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Difference between revisions of "1955 AHSME Problems/Problem 20"

(Created page with "== Problem 20== The expression <math>\sqrt{25-t^2}+5</math> equals zero for: <math> \textbf{(A)}\ \text{no real or imaginary values of }t\qquad\textbf{(B)}\ \text{no real v...")
 
 
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Note: When talking about square roots, it generally means the positive square root, or the "principal root."
 
Note: When talking about square roots, it generally means the positive square root, or the "principal root."
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==See Also==
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{{AHSME 50p box|year=1955|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 11:38, 3 August 2020

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."

See Also

1955 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AHSME Problems and Solutions


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