Question

If $y=x^2+px+q,$ then if the least possible value of $y$ is zero, $q$ is equal to:

$0$

$\dfrac{p^2}{4}$

$\dfrac p2$

$-\dfrac p2$

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

Answer: B

Concepts:quadratic minimum completing the square vertex

Difficulty rating: 1260

Small Hint:

Complete the square in $x^2+px+q$

Big Hint:

The minimum occurs when $x=-\frac{p}{2}$

Solution:

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.

1955 AMC 12 Problem 39

Answer: B

Difficulty rating: 1260

Small Hint:

Big Hint:

Solution:

Problem 39 in Other Years