To solve and check , the given equation has no solution of 9t2−2t−8+tt+2=2

Expert Solution & Answer

Answer to Problem 26WE

The solution of equation are {5,−5}

Explanation of Solution

Given that: : 9t2−2t−8+tt+2=2

First factor the t2−2t−8

t2−2t−8=t2+(2−4)t−8=t2+2t−4t−8=t(t+2)−4(t+2)=(t−4)(t+2)

Rewrite the given equation

9(t+2)(t−4)+tt+2=2

Multiply by (t+2)(t−4) , LCM of (t+2) and (t−4)

(t−4)(t+2)(9(t+2)(t−4)+tt+2)=2(t+2)(t−4)

Apply distributive property

9+t(t−4)=2(t−4)(t+2)9+t2−4t=2(t2+2t−4t−8)9+t2−4t=2t2−4t−16t2−25=0

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

t2−25=0t2=25t=±5t=−5,5

When t equal to -2 and 4, a denominator in the original equation has a value of 0. Therefore,

-2 and 4 are not in the domain of y and cannot be solution.

Check:

For t=−5925−2(−5)−8+−5−5+2=2927+−5−3=213+53=263=22=2LHS=RHSSimilarlyFor t=5925−2(5)−8+55+2=297+57=2147=22=2LHS=RHS

Hence, Solution of equation is {5,−5}

Chapter 5.9, Problem 25WE

Chapter 5.9, Problem 27WE

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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