To solve the equation [tex]\(\sqrt{2x + 1} = 3\)[/tex] for [tex]\(x\)[/tex], we need to follow these steps:
1. Square both sides of the equation to eliminate the square root. Remember that squaring is a reversible operation, but it can introduce extraneous solutions, so we need to verify our answers later.
[tex]\[
(\sqrt{2x + 1})^2 = 3^2
\][/tex]
This simplifies to:
[tex]\[
2x + 1 = 9
\][/tex]
2. Solve the resulting linear equation for [tex]\(x\)[/tex].
[tex]\[
2x + 1 = 9
\][/tex]
Subtract 1 from both sides:
[tex]\[
2x = 8
\][/tex]
Divide both sides by 2:
[tex]\[
x = 4
\][/tex]
3. Verify the solution by substituting [tex]\(x = 4\)[/tex] back into the original equation to check if it holds true.
Substituting [tex]\(x = 4\)[/tex] into [tex]\(\sqrt{2x + 1}\)[/tex]:
[tex]\[
\sqrt{2(4) + 1} = \sqrt{8 + 1} = \sqrt{9} = 3
\][/tex]
Since the left-hand side (LHS) is equal to the right-hand side (RHS):
[tex]\[
3 = 3
\][/tex]
The solution [tex]\(x = 4\)[/tex] satisfies the original equation.
Since [tex]\(x = 4\)[/tex] satisfies the original equation [tex]\(\sqrt{2x + 1} = 3\)[/tex], it is not an extraneous solution.
So, the correct answer is:
[tex]\(x = 4\)[/tex], solution is not extraneous.
