which of the following is an extraneous solution of $sqrt{-3x - 2}=x + 2$?\n$x=-6$\n$x=-1$\n$x = 1$\n$x=6$

Answer:

A. $x = - 6$

Explanation:

Step1: Square both sides of the equation

$(\sqrt{-3x - 2})^2=(x + 2)^2$. So, $-3x-2=x^{2}+4x + 4$.

Step2: Rearrange to form a quadratic equation

$x^{2}+4x+3x + 4 + 2=0$, which simplifies to $x^{2}+7x+6 = 0$.

Step3: Factor the quadratic equation

$(x + 6)(x+1)=0$. So, $x=-6$ or $x=-1$.

Step4: Check the solutions in the original equation

For $x=-6$, $\sqrt{-3(-6)-2}=\sqrt{18 - 2}=\sqrt{16}=4$, and $-6 + 2=-4$. Since $4\neq-4$, $x = - 6$ is extraneous. For $x=-1$, $\sqrt{-3(-1)-2}=\sqrt{3 - 2}=1$, and $-1 + 2=1$. So $x=-1$ is a valid solution.