1951 AHSME Problems/Problem 47

Problem

If $r$ and $s$ are the roots of the equation $ax^2+bx+c=0$ , the value of $\frac{1}{r^{2}}+\frac{1}{s^{2}}$ is:

$\textbf{(A)}\ b^{2}-4ac\qquad\textbf{(B)}\ \frac{b^{2}-4ac}{2a}\qquad\textbf{(C)}\ \frac{b^{2}-4ac}{c^{2}}\qquad\textbf{(D)}\ \frac{b^{2}-2ac}{c^{2}}$ $\textbf{(E)}\ \text{none of these}$

Solution

$r$ and $s$ can be found in terms of $a$ , $b$ , and $c$ by using the quadratic formula; the roots are

$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

It's not hard to check that $r+s=-\frac{b}{a}$ and $rs=\frac{c}{a}$ . Now let's algebraically manipulate what we want to find:

$\frac{1}{r^2}+\frac{1}{s^2}=\frac{r^2+s^2}{r^2s^2}=\frac{(r+s)^2-2rs}{(rs)^2}$

Plugging in the values for $r+s$ and $rs$ gives

$\frac{1}{r^2}+\frac{1}{s^2}=\frac{(-b/a)^2-2(c/a)}{(c/a)^2}=\boxed{\frac{b^2-2ca}{c^2}\textbf{(D)}}$

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 46	Followed by Problem 48
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1951 AHSME Problems/Problem 47

Problem

Solution

See Also