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| − | == Problem ==
| + | #redirect [[2002 AMC 12A Problems/Problem 1]] |
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| − | What is the sum of all of the roots of <math>(2x + 3) (x - 4) + (2x + 3) (x - 6) = 0</math>?
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| − | <math>\text{(A)}\ 7/2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 13</math>
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| − | ==Solution==
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| − | ===Solution 1===
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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>\frac{14}4 = \boxed{\text{(A)}\ 7/2}</math>.
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| − | ===Solution 2===
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| − | Combine terms to get <math>(2x+3)\cdot\Big( (x-4)+(x-6) \Big) = (2x+3)(2x-10)=0</math>, hence the roots are <math>-\frac{3}{2}</math> and <math>5</math>, thus our answer is <math>-\frac{3}{2}+5=\boxed{\text{(A)}\ 7/2}</math>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2002|ab=A|num-b=9|num-a=11}}
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| − | [[Category:Introductory Algebra Problems]]
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