Question

eva 25 - 26: algebra ii 03
solving equations containing two radicals
solving equations containing two radicals
which statement about $sqrt{x-3}-sqrt{x}=3$ is true?
$x = 4$ is an extraneous solution.
$x = 4$ is a true solution.
$x = -1$ is a true solution.
$x = -1$ is an extraneous solution.

Explanation:

Step1: Isolate one radical

$\sqrt{x-3} = 3 + \sqrt{x}$

Step2: Square both sides

$(\sqrt{x-3})^2 = (3 + \sqrt{x})^2$
$x - 3 = 9 + 6\sqrt{x} + x$

Step3: Simplify to isolate remaining radical

$x - 3 - x - 9 = 6\sqrt{x}$
$-12 = 6\sqrt{x}$
$\sqrt{x} = -2$

Step4: Analyze the radical

A square root $\sqrt{x}$ cannot equal a negative number, so there are no true solutions. Now test $x=4$:
Substitute $x=4$ into original equation:
$\sqrt{4-3} - \sqrt{4} = 1 - 2 = -1
eq 3$
So $x=4$ is extraneous. Testing $x=-1$ is invalid since $\sqrt{-1}$ is not a real number.

Answer:

$x = 4$ is an extraneous solution.