Question

cubic and cube root functions and equations
which statement about $sqrt{x-5}-sqrt{x}=5$ is true?
$x = 9$ is a true solution.
$x = 9$ is an extraneous solution.
$x = -3$ is an extraneous solution.
$x = -3$ is a true solution.

Explanation:

Step1: Isolate the radical term

$\sqrt{x - 3} = 5 + \sqrt{x}$

Step2: Square both sides

$(\sqrt{x - 3})^2 = (5 + \sqrt{x})^2$
$x - 3 = 25 + 10\sqrt{x} + x$

Step3: Simplify to isolate radical

$-3 - 25 = 10\sqrt{x}$
$-28 = 10\sqrt{x}$

Step4: Analyze equation validity

$\sqrt{x} = -\frac{28}{10} = -2.8$
Since the square root of a real number cannot be negative, there is no real solution. Now test $x=9$:
Left side: $\sqrt{9-3}-\sqrt{9}=\sqrt{6}-3\approx2.45-3=-0.55
eq5$
So $x=9$ is extraneous. Test $x=-3$: Square roots of negative numbers are not real, so it is not a valid solution.

Answer:

$x = 9$ is an extraneous solution.