Difference between revisions of "2002 AMC 10A Problems/Problem 10"

(New page: == Problem == What is the sum of all of the roots of <math>(2x + 3) (x - 4) + (2x + 3) (x - 6) = 0</math>? <math>\text{(A)}\ 7/2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D...)
 
(Solution)
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==Solution==
 
==Solution==
We expand to get <math>2x^2-8x+3x-12+2x^2-12x+3x-18=0</math> which is <math>4x^2-14x-30=0</math> after combining like terms. Using the quadratic part of Vieta's Formulas, we find the sum of the roots is <math>\boxed{\text{(A)}\ 7/2}</math>.
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We expand to get <math>2x^2-8x+3x-12+2x^2-12x+3x-18=0</math> which is <math>4x^2-14x-30=0</math> after combining like terms. Using the quadratic part of Vieta's Formulas, we find the sum of the roots is <math>\boxed{\text{(A)}\ 7/2}</math>. We could also have factored, which would give a faster solution.

Revision as of 22:32, 26 December 2008

Problem

What is the sum of all of the roots of $(2x + 3) (x - 4) + (2x + 3) (x - 6) = 0$?

$\text{(A)}\ 7/2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 13$

Solution

We expand to get $2x^2-8x+3x-12+2x^2-12x+3x-18=0$ which is $4x^2-14x-30=0$ after combining like terms. Using the quadratic part of Vieta's Formulas, we find the sum of the roots is $\boxed{\text{(A)}\ 7/2}$. We could also have factored, which would give a faster solution.