To solve the equation [tex]\(2 \sqrt{x-5} = 2\)[/tex] for [tex]\(x\)[/tex], let's follow the steps:
1. Isolate the square root term:
Given:
[tex]\[
2 \sqrt{x-5} = 2
\][/tex]
First, divide both sides by 2 to isolate the square root:
[tex]\[
\sqrt{x-5} = 1
\][/tex]
2. Square both sides to eliminate the square root:
Squaring both sides of the equation, we get:
[tex]\[
(\sqrt{x-5})^2 = 1^2
\][/tex]
[tex]\[
x - 5 = 1
\][/tex]
3. Solve for [tex]\(x\)[/tex]:
Add 5 to both sides:
[tex]\[
x = 1 + 5
\][/tex]
[tex]\[
x = 6
\][/tex]
4. Check for extraneous solutions:
Substitute [tex]\(x = 6\)[/tex] back into the original equation to verify if it is a valid solution:
[tex]\[
2 \sqrt{6 - 5} = 2
\][/tex]
[tex]\[
2 \sqrt{1} = 2
\][/tex]
[tex]\[
2 \cdot 1 = 2
\][/tex]
[tex]\[
2 = 2
\][/tex]
Since the left-hand side equals the right-hand side when [tex]\( x = 6 \)[/tex], this solution is valid and not extraneous.
Therefore, for the given options:
- [tex]\(x = 6\)[/tex], solution is not extraneous
- [tex]\(x = 6\)[/tex], solution is extraneous
- [tex]\(x = 11\)[/tex], solution is not extraneous
- [tex]\(x = 11\)[/tex], solution is extraneous
The correct choice is:
[tex]\[ x = 6, \text{ solution is not extraneous} \][/tex]
