2002 AMC 12A Problem 1

Below is the professionally curated solution for Problem 1 of the 2002 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

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Concepts:factoringVieta’s Formulas

Difficulty rating: 890

1.

Compute the sum of all the roots of (2x+3)(x4)+(2x+3)(x6)=0.(2x+3)(x-4)+(2x+3)(x-6)=0.

72\dfrac{7}{2}

44

55

77

1313

Solution:

Factoring out 2x+32x+3 gives (2x+3)[(x4)+(x6)]=(2x+3)(2x10)=0.(2x+3)\big[(x-4)+(x-6)\big] = (2x+3)(2x-10) = 0.

The roots are 32-\dfrac{3}{2} and 5,5, whose sum is 32+5=72.-\dfrac{3}{2} + 5 = \dfrac{7}{2}.

Thus, the correct answer is A.

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