard form.

     3 x 2 −11x−4 = x 2 +x−2 2 x 2 −12x−2 = 0 x 2 −6x−1 = 0 3 x 2 −11x−4 = x 2 +x−2 2 x 2 −12x−2 = 0 x 2 −6x−1 = 0
  2. Identify a, b, a, b, and c. c.

    a=1, b=−6, and c=−1 a=1, b=−6, and c=−1
  3. Write the quadratic formula.

     x= −b± b 2 −4ac 2a x= −b± b 2 −4ac 2a
  4. Substitute.

     x = −( −6 )± ( −6 ) 2 −4( 1 )( −1 ) 2( 1 ) = 6± 36+4 2 = 6± 40 2 = 6± 4 · 10 2 = 6±2 10 2 = 2( 3± 10 ) 2 x = 3± 10 x = −( −6 )± ( −6 ) 2 −4( 1 )( −1 ) 2( 1 ) = 6± 36+4 2 = 6± 40 2 = 6± 4 · 10 2 = 6±2 10 2 = 2( 3± 10 ) 2 x = 3± 10

Practice Set B

Solve each of the following quadratic equations using the quadratic formula.

Exercise 11

2 x 2 +3x−7=0 2 x 2 +3x−7=0

Solution

x= −3± 65 4 x= −3± 65 4

Exercise 12

5 a 2 −2a−1=0 5 a 2 −2a−1=0

Solution

a= 1± 6 5 a= 1± 6 5

Exercise 13

6 y 2 +5=0 6 y 2 +5=0

Solution

no real number solution

Exercise 14

−3 m 2 +2m=0 −3 m 2 +2m=0

Solution

m=0, 2 3 m=0, 2 3

Exercises

For the following problems, solve the equations using the quadratic formula.

Exercise 15

x 2 −2x−3=0 x 2 −2x−3=0

Solution

x=3, −1 x=3, −1

Exercise 16

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

Exercise 17

y 2 −5y+4=0 y 2 −5y+4=0

Solution

y=1, 4 y=1, 4

Exercise 18

a 2 +4a−21=0 a 2 +4a−21=0

Exercise 19

a 2 +12a+20=0 a 2 +12a+20=0

Solution

a=−2, −10 a=−2, −10

Exercise 20

b 2 −4b+4=0 b 2 −4b+4=0

Exercise 21

b 2 +4b+4=0 b 2 +4b+4=0

Solution

b=−2 b=−2

Exercise 22

x 2 +10x+25=0 x 2 +10x+25=0

Exercise 23

2 x 2 −5x−3=0 2 x 2 −5x−3=0

Solution

x=3,− 1 2 x=3,− 1 2

Exercise 24

6 y 2 +y−2=0 6 y 2 +y−2=0

Exercise 25

4 x 2 −2x−1=0 4 x 2 −2x−1=0

Solution

x= 1± 5 4 x= 1± 5 4

Exercise 26

3 y 2 +2y−1=0 3 y 2 +2y−1=0

Exercise 27

5 a 2 −2a−3=0 5 a 2 −2a−3=0

Solution

a=1, − 3 5 a=1, − 3 5

Exercise 28

x 2 −3x+1=0 x 2 −3x+1=0

Exercise 29

x 2 −5x−4=0 x 2 −5x−4=0

Solution

x= 5± 41 2 x= 5± 41 2

Exercise 30

( x+2 )( x−1 )=1 ( x+2 )( x−1 )=1

Exercise 31

( a+4 )( a−5 )=2 ( a+4 )( a−5 )=2

Solution

a= 1± 89 2 a= 1± 89 2

Exercise 32

( x−3 )( x+3 )=7 ( x−3 )( x+3 )=7

Exercise 33

( b−4 )( b+4 )=9 ( b−4 )( b+4 )=9

Solution

b=±5 b=±5

Exercise 34

x 2 +8x=2 x 2 +8x=2

Exercise 35

y 2 =−5y+4 y 2 =−5y+4

Solution

y= −5± 41 2 y= −5± 41 2

Exercise 36

x 2 =−3x+7 x 2 =−3x+7

Exercise 37

x 2 =−2x−1 x 2 =−2x−1

Solution

x=−1 x=−1

Exercise 38

x 2 +x+1=0 x 2 +x+1=0

Exercise 39

a 2 +3a−4=0 a 2 +3a−4=0

Solution

a=−4, 1 a=−4, 1

Exercise 40

y 2 +y=−4 y 2 +y=−4

Exercise 41

b 2 +3b=−2 b 2 +3b=−2

Solution

b=−1, −2 b=−1, −2

Exercise 42

x 2 +6x+8=−x−2 x 2 +6x+8=−x−2

Exercise 43

x 2 +4x=2x−5 x 2 +4x=2x−5

Solution

No real number solution.

Exercise 44

6 b 2 +5b−4= b 2 +b+1 6 b 2 +5b−4= b 2 +b+1

Exercise 45

4 a 2 +7a−2=−2a+a 4 a 2 +7a−2=−2a+a

Solution

−2± 6 2 −2± 6 2

Exercise 46

( 2x+5 )( x−4 )= x 2 −x+2 ( 2x+5 )( x−4 )= x 2 −x+2

Exercise 47

( x−4 ) 2 =3 ( x−4 ) 2 =3

Solution

x=4± 3 x=4± 3

Exercise 48

( x+2 ) 2 =4 ( x+2 ) 2 =4

Exercise 49

( b−6 ) 2 =8 ( b−6 ) 2 =8

Solution

b=6±2 2 b=6±2 2

Exercise 50

( 3−x ) 2 =6 ( 3−x ) 2 =6

Exercise 51

3( x 2 +1 )=2( x+7 ) 3( x 2 +1 )=2( x+7 )

Solution

x= 1± 34 3 x= 1± 34 3

Exercise 52

2( y 2 −3 )=−3( y−1 ) 2( y 2 −3 )=−3( y−1 )

Exercise 53

−4( a 2 +2 )+3=5 −4( a 2 +2 )+3=5

Solution

No real number solution.

Exercise 54

−( x 2 +3x−1 )=2 −( x 2 +3x−1 )=2

Exercises For Review

Exercise 55

((Reference)) Simplify ( x 8 y 7 z 5 x 4 y 6 z 2 ) 2 . ( x 8 y 7 z 5 x 4 y 6 z 2 ) 2 .

Solution

x 8 y 2 z 6 x 8 y 2 z 6

Exercise 56

((Reference)) Write 4 a −6 b 2 c 3 a 5 b −3 4 a −6 b 2 c 3 a 5 b −3 so that only positive exponents appear.

Exercise 57

((Reference)) Find the product: ( 2y+7 )( 3y−1 ). ( 2y+7 )( 3y−1 ).

Solution

6 y 2 +19y−7 6 y 2 +19y−7

Exercise 58

((Reference)) Simplify: 80 − 45 . 80 − 45 .

Exercise 59

((Reference)) Solve x 2 −4x−12=0 x 2 −4x−12=0 by completing the square.

Solution

x=−2, 6 x=−2, 6

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