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Difference between revisions of "1962 AHSME Problems/Problem 11"
(→‎Solution)
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Subtracting <math>p^2-1=4rs</math> from both sides gives <math>1=r^2-2rs+s^2</math>.
 
Subtracting <math>p^2-1=4rs</math> from both sides gives <math>1=r^2-2rs+s^2</math>.
 
Taking square roots, <math>r-s=1 \Rightarrow \boxed{\textbf{(B)}}</math>.
 
Taking square roots, <math>r-s=1 \Rightarrow \boxed{\textbf{(B)}}</math>.
 +
(Another solution is to use the quadratic formula and see that the
 +
roots are <math>\frac{p\pm 1}2</math>, and their difference is 1.)

Revision as of 12:02, 16 April 2014

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.)

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