To determine the restrictions for the given expression:
[tex]\[
\frac{2 x^2-6 x}{x^2+18 x+81} \times \frac{9 x+81}{x^2-9}
\][/tex]
we need to identify where the denominators are equal to zero since division by zero is undefined.
### Step-by-Step Solution:
1. Identify the Denominators:
The denominators in the expression are:
[tex]\[
x^2 + 18x + 81 \quad \text{and} \quad x^2 - 9
\][/tex]
2. Find the Roots of the Denominators:
To find the restrictions, set each denominator equal to zero and solve for [tex]\( x \)[/tex].
- For [tex]\( x^2 + 18x + 81 \)[/tex]:
[tex]\[
x^2 + 18x + 81 = 0
\][/tex]
This is a quadratic equation. We can solve it by factoring:
[tex]\[
(x + 9)(x + 9) = 0
\][/tex]
Therefore:
[tex]\[
x + 9 = 0 \quad \Rightarrow \quad x = -9
\][/tex]
- For [tex]\( x^2 - 9 \)[/tex]:
[tex]\[
x^2 - 9 = 0
\][/tex]
This is a difference of squares, so it factors as:
[tex]\[
(x + 3)(x - 3) = 0
\][/tex]
Therefore:
[tex]\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\][/tex]
and
[tex]\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\][/tex]
3. List All Restrictions:
The values of [tex]\( x \)[/tex] which make any of the denominators zero (and thus undefined for the expression) are:
[tex]\[
x = -9, \quad x = -3, \quad x = 3
\][/tex]
### Conclusion:
The restrictions for the expression are [tex]\( x \neq -9 \)[/tex], [tex]\( x \neq -3 \)[/tex], and [tex]\( x \neq 3 \)[/tex].
Given the options:
- [tex]\( x \neq 3 \)[/tex]
- [tex]\( x \neq -9 \)[/tex]
- [tex]\( x \neq 0 \)[/tex]
- [tex]\( x \neq 9 \)[/tex]
The correct restrictions are:
- [tex]\( x \neq 3 \)[/tex]
- [tex]\( x \neq -9 \)[/tex]
The correct answer is:
- [tex]\( x \neq 3 \)[/tex]
- [tex]\( x \neq -9 \)[/tex]
