Question

If $r$ and $s$ are the roots of the equation $ax^2+bx+c=0,$ the value of $\dfrac1{r^2}+\dfrac1{s^2}$ is:

$b^2-4ac$

$\dfrac{b^2-4ac}{2a}$

$\dfrac{b^2-4ac}{c^2}$

$\dfrac{b^2-2ac}{c^2}$

None of these

Answer: D

Concepts:Vieta’s Formulas algebraic manipulation

Difficulty rating: 1530

Small Hint:

Write the expression over the common denominator $r^2s^2$

Big Hint:

Use $r+s=-\frac{b}{a}$ and $rs=\frac{c}{a}$

Solution:

By Vieta’s formulas, $r+s=-\frac ba,\qquad rs=\frac ca.$ Therefore $\begin{aligned} \frac1{r^2}+\frac1{s^2} &=\frac{(r+s)^2-2rs}{(rs)^2}\\ &=\frac{b^2-2ac}{c^2}. \end{aligned}$

Thus, the correct answer is D.

1951 AMC 12 Problem 47

Answer: D

Difficulty rating: 1530

Small Hint:

Big Hint:

Solution:

Problem 47 in Other Years