1955 AMC 12 Problem 39

Question

If

y=x^2+px+q,

then if the least possible value of

y

is zero,

q

is equal to:

0

\dfrac{p^2}{4}

\dfrac p2

-\dfrac p2

\dfrac{p^2}{4}-q

Complete the square in

x^2+px+q

The minimum occurs when

x=-\frac{p}{2}

Completing the square,

y=\left(x+\frac p2\right)^2+q-\frac{p^2}{4}.

Its least value is

q-\frac{p^2}{4}.

Setting this equal to zero gives

q=\frac{p^2}{4}.

Thus, the correct answer is B.

Problem 39 in other years