1966 AMC 12 Problem 23

Attempt Problem 23 of the 1966 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 1966 AMC 12 solutions, or check the answer key.

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

23.

x

is real and

4y^2+4xy+x+6=0,

then the complete set of values of

x

for which

y

is real, is:

x\leq-2

x\geq3

x\leq2

x\geq3

x\leq-3

x\geq2

-3\leq x\leq2

-2\leq x\leq3

Treat the equation as a quadratic in

y

Require its discriminant

16(x-3)(x+2)

to be nonnegative

As a quadratic in

y,

the equation has discriminant

\begin{aligned} &(4x)^2-16(x+6)\\ &\qquad=16(x-3)(x+2). \end{aligned}

This is nonnegative exactly when

x\leq-2

x\geq3.

Therefore, the correct answer is A.

Problem 23 in other years