2002 AMC 12A Problems/Problem 1

The following problem is from both the 2002 AMC 12A #1 and 2002 AMC 10A #10, so both problems redirect to this page.

Problem

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

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

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 $\frac{14}4 = \boxed{\text{(A)}\ 7/2}$.

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 12A (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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