We have the equation [tex]\( 2 \sqrt{x - 5} = 2 \)[/tex]. Let's solve for [tex]\( x \)[/tex] and determine if the solution is extraneous through a step-by-step process.
1. Isolate the square root term:
[tex]\[
2 \sqrt{x - 5} = 2
\][/tex]
Divide both sides by 2:
[tex]\[
\sqrt{x - 5} = 1
\][/tex]
2. Remove the square root by squaring both sides:
[tex]\[
(\sqrt{x - 5})^2 = 1^2
\][/tex]
This simplifies to:
[tex]\[
x - 5 = 1
\][/tex]
3. Solve for [tex]\( x \)[/tex]:
Add 5 to both sides:
[tex]\[
x = 1 + 5
\][/tex]
Therefore:
[tex]\[
x = 6
\][/tex]
4. Check if the solution is extraneous:
Substitute [tex]\( x = 6 \)[/tex] back into the original equation to verify:
[tex]\[
2 \sqrt{6 - 5} = 2
\][/tex]
Simplify inside the square root:
[tex]\[
2 \sqrt{1} = 2
\][/tex]
Which simplifies to:
[tex]\[
2 \cdot 1 = 2
\][/tex]
[tex]\[
2 = 2
\][/tex]
Since both sides of the original equation are equal when [tex]\( x = 6 \)[/tex], the solution satisfies the original equation and is therefore not extraneous.
Thus, the correct answer is:
- [tex]\( x = 6 \)[/tex], solution is not extraneous
