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Chapter 8: Problem 15

Values that make the denominators equal zero cannot be solutions of an equation. Find \(a l l\) of the values that make the denominators zero and which, therefore, cannot be solutions of each equation. Do NOT solve the equation. $$ \frac{9 h}{h^{2}-5 h-36}+\frac{1}{h+4}=\frac{h+7}{3 h-27} $$

Short Answer

Expert verified
The restricted values for h are \(-4\) and \(9\). These values cannot be solutions of the equation.

Step by step solution

01

Find the zeros of the first denominator

The first denominator is \(h^2-5h-36\). To find the values of h that make this expression equal to zero, we need to factor it and solve for h. \((h^2-5h-36)=0\) can be factored as \((h-9)(h+4)=0\). Therefore, the zeros of this denominator are h=9 and h=-4.
02

Find the zeros of the second denominator

The second denominator is \(h+4\). To find the values of h that make this expression equal to zero, solve for h: \((h+4)=0\). Thus, the value of h that makes this denominator equal to zero is h=-4.
03

Find the zeros of the third denominator

The third denominator is \(3h-27\). To find the values of h that make this expression equal to zero, solve for h: \((3h-27)=0\). Dividing both sides by 3, we get h=9.
04

Compile the restricted values and simplify

We have found the zeros of all three denominators. The restricted values for h are {-4, 9}, which means the values h=-4 and h=9 cannot be solutions of the given equation. So, the restricted values for h are \(-4\) and \(9\). These values cannot be solutions of the equation.

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Most popular questions from this chapter

Solve for the indicated variable. $$h=\frac{3 A}{r+s} \text { for } s$$

Solve each equation. $$\frac{4}{x^{2}+2 x-15}=\frac{8}{x-3}+\frac{2}{x+5}$$

Perform the indicated operations. $$\frac{1}{x+y}+\frac{x}{x^{2}-y^{2}}-\frac{4}{2 x-2 y}$$

Perform the indicated operations. $$\frac{3 v-4}{6 v^{2}-v-5}-\frac{v-2}{3 v^{2}+v-4}$$

Solve for the indicated variable. $$V=\frac{n R T}{P} \text { for } P$$
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