which statement about $sqrt{x - 5}-sqrt{x}=5$ is true?\n$x = - 3$ is a true solution.\n$x = - 3$ is an…

which statement about $sqrt{x - 5}-sqrt{x}=5$ is true?\n$x = - 3$ is a true solution.\n$x = - 3$ is an extraneous solution.\n$x = 9$ is a true solution.\n$x = 9$ is an extraneous solution.

which statement about $sqrt{x - 5}-sqrt{x}=5$ is true?\n$x = - 3$ is a true solution.\n$x = - 3$ is an extraneous solution.\n$x = 9$ is a true solution.\n$x = 9$ is an extraneous solution.

Answer

Answer:

x = 9 is an extraneous solution.

Explanation:

Step1: Isolate one square - root term

(\sqrt{x - 5}=5+\sqrt{x})

Step2: Square both sides

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

Step3: Simplify the equation

(x - 5 - x=25 + 10\sqrt{x}+x - x) (-5=25 + 10\sqrt{x}) (10\sqrt{x}=-30) (\sqrt{x}=-3) (not possible in the real - number system as the square root of a non - negative number is non - negative) Let's check the proposed solutions: For (x=-3), in the original equation (\sqrt{-3 - 5}-\sqrt{-3}), we have square roots of negative numbers, which are not real numbers in the real - number system. For (x = 9), the left - hand side of the original equation is (\sqrt{9 - 5}-\sqrt{9}=\sqrt{4}-\sqrt{9}=2 - 3=-1\neq5). So (x = 9) is an extraneous solution.