Let's solve the given equation [tex]\(\sqrt{2x + 1} = 3\)[/tex] step by step.
1. First, we need to eliminate the square root. We can do this by squaring both sides of the equation:
[tex]\[
(\sqrt{2x + 1})^2 = 3^2
\][/tex]
2. Simplifying both sides, we get:
[tex]\[
2x + 1 = 9
\][/tex]
3. Next, solve for [tex]\(x\)[/tex]. To do this, isolate the [tex]\(x\)[/tex] term:
[tex]\[
2x + 1 - 1 = 9 - 1
\][/tex]
[tex]\[
2x = 8
\][/tex]
4. Divide both sides by 2 to solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{8}{2}
\][/tex]
[tex]\[
x = 4
\][/tex]
So, we have found that [tex]\(x = 4\)[/tex].
5. Now, we need to check if this solution is extraneous. An extraneous solution is one that does not satisfy the original equation.
Substitute [tex]\(x = 4\)[/tex] back into the original equation [tex]\(\sqrt{2x + 1} = 3\)[/tex]:
[tex]\[
\sqrt{2(4) + 1} = \sqrt{8 + 1} = \sqrt{9}
\][/tex]
6. Simplifying the right hand side:
[tex]\[
\sqrt{9} = 3
\][/tex]
Since both sides of the original equation are equal when [tex]\(x = 4\)[/tex], the solution is valid.
Therefore, the solution is:
[tex]\(x = 4\)[/tex], solution is not extraneous
