1962 AMC 12 Problem 11

Attempt Problem 11 of the 1962 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 1962 AMC 12 solutions, or check the answer key.

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

11.

The difference between the larger root and the smaller root of x2−px+p2−14=0 x^2-px+\frac{p^2-1}{4}=0 is:
00
11
22
pp
p+1p+1
Answer: B
Concepts:quadraticalgebraic manipulation
Small hint:
Compute the discriminant of the quadratic
Big hint:
The two roots are p±Δ2\frac{p\pm\sqrt{\Delta}}{2}
Solution:
The discriminant is p2−4(p2−14)=1. p^2-4\left(\frac{p^2-1}{4}\right)=1. The roots are p+12\frac{p+1}{2} and p−12,\frac{p-1}{2}, whose difference is 1.1.
Therefore, the correct answer is B.
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