Difference between revisions of "2002 AMC 12A Problems/Problem 1"
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We see a <math>(2x+3)</math> term in both addends so we can factor the equation to <math>(2x+3)((x-4)+(x-6))=(2x+3)(2x-10)=0.</math> The two roots become obvious: they are <math>-\frac{3}{2}</math> and <math>\frac{10}{2}.</math> Adding these two gives <math>\boxed{\textbf{(A) }7/2}.</math> | We see a <math>(2x+3)</math> term in both addends so we can factor the equation to <math>(2x+3)((x-4)+(x-6))=(2x+3)(2x-10)=0.</math> The two roots become obvious: they are <math>-\frac{3}{2}</math> and <math>\frac{10}{2}.</math> Adding these two gives <math>\boxed{\textbf{(A) }7/2}.</math> | ||
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| + | -starwarmonkey | ||
==Video Solution by Daily Dose of Math== | ==Video Solution by Daily Dose of Math== | ||
Latest revision as of 15:22, 26 August 2024
- The following problem is from both the 2002 AMC 12A #1 and 2002 AMC 10A #10, so both problems redirect to this page.
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
We see a 
term in both addends so we can factor the equation to 
The two roots become obvious: they are 
and 
Adding these two gives 
-starwarmonkey
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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. 
