Difference between revisions of "2002 AMC 10A Problems/Problem 10"
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===Solution 2=== | ===Solution 2=== | ||
| − | Combine terms to get <math>(2x+3)(2x-10)=0</math>, hence the roots are <math>-\frac{3}{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== | ||
Revision as of 11:51, 24 January 2009
Problem
What is the sum of all of 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 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 | ||
