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

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==Solution==
 
==Solution==
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{{solution}}
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Call the two roots <math>r</math> and <math>s</math>, with <math>r \ge s</math>.
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By Vieta's formulas, <math>p=r+s</math> and <math>(p^2-1)/4=rs.</math>
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(Multiplying both sides of the second equation by 4 gives <math>p^2-1=4rs</math>.)
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The value we need to find, then, is <math>r-s</math>.
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Since <math>p=r+s</math>, <math>p^2=r^2+2rs+s^2</math>.
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Subtracting <math>p^2-1=4rs</math> from both sides gives <math>1=r^2-2rs+s^2</math>.
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Taking square roots, <math>r-s=1 \Rightarrow \boxed{\textbf{(B)}}</math>.
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(Another solution is to use the quadratic formula and see that the
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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 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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