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-1

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

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

By Vieta's Formula, $r+s=-\frac{b}{a}$ 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

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

Solution-2

here $r$ and $s$ are the roots of the equation $ax^2+bx+c=0$,
now we can convert this equation with roots $r^2$ and $s^2$.
let, $y=x^2$ then above equation becomes
\[ay+c=-b\sqrt y\] square on both sides we get \[a^2y^2 +(2ac-b^2)y +c^2 =0\]
Again we can change this equation with roots $\frac{1}{r^2}$ and $\frac{1}{s^2}$ .

let $z=\frac{1}{y}$ then, $a^2\frac{1}{z^2}+\frac{2ac-b^2}{z}+c^2=0$ then

       \[c^2z^2+z(2ac-b^2)+a^2=0\]

then the sum of roots of the above equation is $\frac{1}{r^2}+\frac{1}{s^2}=\frac{b^2-2ac}{c^2}$

hence, \[\frac{1}{r^2}+\frac{1}{s^2}=\boxed{\frac{b^2-2ca}{c^2}\textbf{(D)}}\]

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 46
Followed by
Problem 48
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