# 1955 AHSME Problems/Problem 20

## Problem 20

The expression $\sqrt{25-t^2}+5$ equals zero for:

$\textbf{(A)}\ \text{no real or imaginary values of }t\qquad\textbf{(B)}\ \text{no real values of }t\text{ only}\\ \textbf{(C)}\ \text{no imaginary values of }t\text{ only}\qquad\textbf{(D)}\ t=0\qquad\textbf{(E)}\ t=\pm 5$

## Solution

We can make the following equation: $\sqrt{25-t^2}+5 = 0$, which means that $\sqrt{25-t^2} = -5$ Square both sides, and we get $25 - t^2 = 25$, which would mean that $t = 0$, but that is an extraneous solution and doesn't work in the original equation.

Therefore, the correct answer is $\textbf{(A)}$

Note: When talking about square roots, it generally means the positive square root, or the "principal root."

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