2002 AMC 12A Problem 1 — Solution | LIVE by Po-Shen Loh

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2002 AMC 12A Problem 1

Attempt Problem 1 of the 2002 AMC 12A 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 2002 AMC 12A solutions, or check the answer key.

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

1.

Compute the sum of all the roots of $\begin{aligned} &(2x+3)(x-4) \\ &\quad {}+(2x+3)(x-6)=0. \end{aligned}$

$\dfrac{7}{2}$

$4$

$5$

$7$

$13$

Answer: A

Concepts:factoring Vieta’s Formulas

Difficulty rating: 890

Solution:

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

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

Thus, the correct answer is A.

Problem 2#2 →

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