To solve the equation [tex]\(\sqrt{3x + 16} = x + 2\)[/tex], let's go through the steps methodically.
1. Square both sides to eliminate the square root:
[tex]\[
(\sqrt{3x + 16})^2 = (x + 2)^2
\][/tex]
This gives:
[tex]\[
3x + 16 = x^2 + 4x + 4
\][/tex]
2. Rearrange the equation to set it to zero:
[tex]\[
x^2 + 4x + 4 - 3x - 16 = 0
\][/tex]
Simplify the terms:
[tex]\[
x^2 + x - 12 = 0
\][/tex]
3. Factor the quadratic equation:
[tex]\[
x^2 + x - 12 = (x - 3)(x + 4) = 0
\][/tex]
4. Find the roots:
[tex]\[
x - 3 = 0 \quad \text{or} \quad x + 4 = 0
\][/tex]
Thus,
[tex]\[
x = 3 \quad \text{or} \quad x = -4
\][/tex]
5. Verify the roots by substituting them back into the original equation to check for extraneous solutions:
- For [tex]\(x = 3\)[/tex]:
[tex]\[
\sqrt{3(3) + 16} = 3 + 2
\][/tex]
[tex]\[
\sqrt{9 + 16} = 5
\][/tex]
[tex]\[
\sqrt{25} = 5
\][/tex]
This is true, so [tex]\(x = 3\)[/tex] is a valid solution.
- For [tex]\(x = -4\)[/tex]:
[tex]\[
\sqrt{3(-4) + 16} = -4 + 2
\][/tex]
[tex]\[
\sqrt{-12 + 16} = -2
\][/tex]
[tex]\[
\sqrt{4} = -2
\][/tex]
Since [tex]\(\sqrt{4} = 2\)[/tex] and not [tex]\(-2\)[/tex], [tex]\(x = -4\)[/tex] is not a valid solution.
After checking, we find that the only valid solution is:
[tex]\[
\{3\}
\][/tex]
Therefore, the correct answer is:
3) [tex]\(\{3\}\)[/tex]
