Let's solve the equation [tex]\(\sqrt{8x + 1} = 5\)[/tex] step-by-step and identify if the solution is extraneous or not.
1. Isolate the square root term:
[tex]\[
\sqrt{8x + 1} = 5
\][/tex]
2. Square both sides to eliminate the square root:
[tex]\[
(\sqrt{8x + 1})^2 = 5^2
\][/tex]
This simplifies to:
[tex]\[
8x + 1 = 25
\][/tex]
3. Solve for [tex]\(x\)[/tex]:
Subtract 1 from both sides:
[tex]\[
8x = 24
\][/tex]
Divide both sides by 8:
[tex]\[
x = 3
\][/tex]
4. Check if the solution is extraneous by substituting [tex]\(x = 3\)[/tex] back into the original equation:
[tex]\[
\sqrt{8(3) + 1} = 5
\][/tex]
Simplify inside the square root:
[tex]\[
\sqrt{24 + 1} = 5
\][/tex]
[tex]\[
\sqrt{25} = 5
\][/tex]
Since [tex]\(\sqrt{25} = 5\)[/tex] is true, the solution [tex]\(x = 3\)[/tex] satisfies the original equation.
Therefore, the solution [tex]\(x = 3\)[/tex] is not extraneous.
So, the correct answer is:
[tex]\[
x = 3, \text{ solution is not extraneous}
\][/tex]
