If a and b are positive real numbers and each of the equations x^2+ax+2b&=0, x^2+2bx+a&=0 has real roots, then the smallest possible value of a+b is

1984 AMC 12 Problem 24

Attempt Problem 24 of the 1984 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 1984 AMC 12 solutions, or check the answer key.

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

24.

a

and

b

are positive real numbers and each of the equations

\begin{aligned} x^2+ax+2b&=0,\\ x^2+2bx+a&=0 \end{aligned}

has real roots, then the smallest possible value of

a+b

2

3

4

5

6

Require both quadratic discriminants to be nonnegative

Combine

a^2\ge8b

and

b^2\ge a

to bound

a

and

b

The two discriminants give

a^2\ge8b

and

b^2\ge a.

Therefore

a^4\ge64b^2\ge64a.

Since

a\gt0,

this yields

a\ge4,

and then

b^2\ge a\ge4

gives

b\ge2.

Thus

a+b\ge6.

Equality occurs at

a=4,b=2,

for which both discriminants are zero.

Therefore, the correct answer is E.