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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Expert Solution & Answer
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Chapter 3.6, Problem 39E
Solution

To calculate: The value of x for the equation x2+4=x+5 and also check the extraneous solution.

The resultant answer is x=2110 .

Given information:

The equation is x2+4=x+5 .

Formula used:

Raise each side of the equation to the same power.

Calculation:

Write the original equation:

  x2+4=x+5

Square each side of the equation,

  x2+42=x+52x2+4=x2+10x+25x2x210x=25410x=21

Divide both sides by -10,

  10x10=2110x=2110

Now check x=2110 in the original equation.

  x2+4=x+5

Now substitute 2110 for x and simplify,

  21102+4=?2110+5441100+4=?21+5010441+400100=?21+5010841100=?29102910=2910

Therefore, the only solution is 2110 .

Chapter 3 Solutions

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Ch. 3.1 - Prob. 11GPCh. 3.1 - Prob. 12GPCh. 3.1 - Prob. 13GPCh. 3.1 - Prob. 14GPCh. 3.1 - Prob. 15GPCh. 3.1 - Prob. 16GPCh. 3.1 - Prob. 17GPCh. 3.1 - Prob. 18GPCh. 3.1 - Prob. 19GPCh. 3.1 - Prob. 1ECh. 3.1 - Prob. 2ECh. 3.1 - Prob. 3ECh. 3.1 - Prob. 4ECh. 3.1 - Prob. 5ECh. 3.1 - Prob. 6ECh. 3.1 - Prob. 7ECh. 3.1 - Prob. 8ECh. 3.1 - Prob. 9ECh. 3.1 - Prob. 10ECh. 3.1 - Prob. 11ECh. 3.1 - Prob. 12ECh. 3.1 - Prob. 13ECh. 3.1 - Prob. 14ECh. 3.1 - Prob. 15ECh. 3.1 - Prob. 16ECh. 3.1 - Prob. 17ECh. 3.1 - Prob. 18ECh. 3.1 - Prob. 19ECh. 3.1 - Prob. 20ECh. 3.1 - Prob. 21ECh. 3.1 - Prob. 22ECh. 3.1 - Prob. 23ECh. 3.1 - Prob. 24ECh. 3.1 - Prob. 25ECh. 3.1 - Prob. 26ECh. 3.1 - Prob. 27ECh. 3.1 - Prob. 28ECh. 3.1 - Prob. 29ECh. 3.1 - Prob. 30ECh. 3.1 - Prob. 31ECh. 3.1 - Prob. 32ECh. 3.1 - Prob. 33ECh. 3.1 - Prob. 34ECh. 3.1 - Prob. 35ECh. 3.1 - Prob. 36ECh. 3.1 - Prob. 37ECh. 3.1 - Prob. 38ECh. 3.1 - Prob. 39ECh. 3.1 - Prob. 40ECh. 3.1 - Prob. 41ECh. 3.1 - Prob. 42ECh. 3.1 - Prob. 43ECh. 3.1 - Prob. 44ECh. 3.1 - 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