Algebra and Trigonometry: Structure and Method, Book 2

2000th Edition

ISBN: 9780395977255

Author: MCDOUGAL LITTEL

Publisher: McDougal Littell

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Question

Chapter 5.9, Problem 23WE

To determine

To solve and check , the given equation has no solution of 5u2+u−6=2−u−3u−2

Expert Solution & Answer

Answer to Problem 23WE

The answer is {−4}

Explanation of Solution

Given that: : 5u2+u−6=2−u−3u−2

First factor the u2+u−6

u2+u−6=u2+(3−2)u−6=u2+3u−2u−6=u(u+3)−2(u+3)=(u+3)(u−2)

Rewrite the given equation

5(u+3)(u−2)=2−u−3u−2

Multiply by (u+3)(u−2) , LCM of (u+3) and (u−2)

(u+3)(u−2)(5(u+3)(u−2))=(u+3)(u−2)(2−u−3u−2)

Apply distributive property

5=2(u+3)(u−2)−(u−3)(u+3)5=2(u2−2u+3u−6)−(u2−3u+3u−9)5=2(u2+u−6)−(u2−9)5=2u2+2u−12−u2+95=u2+2u−3

Add -5 to both sides

5−5=u2+2u−3−5u2+2u−8=0

Apply the factor method and Apply zero product property to find the solution of equation

u2+2u−8=0=0u2+(4−2)u−8=0u2+4u−2u−8=0u(u+4)−2(u−8)=0(u+4)(u−2)=0u=−4,2

When u equal to -3 and 2, a denominator in the original equation has a value of 0. Therefore,

-3 and 2 are not in the domain of x and cannot be solution.

Check:

For u=−45(−4+3)(−4−2)=2−−4−3−4−25(−1)(−6)=2−−7−656=56LHS=RHS

For u=25(2+3)(2−2)=2−2−32−25(0)≠2−−10→ undefined

Therefore, {u=−4} is defined the solution.

Hence, the solution of equation is {−4}

Chapter 5.9, Problem 22WE

Chapter 5.9, Problem 24WE

Chapter 5 Solutions

Algebra and Trigonometry: Structure and Method, Book 2

Ch. 5.1 - Prob. 1OE Ch. 5.1 - Prob. 2OE Ch. 5.1 - Prob. 3OE Ch. 5.1 - Prob. 4OE Ch. 5.1 - Prob. 5OE Ch. 5.1 - Prob. 6OE Ch. 5.1 - Prob. 7OE Ch. 5.1 - Prob. 8OE Ch. 5.1 - Prob. 9OE Ch. 5.1 - Prob. 10OE

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Knowledge Booster

To solve and check , the given equation has no solution of 5 u 2 + u − 6 = 2 − u − 3 u − 2

To solve and check , the given equation has no solution of 5u2+u−6=2−u−3u−2

Answer to Problem 23WE

Explanation of Solution

Chapter 5 Solutions

Additional Math Textbook Solutions