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2002 AMC 12A Problems/Problem 1

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

Contents

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 (ProblemsResources)
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
2002 AMC 10A (ProblemsResources)
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
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