Question

select the correct answer.
which equation has an extraneous solution?
a. (sqrt5{x} = -2)
b. (sqrt3{x + 1} = 10)
c. (sqrt{x} = -5)
d. (sqrt4{x - 5} = 3)

Explanation:

Step1: Recall extraneous solution concept

An extraneous solution is a solution that emerges during solving but doesn't satisfy the original equation. For radical equations, we consider the domain (real - valued outputs of radicals):

• For an even - indexed radical (like square root, fourth root, sixth root, etc.), the expression under the radical (radicand) must be non - negative, and the value of the radical itself (the output) is non - negative.
• For an odd - indexed radical (like cube root, fifth root, etc.), the radicand can be any real number, and the radical can take any real value (positive, negative, or zero).

Step2: Analyze Option A

The equation is \(\sqrt[5]{x}=-2\). The fifth root is an odd - indexed radical. To solve for \(x\), we raise both sides to the 5th power: \(x = (-2)^5=-32\).
Now, substitute \(x = - 32\) back into the original equation: \(\sqrt[5]{-32}=\sqrt[5]{(-2)^5}=-2\), which matches the right - hand side. So, this solution is valid.

Step3: Analyze Option B

The equation is \(\sqrt[3]{x + 1}=10\). The cube root is an odd - indexed radical. Raise both sides to the 3rd power: \(x + 1=10^3 = 1000\), so \(x=999\).
Substitute \(x = 999\) back into the original equation: \(\sqrt[3]{999 + 1}=\sqrt[3]{1000}=10\), which matches the right - hand side. So, this solution is valid.

Step4: Analyze Option C

The equation is \(\sqrt{x}=-5\). The square root is an even - indexed radical. By the definition of the real - valued square root, \(\sqrt{x}\geq0\) for all real \(x\) in its domain (\(x\geq0\)).
If we try to solve it, we square both sides: \(x = (-5)^2 = 25\).
Now, substitute \(x = 25\) back into the original equation: \(\sqrt{25}=5
eq - 5\). So, \(x = 25\) is an extraneous solution.

Step5: Analyze Option D

The equation is \(\sqrt[4]{x - 5}=3\). The fourth root is an even - indexed radical. The radicand \(x - 5\geq0\) (so \(x\geq5\)) and the value of the fourth root is non - negative.
Raise both sides to the 4th power: \(x-5 = 3^4=81\), so \(x=81 + 5=86\).
Substitute \(x = 86\) back into the original equation: \(\sqrt[4]{86 - 5}=\sqrt[4]{81}=\sqrt[4]{3^4}=3\), which matches the right - hand side. So, this solution is valid.

Answer:

C. \(\sqrt{x}=-5\)