1971 AHSME Problems/Problem 20

Problem

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

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

Solution 1

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 $\boxed{\textbf{(E) }\text{None of these}}.$

-edited by coolmath34

Solution 2

The given equation can be rewritten as $x^2+2hx-3=0$ . By Vieta's Formulas, we know that the sum of the roots is $-2h$ . Thus, by Newton Sums, we have the following equation: \begin{align*} 10+(2h)(-2h)+2(-3) &= 0 \\ 10-4h^2-6 &= 0 \\ -4h^2 &= -4 \\ h^2 &= 1 \\ |h| &= 1 \\ \end{align*} Thus, our answer is $\boxed{\textbf{(E) }\text{None of these}}$ .

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1971 AHSME Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

See Also