1958 AMC 12 Problem 30

Attempt Problem 30 of the 1958 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 1958 AMC 12 solutions, or check the answer key.

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30.

If xy=bxy=b and 1x2+1y2=a,\dfrac1{x^2}+\dfrac1{y^2}=a, then (x+y)2(x+y)^2 equals:
(a+2b)2(a+2b)^2
a2+b2a^2+b^2
b(ab+2)b(ab+2)
ab(b+2)ab(b+2)
1a+2b\dfrac1a+2b
Answer: C
Concepts:rational equationsymmetry (algebra)algebraic manipulation
Small hint:
Combine 1x2+1y2\frac{1}{x^2}+\frac{1}{y^2} using the common denominator x2y2x^2y^2
Big hint:
Use xy=bxy=b to find x2+y2x^2+y^2, then add 2xy2xy
Solution:
Since xy=b,xy=b, a=x2+y2x2y2=x2+y2b2, a=\frac{x^2+y^2}{x^2y^2} =\frac{x^2+y^2}{b^2}, so x2+y2=ab2.x^2+y^2=ab^2. Therefore (x+y)2=x2+y2+2xy=ab2+2b=b(ab+2). \begin{aligned} (x+y)^2 &=x^2+y^2+2xy\\ &=ab^2+2b\\ &=b(ab+2). \end{aligned}
Therefore, the correct answer is C.
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