Question

solve the radical equation.
\sqrt{25 - 40x} = x - 10
which is the true solution?
both $-5$ and $-15$ are true solutions.
there are no true solutions.
$-15$
$-5$

Explanation:

Step1: Square both sides to eliminate radical

$$(\sqrt{25-40x})^2=(x-10)^2$$
$$25-40x=x^2-20x+100$$

Step2: Rearrange to standard quadratic form

$$x^2-20x+100-25+40x=0$$
$$x^2+20x+75=0$$

Step3: Factor the quadratic equation

$$(x+5)(x+15)=0$$
$$x=-5 \text{ or } x=-15$$

Step4: Check solutions in original equation

For $x=-5$:
Left side: $\sqrt{25-40(-5)}=\sqrt{25+200}=\sqrt{225}=15$
Right side: $-5-10=-15$
$15
eq -15$, so $x=-5$ is extraneous.

For $x=-15$:
Left side: $\sqrt{25-40(-15)}=\sqrt{25+600}=\sqrt{625}=25$
Right side: $-15-10=-25$
$25
eq -25$, so $x=-15$ is extraneous.

Answer:

There are no true solutions.