1953 AMC 12 Problem 44

Attempt Problem 44 of the 1953 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 1953 AMC 12 solutions, or check the answer key.

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

44.

In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and obtains 88 and 22 for the roots. Another student makes a mistake only in the coefficient of the first degree term and finds −9-9 and −1-1 for the roots. The correct equation was:
x2−10x+9=0x^2-10x+9=0
x2+10x+9=0x^2+10x+9=0
x2−10x+16=0x^2-10x+16=0
x2−8x−9=0x^2-8x-9=0
none of these
Answer: A
Concepts:quadratic rootsVieta formulascoefficients
Small hint:
The first student’s correct linear coefficient is determined by the sum 8+28+2
Big hint:
The second student’s correct constant term is determined by the product (−9)(−1)(-9)(-1)
Solution:
The first student changed only the constant term, so the correct coefficient of xx is the one in the monic quadratic with roots 88 and 2:2: it is −(8+2)=−10.-(8+2)=-10. The second student changed only that linear coefficient, so the correct constant is (−9)(−1)=9.(-9)(-1)=9. Therefore the correct equation is x2−10x+9=0. x^2-10x+9=0.
Thus, the correct answer is A.
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