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Given the equation [tex]2 \sqrt{x-5}=2[/tex], solve for [tex]x[/tex] and identify if it is an extraneous solution.

A. [tex]x=6[/tex], solution is not extraneous
B. [tex]x=6[/tex], solution is extraneous
C. [tex]x=11[/tex], solution is not extraneous
D. [tex]x=11[/tex], solution is extraneous



Answer :

GinnyAnswer
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

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