1976 AMC 12 Problem 6

Attempt Problem 6 of the 1976 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 1976 AMC 12 solutions, or check the answer key.

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c

is a real number and the negative of one of the solutions of

x^2-3x+c=0

is a solution of

x^2+3x-c=0,

then the solutions of

x^2-3x+c=0

are

1,

2

-1,

-2

0,

3

0,

-3

\frac32,

\frac32

Call the first root

r

and substitute

-r

into the second equation

The two resulting equations differ only in the sign of

c

For such a root

r,

\begin{aligned} r^2-3r+c&=0,\\ (-r)^2+3(-r)-c&=0. \end{aligned}

Subtracting gives

2c=0.

Thus the first equation is

x(x-3)=0,

with roots

0

and

3.

Therefore, the correct answer is C.

Problem 6 in other years