Difference between revisions of "1951 AHSME Problems/Problem 47"

Revision as of 11:27, 5 July 2013

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
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All AHSME Problems and Solutions

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Difference between revisions of "1951 AHSME Problems/Problem 47"

Revision as of 11:27, 5 July 2013

Problem

Solution

See Also