Answer
$\{3\}$
Work Step by Step
Square both sides.
$\left(\sqrt{15-2x}\right)^2=(x)^2$
$15-2x=x^2$
Add $2x-15$ to both sides.
$15-2x+2x-15=x^2+2x-15$
Simplify.
$0=x^2+2x-15$
Rewrite $2x$ as $5x-3x$.
$0=x^2+5x-3x-15$
Group the terms.
$0=(x^2+5x)+(-3x-15)$
Factor out $x$ in the first group and $-3$ in the second group.
$0=x(x+5)-3(x+5)$
Factor out $(x+5)$.
$0=(x+5)(x-3)$
Use Zero-Product Property which states that if $ab=0$, then either $a=0$, $b=0$, or both are $0$..
$x+5=0\quad $ or $\quad x-3=0$
Solve for $x$ in each:equation:
$x=-5\quad $ or $\quad x=3$
Check:
For $x=-5$:
$\sqrt{15-2(-5)}=-5$
$\sqrt{15+10}=-5$
$\sqrt{25}=-5$
$\sqrt{5^2}=-5$
$5=-5$ False. This means $-5$ is an extraneous solution.
For $x=3$:
$\sqrt{15-2(3)}=3$
$\sqrt{15-6}=3$
$\sqrt{9}=3$
$\sqrt{3^2}=3$
$3=3$ True.
Hence, the solution set is $\{3\}$.
