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.)

See Also

1962 AHSC (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png