1953 AHSME Problems/Problem 23

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

$\textbf{(A)}$ an extraneous root between $-5$ and $-1$

$\textbf{(B)}$ an extraneous root between $-10$ and $-6$

$\textbf{(C)}$ a true root between $20$ and $25$

$\textbf{(D)}$ two true roots

$\textbf{(E)}$ two extraneous roots

We multiply both sides by $\sqrt{x+10}$ to get the equation $x+4=5\sqrt{x+10}$ . We square both sides to get $x^2+8x+16=25x+250$ , or $x^2-17x-234=0$ . We can factor the quadratic as $(x+9)(x-26)=0$ , giving us roots of $-9$ and $26$ . We plug these values in to find that $-9$ is an extraneous root and that $26$ is a true root, giving an answer of $\boxed{B}$ .

Page

Toolbox

Search

1953 AHSME Problems/Problem 23