Difference between revisions of "1962 AHSME Problems/Problem 11"

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(Another solution is to use the quadratic formula and see that the
 
(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.)
 
roots are <math>\frac{p\pm 1}2</math>, and their difference is 1.)
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==See Also==
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{{AHSME 40p box|year=1962|before=Problem 10|num-a=12}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:15, 3 October 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.)

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AHSME Problems and Solutions

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