Question

question
solve for all possible values of x.
$sqrt{x + 12} = x + 6$

Explanation:

Step1: Square both sides to eliminate root

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

Step2: Rearrange to quadratic form

$$x^2+12x+36-x-12=0$$
$$x^2+11x+24=0$$

Step3: Factor the quadratic equation

$$(x+3)(x+8)=0$$

Step4: Solve for x candidates

$$x+3=0 \quad \text{or} \quad x+8=0$$
$$x=-3 \quad \text{or} \quad x=-8$$

Step5: Verify candidates in original equation

For $x=-3$: $\sqrt{-3+12}=-3+6$ → $\sqrt{9}=3$ → $3=3$, valid.
For $x=-8$: $\sqrt{-8+12}=-8+6$ → $\sqrt{4}=-2$ → $2=-2$, invalid.

Answer:

$x=-3$