To solve the equation [tex]\(\sqrt{2x + 8} = 6\)[/tex] for [tex]\(x\)[/tex] and determine if the solution is extraneous, follow these detailed steps:
1. Eliminate the square root by squaring both sides of the equation:
[tex]\[
(\sqrt{2x + 8})^2 = 6^2
\][/tex]
This simplifies to:
[tex]\[
2x + 8 = 36
\][/tex]
2. Solve for [tex]\(x\)[/tex] by isolating it on one side of the equation:
- Subtract 8 from both sides:
[tex]\[
2x = 36 - 8
\][/tex]
Simplifying this, you get:
[tex]\[
2x = 28
\][/tex]
- Divide both sides by 2:
[tex]\[
x = \frac{28}{2}
\][/tex]
Simplifying further, you get:
[tex]\[
x = 14
\][/tex]
3. Check for extraneous solutions by substituting the value of [tex]\(x\)[/tex] back into the original equation to ensure it satisfies the equation:
- Substitute [tex]\(x = 14\)[/tex] back into the original equation:
[tex]\[
\sqrt{2(14) + 8} = \sqrt{28 + 8} = \sqrt{36}
\][/tex]
This simplifies to:
[tex]\[
\sqrt{36} = 6
\][/tex]
- Since [tex]\(\sqrt{36} = 6\)[/tex] holds true, the solution is not extraneous.
Therefore, the solution to the equation [tex]\(\sqrt{2x + 8} = 6\)[/tex] is [tex]\(x = 14\)[/tex], and it is not an extraneous solution.
The correct answer is:
[tex]\[ \boxed{14 ; solution is not extraneous} \][/tex]
