select the correct answer.\nwhat is the solution to the equation?\n$sqrt{-2x - 5}-4 = x$\na. -7 and -3\nb. 3…

Explanation:

Step1: Isolate the square - root term

Add 4 to both sides of the equation $\sqrt{-2x - 5}-4=x$ to get $\sqrt{-2x - 5}=x + 4$.

Step2: Square both sides

Squaring both sides gives $-2x-5=(x + 4)^2$. Expand the right - hand side: $(x + 4)^2=x^{2}+8x + 16$. So, $-2x-5=x^{2}+8x + 16$.

Step3: Rearrange to form a quadratic equation

Move all terms to one side to get $x^{2}+8x+2x+16 + 5=0$, which simplifies to $x^{2}+10x+21 = 0$.

Step4: Factor the quadratic equation

Factor $x^{2}+10x + 21$ as $(x + 3)(x+7)=0$.

Step5: Solve for x

Set each factor equal to zero: $x+3 = 0$ gives $x=-3$ and $x + 7=0$ gives $x=-7$.

Step6: Check for extraneous solutions

For $x=-7$: Substitute into the original equation $\sqrt{-2(-7)-5}-4=-7$. $\sqrt{14 - 5}-4=-7$, $\sqrt{9}-4=-7$, $3 - 4=-1\neq-7$. So, $x=-7$ is an extraneous solution. For $x=-3$: Substitute into the original equation $\sqrt{-2(-3)-5}-4=-3$. $\sqrt{6 - 5}-4=-3$, $\sqrt{1}-4=-3$, $1 - 4=-3$. So, $x=-3$ is the solution.

Answer:

C. -3