Question

9 multiple choice 1 point
which statement is true about the solutions of $x + 1 = \sqrt{7x + 15}$?
the solution is x = -7, and the extraneous solution is 2
the solution is x = 2 and the extraneous solution is x = -7
the solution is x = 7, and the extraneous solution is -2
the solutions are x = -2 and x = 7, and there are no extraneous solutions.

Explanation:

Step1: Square both sides to eliminate root

$$(x+1)^2 = (\sqrt{7x+15})^2$$
$$x^2 + 2x + 1 = 7x + 15$$

Step2: Rearrange to quadratic equation

$$x^2 + 2x + 1 - 7x - 15 = 0$$
$$x^2 - 5x - 14 = 0$$

Step3: Factor the quadratic

$$(x-7)(x+2) = 0$$
$$x=7 \text{ or } x=-2$$

Step4: Verify solutions in original equation

For $x=7$: $7+1 = \sqrt{7*7+15} \implies 8 = \sqrt{64} \implies 8=8$, valid.
For $x=-2$: $-2+1 = \sqrt{7*(-2)+15} \implies -1 = \sqrt{1} \implies -1=1$, invalid (extraneous).

Answer:

C. The solution is x = 7, and the extraneous solution is -2