Difference between revisions of "2002 AMC 12A Problems/Problem 1"
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| + | {{duplicate|[[2002 AMC 12A Problems|2009 AMC 12A #1]] and [[2002 AMC 10A Problems|2009 AMC 10A #10]]}} | ||
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== Problem == | == Problem == | ||
Compute the sum of all the roots of | Compute the sum of all the roots of | ||
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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> | ||
== Solution == | == Solution == | ||
| − | + | ===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>. | |
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| − | <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>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=A|before=First Question|num-a=2}} | {{AMC12 box|year=2002|ab=A|before=First Question|num-a=2}} | ||
| + | {{AMC10 box|year=2002|ab=A|num-b=9|num-a=11}} | ||
Revision as of 07:48, 18 February 2009
- The following problem is from both the 2009 AMC 12A #1 and 2009 AMC 10A #10, so both problems redirect to this page.
Problem
Compute the sum of all the roots of
Solution
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 | ||
