1958 AMC 12 Problem 41

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

For the new monic quadratic,

p

is the negative of the sum

r^2+s^2

Write

r^2+s^2=(r+s)^2-2rs

and use Vieta’s formulas for the original quadratic

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.

Problem 41 in other years