Difference between revisions of "2002 AMC 12A Problems/Problem 1"
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<math> \mathrm{(A) \ } \frac{7}{2}\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 13 </math> | <math> \mathrm{(A) \ } \frac{7}{2}\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 13 </math> | ||
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− | + | ==Solution 1== | |
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>. | 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>. | ||
− | + | ==Solution 2== | |
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>. | 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>. | ||
Revision as of 19:12, 1 May 2021
- The following problem is from both the 2002 AMC 12A #1 and 2002 AMC 10A #10, so both problems redirect to this page.
Contents
[hide]Problem
Compute the sum of all the roots of
Solution 1
We expand to get which is after combining like terms. Using the quadratic part of Vieta's Formulas, we find the sum of the roots is .
Solution 2
Combine terms to get , hence the roots are and , thus our answer is .
See also
2002 AMC 12A (Problems • Answer Key • Resources) | |
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 (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.