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

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

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

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

The nature of solutions

Solutions of x 2 = K

For quadratic equations of the form x 2 = K ,

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

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 .

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25 a 2 = 36 a 2 = 36 25 a = ± 36 25 a = ± 6 5
C h e c k : 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 .

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4 m 2 32 = 0 4 m 2 = 32 m 2 = 32 4 m 2 = 8 m = ± 8 m = ± 2 2
C h e c k : 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 .

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Solve 5 x 2 15 y 2 z 7 = 0 for 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 3 z

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Use a calculator. Calculator problem.  Solve 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
On the Calculator
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

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

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Practice set a

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

x 2 144 = 0

x = ± 12

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9 y 2 121 = 0

y = ± 11 3

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

n = ± m p 4 6

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

q = ± 3

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

m = ± 11.74

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Sample set b

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

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

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

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Practice set b

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

( a + 6 ) 2 = 64

a = 2 , 14

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( m 4 ) 2 = 15

m = 4 ± 15

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( y 7 ) 2 = 49

y = 0 , 14

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( k 1 ) 2 = 12

k = 1 ± 2 3

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( x 11 ) 2 = 0

x = 11

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Exercises

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

a 2 8 = 0

a = ± 2 2

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x 2 10 = 0

x = ± 10

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3 x 2 27 = 0

x = ± 3

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For the following problems, solve for the indicated variable.

x 2 = 9 b 2 , for x

x = ± 3 b

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k 2 = m 2 n 2 , for k

k = ± m n

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

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

y = ± a n

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

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x 2 z 2 = 0 , for x

x = ± z

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5 a 2 10 b 2 = 0 , for a

a = b 2 , b 2

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

( x 2 ) 2 = 9

x = 5 , 1

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( a 5 ) 2 = 36

x = 11 , 1

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( a + 9 ) 2 = 1

a = 8 , 10

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( x + 4 ) 2 = 5

a = 4 ± 5

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( x + 1 ) 2 = a , for x

x = 1 ± a

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

y = 2 ± a

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

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( x a ) 2 = b 2 , for x

x = a ± b

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

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Use a calculator.  calculator problems

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

8 a 2 168 = 0

a = ± 4.58

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0.03 y 2 = 1.6

y = ± 7.30

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1.001 x 2 0.999 = 0

x = ± 1.00

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Exercises for review

( [link] ) Graph the linear inequality 3 ( x + 2 ) < 2 ( 3 x + 4 ) .

A horizontal line with arrows on both ends.

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( [link] ) Solve the fractional equation x 1 x + 4 = x + 3 x 1 .

x = 11 9

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( [link] ) Find the product: 32 x 3 y 5 2 x 3 y 3 .

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( [link] ) Solve x 2 4 x = 0.

x = 0 , 4

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( [link] ) Solve y 2 8 y = 12.

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Questions & Answers

Preparation and Applications of Nanomaterial for Drug Delivery
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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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