1971 AHSME Problems/Problem 20


The sum of the squares of the roots of the equation $x^2+2hx=3$ is $10$. The absolute value of $h$ is equal to

$\textbf{(A) }-1\qquad \textbf{(B) }\textstyle\frac{1}{2}\qquad \textbf{(C) }\textstyle\frac{3}{2}\qquad \textbf{(D) }2\qquad \textbf{(E) }\text{None of these}$


We can rewrite the equation as $x^2 + 2hx - 3 = 0.$ By Vieta's Formulas, the sum of the roots is $-2h$ and the product of the roots is $-3.$

Let the two roots be $r$ and $s.$ Note that \[r^2 + s^2 = (r+s)^2 - 2rs = (-2h)^2 -2(-3)\]

Therefore, $4h^2 + 6 = 10$ and $h = \pm 1.$ This doesn't match any of the answer choices, so the answer is $\textbf{(E)}.$

-edited by coolmath34