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Solving Quadratic Equations Using the Method of Extraction of Roots

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 solve quadratic equations using the method of extraction of roots, be able to determine the nature of the solutions to a quadratic equation.

Overview

  • The Method Of Extraction Of Roots
  • The Nature Of Solutions

The Method Of Extraction Of Roots

Extraction of Roots

Quadratic equations of the form x 2 K=0 x 2 K=0 can be solved by the method of extraction of roots by rewriting it in the form x 2 =K. x 2 =K.

To solve x 2 =K, x 2 =K, we are required to find some number, x, x, that when squared produces K. K. This number, x, x, must be a square root of K. K. If K K is greater than zero, we know that it possesses two square roots, K K and K . K . We also know that

( K ) 2 =( K )( K )=K and ( K ) 2 =( K )( K ) =K ( K ) 2 =( K )( K )=K and ( K ) 2 =( K )( K ) =K

We now have two replacements for x x that produce true statements when substituted into the equation. Thus, x= K x= K and x= K x= K are both solutions to x 2 =K. x 2 =K. We use the notation x=± K x=± K to denote both the principal and the secondary square roots.

The Nature of Solutions

Solutions of x 2 =K x 2 =K

For quadratic equations of the form x 2 =K, x 2 =K,

  1. If K K is greater than or equal to zero, the solutions are ± K . ± K .
  2. If K K is negative, no real number solutions exist.
  3. If K K is zero, the only solution is 0.

Sample Set A

Solve each of the following quadratic equations using the method of extraction of roots.

Example 1

x 2 49 = 0. Rewrite . x 2 = 49 x = ± 49 x = ±7 C h e c k : (7) 2 = 49 Is this correct? ( 7 ) 2 = 49 Is this correct 49=49 Yes, this is correct . 49=49 Yes, this is correct. x 2 49 = 0. Rewrite . x 2 = 49 x = ± 49 x = ±7 C h e c k : (7) 2 = 49 Is this correct? ( 7 ) 2 = 49 Is this correct 49=49 Yes, this is correct . 49=49 Yes, this is correct.

Example 2

25 a 2 = 36 a 2 = 36 25 a = ± 36 25 a = ± 6 5 25 a 2 = 36 a 2 = 36 25 a = ± 36 25 a = ± 6 5
Check: 25 ( 6 5 ) 2 = 36 Is this correct? 25 ( 6 5 ) 2 = 36 Is this correct? 25 ( 36 25 ) 2 = 36 Is this correct? 25( 36 25 ) = 36 Is this correct? 36 = 36 Yes, this is correct. 36 = 36 Yes, this is correct. Check: 25 ( 6 5 ) 2 = 36 Is this correct? 25 ( 6 5 ) 2 = 36 Is this correct? 25 ( 36 25 ) 2 = 36 Is this correct? 25( 36 25 ) = 36 Is this correct? 36 = 36 Yes, this is correct. 36 = 36 Yes, this is correct.

Example 3

4 m 2 32 = 0 4 m 2 = 32 m 2 = 32 4 m 2 = 8 m = ± 8 m = ±2 2 4 m 2 32 = 0 4 m 2 = 32 m 2 = 32 4 m 2 = 8 m = ± 8 m = ±2 2
Check: 4 (2 2 ) 2 = 32 Is this correct? 4 (2 2 ) 2 = 32 Is this correct? 4[ 2 2 ( 2 ) 2 ] = 32 Is this correct? 4[ (2) 2 ( 2 ) 2 ] = 32 Is this correct? 4[4·2] = 32 Is this correct? 4[4·2] = 32 Is this correct? 4·8 = 32 Is this correct? 4·8 = 32 Is this correct? 32 = 32 Yes, this is correct. 32 = 32 Yes, this is correct. Check: 4 (2 2 ) 2 = 32 Is this correct? 4 (2 2 ) 2 = 32 Is this correct? 4[ 2 2 ( 2 ) 2 ] = 32 Is this correct? 4[ (2) 2 ( 2 ) 2 ] = 32 Is this correct? 4[4·2] = 32 Is this correct? 4[4·2] = 32 Is this correct? 4·8 = 32 Is this correct? 4·8 = 32 Is this correct? 32 = 32 Yes, this is correct. 32 = 32 Yes, this is correct.

