Question

The roots of $Ax^2+Bx+C=0$ are $r$ and $s.$ For the roots of

$x^2+px+q=0$

to be $r^2$ and $s^2,$ $p$ must equal:

$\dfrac{B^2-4AC}{A^2}$

$\dfrac{B^2-2AC}{A^2}$

$\dfrac{2AC-B^2}{A^2}$

$B^2-2C$

$2C-B^2$

Answer: C

Concepts:Vieta’s Formulas quadratic symmetry (algebra)

Difficulty rating: 1590

Small Hint:

For the new monic quadratic, $p$ is the negative of the sum $r^2+s^2$

Big Hint:

Write $r^2+s^2=(r+s)^2-2rs$ and use Vieta’s formulas for the original quadratic

Solution:

For the original quadratic, $r+s=-\frac BA,\qquad rs=\frac CA.$ Hence $\begin{aligned} r^2+s^2 &=(r+s)^2-2rs\\ &=\frac{B^2-2AC}{A^2}. \end{aligned}$ The coefficient $p$ is the negative of this sum, so $p=\frac{2AC-B^2}{A^2}.$

Therefore, the correct answer is C.

1958 AMC 12 Problem 41

Answer: C

Difficulty rating: 1590

Small Hint:

Big Hint:

Solution:

Problem 41 in Other Years