1970 AMC 12 Problem 14

Attempt Problem 14 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.

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14.

Consider

x^2+px+q=0,

where

p

and

q

are positive numbers. If the roots of this equation differ by

1,

then

p

equals:

\sqrt{4q+1}

q-1

-\sqrt{4q+1}

q+1

\sqrt{4q-1}

Use the quadratic formula to express the difference of the roots

The difference has magnitude

\sqrt{p^2-4q}

The two roots differ in absolute value by

\sqrt{p^2-4q}.

Hence

p^2-4q=1,

p^2=4q+1.

Since

p

is positive,

p=\sqrt{4q+1}.

Therefore, the correct answer is A.

Problem 14 in other years