Example 4

Solve 5 x 2 15 y 2 z 7 =0 5 x 2 15 y 2 z 7 =0 for x. x.
5 x 2 = 15 y 2 z 7 Divide both sides by 5. x 2 = 3 y 2 z 7 x = ± 3 y 2 z 7 x = ±y z 3 3z 5 x 2 = 15 y 2 z 7 Divide both sides by 5. x 2 = 3 y 2 z 7 x = ± 3 y 2 z 7 x = ±y z 3 3z

Example 5

Use a calculator. Calculator problem.  Solve 14 a 2 235 = 0. 14 a 2 235 = 0. Round to the nearest hundredth.
14 a 2 235 = 0. Rewrite . 14 a 2 = 235 Divide both sides by 14 . a 2 = 235 14 14 a 2 235 = 0. Rewrite . 14 a 2 = 235 Divide both sides by 14 . a 2 = 235 14
On the Calculator On the Calculator
Type 235 Press ÷ Type 14 Press = Press Display reads: 4.0970373 Type 235 Press ÷ Type 14 Press = Press Display reads: 4.0970373
Rounding to the nearest hundredth produces 4.10. We must be sure to insert the ± ± symbol. a ± 4.10 a ± 4.10

Example 6

k 2 = 64 k = ± 64 k 2 = 64 k = ± 64
The radicand is negative so no real number solutions exist.

Practice Set A

Solve each of the following quadratic equations using the method of extraction of roots.

Exercise 1

x 2 144=0 x 2 144=0

Solution

x=±12 x=±12

Exercise 2

9 y 2 121=0 9 y 2 121=0

Solution

y=± 11 3 y=± 11 3

Exercise 3

6 a 2 =108 6 a 2 =108

Solution

a=±3 2 a=±3 2

Exercise 4

Solve 4 n 2 =24 m 2 p 8 4 n 2 =24 m 2 p 8 for n. n.

Solution

n=±m p 4 6 n=±m p 4 6

Exercise 5

Solve 5 p 2 q 2 =45 p 2 5 p 2 q 2 =45 p 2 for q. q.

Solution

q=±3 q=±3

Exercise 6

Use a calculator. Solve 16 m 2 2206=0. 16 m 2 2206=0. Round to the nearest hundredth.

Solution

m=±11.74 m=±11.74

Exercise 7

h 2 =100 h 2 =100

Sample Set B

Solve each of the following quadratic equations using the method of extraction of roots.

Example 7

(x+2) 2 = 81 x+2 = ± 81 x+2 = ±9 Subtract 2 from both sides. x = 2±9 x = 2+9 and x = 29 x = 7 x = 11 (x+2) 2 = 81 x+2 = ± 81 x+2 = ±9 Subtract 2 from both sides. x = 2±9 x = 2+9 and x = 29 x = 7 x = 11

Example 8

( a+3 ) 2 = 5 a+3 = ± 5 Subtract 3 from both sides. a = 3± 5 ( a+3 ) 2 = 5 a+3 = ± 5 Subtract 3 from both sides. a = 3± 5

Practice Set B

Solve each of the following quadratic equations using the method of extraction of roots.

Exercise 8

( a+6 ) 2 =64 ( a+6 ) 2 =64

Solution

a=2,14 a=2,14

Exercise 9

( m4 ) 2 =15 ( m4 ) 2 =15

Solution

m=4± 15 m=4± 15

Exercise 10

( y7 ) 2 =49 ( y7 ) 2 =49

Solution

y=0,14 y=0,14

Exercise 11

( k1 ) 2 =12 ( k1 ) 2 =12

Solution

k=1±2 3 k=1±2 3

Exercise 12

( x11 ) 2 =0 ( x11 ) 2 =0

Solution

x=11 x=11

Exercises

For the following problems, solve each of the quadratic equations using the method of extraction of roots.

Exercise 13

x 2 =36 x 2 =36

Solution

x=±6 x=±6

Exercise 14

x 2 =49 x 2 =49

Exercise 15

a 2 =9 a 2 =9

Solution

a=±3 a=±3

Exercise 16

a 2 =4 a 2 =4

Exercise 17

b 2 =1 b 2 =1

Solution

b=±1 b=±1

Exercise 18

a 2 =1 a 2 =1

Exercise 19

x 2 =25 x 2 =25

Solution

x=±5 x=±5

Exercise 20

x 2 =81 x 2 =81

Exercise 21

a 2 =5 a 2 =5

Solution

a=± 5 a=± 5

Exercise 22

a 2 =10 a 2 =10

Exercise 23

b 2 =12 b 2 =12

Solution

b=±2 3 b=±2 3

Exercise 24

b 2 =6 b 2 =6

Exercise 25

y 2 =3 y 2 =3

Solution

y=± 3 y=± 3

Exercise 26

y 2 =7 y 2 =7

Exercise 27

a 2 8=0 a 2 8=0

Solution

a=±2 2 a=±2 2

Exercise 28

a 2 3=0 a 2 3=0

Exercise 29

a 2 5=0 a 2 5=0

Solution

a=± 5 a=± 5

Exercise 30

y 2 1=0 y 2 1=0

Exercise 31

x 2 10=0 x 2 10=0

Solution

x=± 10 x=± 10

Exercise 32

x 2 11=0 x 2 11=0

Exercise 33

3 x 2 27=0 3 x 2 27=0

Solution

x=±3 x=±3

Exercise 34

5 b 2 5=0 5 b 2 5=0

Exercise 35

2 x 2 =50 2 x 2 =50

Solution

x=±5 x=±5

Exercise 36

4 a 2 =40 4 a 2 =40

Exercise 37

2 x 2 =24 2 x 2 =24

Solution

x=±2 3 x=±2 3

For the following problems, solve for the indicated variable.

