Difference between revisions of "1958 AHSME Problems/Problem 41"

Revision as of 14:10, 22 February 2018

Problem

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:

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

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 $-\left(\frac{B}{A}\right)^2 - 2\frac{C}{A} = -p$ .

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 $\fbox{C}$ .

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 40	Followed by Problem 42
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 Simplifying, <math>\frac{B^2 - 2CA}{A^2} = -p</math>.
-Finally, multiply both sides by <math>p = \frac{2CA - B^2}{A^2}</math>, making the answer <math>\fbox{C}</math>.
+Finally, multiply both sides by <math>-1</math> to get <math>p = \frac{2CA - B^2}{A^2}</math>, making the answer <math>\fbox{C}</math>.
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Difference between revisions of "1958 AHSME Problems/Problem 41"

Revision as of 14:10, 22 February 2018

Problem

Solution

See Also