Competition problem — Algebra — Maths Olympiad Prep

Library / /206 of 520

Algebra Difficulty 2.8 Junior Find the answer

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:
$\textbf{(A)}\ \frac{B^2 - 4AC}{A^2}$ $\textbf{(B)}\ \frac{B^2 - 2AC}{A^2}$ $\textbf{(C)}\ \frac{2AC - B^2}{A^2}$ $\\ \textbf{(D)}\ B^2 - 2C$ $\textbf{(E)}\ 2C - B^2$

Your answer

Multiple choice: answer with the letter of the option you want.

Solution

By Vieta's, $r + s = -\frac{B}{A}$ , $rs = \frac{C}{A}$ , and $r^2 + s^2 = -p$ . Note that $(r+s)^2 = r^2 + s^2 + 2rs$ .
Therefore, $(r + s)^2 - 2rs = r^2 + s^2$ , or $-\left(\frac{B}{A}\right)^2 - \frac{2C}{A} = -p$ .
Simplifying, $\frac{B^2 - 2CA}{A^2} = -p$ .
Finally, multiply both sides by $-1$ to get $p = \frac{2CA - B^2}{A^2}$ , making the answer $\fbox{C}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.