To solve the equation [tex]\(\sqrt{2x + 1} = 3\)[/tex] for [tex]\(x\)[/tex] and determine if the solution is extraneous, let's follow these steps:
1. Isolate the square root and eliminate it by squaring both sides:
[tex]\[
\sqrt{2x + 1} = 3
\][/tex]
Squaring both sides, we get:
[tex]\[
(\sqrt{2x + 1})^2 = 3^2
\][/tex]
This simplifies to:
[tex]\[
2x + 1 = 9
\][/tex]
2. Solve the resulting linear equation:
Subtract 1 from both sides:
[tex]\[
2x = 8
\][/tex]
Divide both sides by 2:
[tex]\[
x = 4
\][/tex]
So the solution found is:
[tex]\[
x = 4
\][/tex]
3. Check for extraneous solutions:
Substitute [tex]\(x = 4\)[/tex] back into the original equation to verify:
[tex]\[
\sqrt{2(4) + 1} = \sqrt{8 + 1} = \sqrt{9} = 3
\][/tex]
Since the left-hand side equals the right-hand side (3 = 3), the solution [tex]\(x = 4\)[/tex] is valid.
4. Conclusion:
The solution [tex]\(x = 4\)[/tex] is not extraneous.
Therefore, the correct answer is:
[tex]\[
x = 4, \text{ solution is not extraneous}
\][/tex]
