2002 AMC 10A Problems/Problem 10

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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

Solution 1

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 $\boxed{\text{(A)}\ 7/2}$.

Solution 2

Combine terms to get $(2x+3)\cdot\Big( (x-4)+(x-6) \Big) = (2x+3)(2x-10)=0$, hence the roots are $-\frac{3}{2}$ and $5$, thus our answer is $-\frac{3}{2}+5=\boxed{\text{(A)}\ 7/2}$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AMC 10 Problems and Solutions