Holt Mcdougal Larson Algebra 2: Student Edition 2012
Holt Mcdougal Larson Algebra 2: Student Edition 2012
1st Edition
ISBN: 9780547647159
Author: HOLT MCDOUGAL
Publisher: HOLT MCDOUGAL
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Chapter 3.6, Problem 55E
Solution

To state: A radical equation that has two extraneous solutions.

The equation is 20x2−44=2x and its resultant extraneous solutions are −1 and −12 .

Given information:

The statement is “Give an example of a radical equation that has two extraneous solutions.”

Explanation:

Let’s consider a radical equation 20x2−44=2x as an example.

Now raise both sides of the equation to the power 4,

  20x2−444=2x420x2−4=16x4

Now right the equation in standard form,

  16x4−20x2+4=0

Factor the equation,

  42x−12x+1x−1x+1=0

Apply the zero-product property and find the values of x ,

  2x−1=0, x=122x+1=0, x=−12x−1=0, x=1x+1=0, x=−1

Now check the solutions by substituting them in original equation,

  20x2−44=2x

Substitute 12 for x into the equation,

  20122−44=?2122014−44=?15−44=?114=?11=1

Substitute −12 for x into the equation,

  20−122−44=?2−122014−44=?−15−44=?−114=?−11≠−1

Substitute 1 for x into the equation,

  2012−44=?21201−44=?220−44=?2164=?22=2

Substitute −1 for x into the equation,

  20−12−44=?2−1201−44=?−220−44=?−2164=?−22≠2

The solutions are 12 and 1 . The apparent solutions −12 and −1 are extraneous.

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Chapter 3.6, Problem 54E
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Chapter 3.6, Problem 56PS

Chapter 3 Solutions

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Ch. 3.1 - Prob. 1GPCh. 3.1 - Prob. 2GPCh. 3.1 - Prob. 3GPCh. 3.1 - Prob. 4GPCh. 3.1 - Prob. 5GPCh. 3.1 - Prob. 6GPCh. 3.1 - Prob. 7GPCh. 3.1 - Prob. 8GPCh. 3.1 - Prob. 9GPCh. 3.1 - Prob. 10GP
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