Solve for [tex]x[/tex], given the equation [tex]\sqrt{x-5} + 7 = 11[/tex].

A. [tex]x = 21[/tex], solution is extraneous
B. [tex]x = 21[/tex], solution is not extraneous
C. [tex]x = 81[/tex], solution is extraneous
D. [tex]x = 81[/tex], solution is not extraneous



Answer :

Let's solve the given equation step by step:

The equation is:
[tex]\[ \sqrt{x - 5} + 7 = 11 \][/tex]

First, isolate the square root term by subtracting 7 from both sides of the equation:
[tex]\[ \sqrt{x - 5} = 11 - 7 \][/tex]

Simplify the right side:
[tex]\[ \sqrt{x - 5} = 4 \][/tex]

Next, eliminate the square root by squaring both sides of the equation:
[tex]\[ (\sqrt{x - 5})^2 = 4^2 \][/tex]

Simplify both sides:
[tex]\[ x - 5 = 16 \][/tex]

Now, solve for [tex]\( x \)[/tex] by adding 5 to both sides:
[tex]\[ x = 16 + 5 \][/tex]

Thus, the solution is:
[tex]\[ x = 21 \][/tex]

Now, we need to check if this solution is extraneous. To do this, we substitute [tex]\( x = 21 \)[/tex] back into the original equation:
[tex]\[ \sqrt{21 - 5} + 7 = 11 \][/tex]

Simplify inside the square root:
[tex]\[ \sqrt{16} + 7 = 11 \][/tex]

Since [tex]\(\sqrt{16} = 4\)[/tex], we get:
[tex]\[ 4 + 7 = 11 \][/tex]

This simplifies to:
[tex]\[ 11 = 11 \][/tex]

Since the left side equals the right side, the solution [tex]\( x = 21 \)[/tex] is correct and not extraneous.

Therefore, the correct solution is:
[tex]\[ x = 21, \text{ solution is not extraneous} \][/tex]