1970 AMC 12 Problem 6

Attempt Problem 6 of the 1970 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1970 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

The smallest value of

x^2+8x

for real values of

x

is:

-16.25

-16

-15

-8

None of these

Complete the square

Rewrite the expression as

(x+4)^2-16

Completing the square gives

x^2+8x=(x+4)^2-16.

Since

(x+4)^2\ge0,

the minimum occurs at

x=-4

and equals

-16.

Therefore, the correct answer is B.

Problem 6 in other years