Question

solving the equation $x = \sqrt{2x + 15}$ algebraically results to the solution set $\\{-3, 5\\}$

• $-3$ is an extraneous solution because it does not make the equation true
• $-3$ is an extraneous solution because it does make the equation true
• $5$ is an extraneous solution because it does make the equation true
• $5$ is an extraneous solution because it does not make the equation true

Explanation:

Step1: Test $x=-3$ in the equation

Substitute $x=-3$ into $x=\sqrt{2x+15}$:
Left side: $-3$
Right side: $\sqrt{2(-3)+15}=\sqrt{-6+15}=\sqrt{9}=3$
$-3
eq 3$, so $x=-3$ is extraneous.

Step2: Test $x=5$ in the equation

Substitute $x=5$ into $x=\sqrt{2x+15}$:
Left side: $5$
Right side: $\sqrt{2(5)+15}=\sqrt{10+15}=\sqrt{25}=5$
$5=5$, so $x=5$ is valid.

Answer:

-3 is an extraneous solution because it does not make the equation true