Difference between revisions of "2002 AMC 12A Problems/Problem 1"
Dairyqueenxd (talk | contribs) (→Solution 1) |
Dairyqueenxd (talk | contribs) (→Solution 2) |
||
| Line 11: | Line 11: | ||
==Solution 2== | ==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{\ | + | 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{\textbf{(A) } 7/2}</math>. |
== See also == | == See also == | ||
Revision as of 12:06, 8 November 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
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. 
