Question

solve the given radical equation. check all proposed solutions. $sqrt{3x + 22}=x + 6$ select the correct choice below and, if necessary, fill in the answer box to cor a. the solution set is (use a comma to separate answers as needed. simplify your answer.) b. the solution set is the empty set

Explanation:

Step1: Square both sides

$(\sqrt{3x + 22})^2=(x + 6)^2$
$3x+22=x^{2}+12x + 36$

Step2: Rearrange to quadratic - form

$x^{2}+12x+36-3x - 22 = 0$
$x^{2}+9x + 14=0$

Step3: Factor the quadratic equation

$(x + 2)(x+7)=0$

Step4: Solve for x

$x+2 = 0$ gives $x=-2$; $x + 7=0$ gives $x=-7$

Step5: Check solutions

For $x=-2$: $\sqrt{3(-2)+22}=\sqrt{-6 + 22}=\sqrt{16}=4$, and $-2 + 6=4$, so $x=-2$ is a solution.
For $x=-7$: $\sqrt{3(-7)+22}=\sqrt{-21 + 22}=1$, and $-7 + 6=-1$, so $x=-7$ is not a solution.

Answer:

A. The solution set is $\{-2\}$