Exercise 38

x 2 =4 a 2 , x 2 =4 a 2 , for x x

Exercise 39

x 2 =9 b 2 , x 2 =9 b 2 , for x x

Solution

x=±3b x=±3b

Exercise 40

a 2 =25 c 2 , a 2 =25 c 2 , for a a

Exercise 41

k 2 = m 2 n 2 , k 2 = m 2 n 2 , for k k

Solution

k=±mn k=±mn

Exercise 42

k 2 = p 2 q 2 r 2 , k 2 = p 2 q 2 r 2 , for k k

Exercise 43

2 y 2 =2 a 2 n 2 , 2 y 2 =2 a 2 n 2 , for y y

Solution

y=±an y=±an

Exercise 44

9 y 2 =27 x 2 z 4 , 9 y 2 =27 x 2 z 4 , for y y

Exercise 45

x 2 z 2 =0, x 2 z 2 =0, for x x

Solution

x=±z x=±z

Exercise 46

x 2 z 2 =0, x 2 z 2 =0, for z z

Exercise 47

5 a 2 10 b 2 =0, 5 a 2 10 b 2 =0, for a a

Solution

a=b 2 ,b 2 a=b 2 ,b 2

For the following problems, solve each of the quadratic equations using the method of extraction of roots.

Exercise 48

( x1 ) 2 =4 ( x1 ) 2 =4

Exercise 49

( x2 ) 2 =9 ( x2 ) 2 =9

Solution

x=5,1 x=5,1

Exercise 50

( x3 ) 2 =25 ( x3 ) 2 =25

Exercise 51

( a5 ) 2 =36 ( a5 ) 2 =36

Solution

x=11,1 x=11,1

Exercise 52

( a+3 ) 2 =49 ( a+3 ) 2 =49

Exercise 53

( a+9 ) 2 =1 ( a+9 ) 2 =1

Solution

a=8,10 a=8,10

Exercise 54

( a6 ) 2 =3 ( a6 ) 2 =3

Exercise 55

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

Solution

a=4± 5 a=4± 5

Exercise 56

( b+6 ) 2 =7 ( b+6 ) 2 =7

Exercise 57

( x+1 ) 2 =a, ( x+1 ) 2 =a, for x x

Solution

x=1± a x=1± a

Exercise 58

( y+5 ) 2 =b, ( y+5 ) 2 =b, for y y

Exercise 59

( y+2 ) 2 = a 2 , ( y+2 ) 2 = a 2 , for y y

Solution

y=2±a y=2±a

Exercise 60

( x+10 ) 2 = c 2 , ( x+10 ) 2 = c 2 , for x x

Exercise 61

( xa ) 2 = b 2 , ( xa ) 2 = b 2 , for x x

Solution

x=a±b x=a±b

Exercise 62

( x+c ) 2 = a 2 , ( x+c ) 2 = a 2 , for x x

Use a calculator. Calculator Problems

For the following problems, round each result to the nearest hundredth.

Exercise 63

8 a 2 168=0 8 a 2 168=0

Solution

a=±4.58 a=±4.58

Exercise 64

6 m 2 5=0 6 m 2 5=0

Exercise 65

0.03 y 2 =1.6 0.03 y 2 =1.6

Solution

y=±7.30 y=±7.30

Exercise 66

0.048 x 2 =2.01 0.048 x 2 =2.01

Exercise 67

1.001 x 2 0.999=0 1.001 x 2 0.999=0

Solution

x=±1.00 x=±1.00

Exercises For Review

Exercise 68

((Reference)) Graph the linear inequality 3( x+2 )<2( 3x+4 ). 3( x+2 )<2( 3x+4 ).

A horizontal line with arrows on both ends.

Exercise 69

((Reference)) Solve the fractional equation x1 x+4 = x+3 x1 . x1 x+4 = x+3 x1 .

Solution

x= 11 9 x= 11 9

Exercise 70

((Reference)) Find the product: 32 x 3 y 5 2 x 3 y 3 . 32 x 3 y 5 2 x 3 y 3 .

Exercise 71

((Reference)) Solve x 2 4x=0. x 2 4x=0.

Solution

x=0,4 x=0,4

Exercise 72

((Reference)) Solve y 2 8y=12. y 2 8y=12.

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