1962 AHSME Problems/Problem 11

Problem

The difference between the larger root and the smaller root of $x^2 - px + (p^2 - 1)/4 = 0$ is:

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ p\qquad\textbf{(E)}\ p+1$

Solution

Call the two roots $r$ and $s$ , with $r \ge s$ . By Vieta's formulas, $p=r+s$ and $(p^2-1)/4=rs.$ (Multiplying both sides of the second equation by 4 gives $p^2-1=4rs$ .) The value we need to find, then, is $r-s$ . Since $p=r+s$ , $p^2=r^2+2rs+s^2$ . Subtracting $p^2-1=4rs$ from both sides gives $1=r^2-2rs+s^2$ . Taking square roots, $r-s=1 \Rightarrow \boxed{\textbf{(B)}}$ . (Another solution is to use the quadratic formula and see that the roots are $\frac{p\pm 1}2$ , and their difference is 1.)

1962 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1962 AHSME Problems/Problem 11

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