Question

\[ g(x)=\frac{25-x^{2}}{x^{2}+5 x} \] (a) Find the domain of \( g(x) \). \( x \neq-5, x \neq 5 \) \( x \neq 25 \) \( x \neq 25, x \neq 0 \) \( x \neq-5, x \neq 0 \) \( x \neq-5 \) \( x \neq 0 \) (b) How many solutions does each equation have? \[ \begin{array}{l} g(x)=0 \\ g(x)=1 \\ g(x)=-1 \end{array} \] \( \square \) \( \square \) \( \square \)

          \[
g(x)=\frac{25-x^{2}}{x^{2}+5 x}
\]
(a) Find the domain of \( g(x) \).
\( x \neq-5, x \neq 5 \)
\( x \neq 25 \)
\( x \neq 25, x \neq 0 \)
\( x \neq-5, x \neq 0 \)
\( x \neq-5 \)
\( x \neq 0 \)
(b) How many solutions does each equation have?
\[
\begin{array}{l}
g(x)=0 \\
g(x)=1 \\
g(x)=-1
\end{array}
\]
\( \square \)
\( \square \)
\( \square \)

g(x)=(25-x^2)/(x^2+5 x)

(a) Find the domain of g(x).
x ≠-5, x ≠ 5
x ≠ 25
x ≠ 25, x ≠ 0
x ≠-5, x ≠ 0
x ≠-5
x ≠ 0
(b) How many solutions does each equation have?

g(x)=0

g(x)=1

g(x)=-1

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Added by Alexis B.

Precalculus

Robert Blitzer 6th Edition

Chapter 1

Instant Answer

Solved by Expert Keondre Parker

Step 1

The function \( g(x) = \frac{25-x^2}{x^2 + 5x} \) is defined for all \( x \) except where the denominator is zero. To find these values, solve the equation: \[ x^2 + 5x = 0 \] Factoring out \( x \), we get: \[ x(x + 5) = 0 \] Setting each factor equal to zero Show more…

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\[ g(x)=\frac{25-x^{2}}{x^{2}+5 x} \] (a) Find the domain of \( g(x) \). \( x \neq-5, x \neq 5 \) \( x \neq 25 \) \( x \neq 25, x \neq 0 \) \( x \neq-5, x \neq 0 \) \( x \neq-5 \) \( x \neq 0 \) (b) How many solutions does each equation have? \[ \begin{array}{l} g(x)=0 \\ g(x)=1 \\ g(x)=-1 \end{array} \] \( \square \) \( \square \) \( \square \)