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
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Page

Toolbox

Search

1962 AHSME Problems/Problem 11

Problem

Solution

See Also