1953 AHSME Problems/Problem 23

Problem

The equation $\sqrt {x + 10} - \frac {6}{\sqrt {x + 10}} = 5$ has:

$\textbf{(A)}\ \text{an extraneous root between } - 5\text{ and } - 1 \\ \textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6\\ \textbf{(C)}\ \text{a true root between }20\text{ and }25\qquad \textbf{(D)}\ \text{two true roots}\\ \textbf{(E)}\ \text{two extraneous roots}$

Solution

We can multiply both sides by $\sqrt{x+10}$ to get $x+4=5\sqrt{x+10}$ . We can now square both sides to get $x^2+8x+16=25x+250$ , which yields $x^2-17x-234=0$ . We can factor the quadratic as $(x+9)(x-26)=0$ , giving us roots of $-9$ and $26$ . Plugging in these values, we find that $-9$ is an extraneous root and $26$ is a true root. This gives us the final answer of $\boxed{\textbf{(B)}\ \text{an extraneous root between }-10\text{ and }-6}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1953 AHSME Problems/Problem 23

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