Difference between revisions of "2003 AMC 10A Problems/Problem 5"
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| − | We can use the sum and product of a quadratic: | + | We can use the sum and product of a quadratic (a.k.a Vieta): |
<math>(d-1)(e-1)=de-(d+e)+1 \implies\text{product}-\text{sum}+1 \implies \dfrac{c}{a}-\left(-\dfrac{b}{a}\right)+1 \implies \dfrac{b+c}{a}+1= \dfrac{5}{-5}+1=\boxed{\mathrm{(B)}\ 0}</math> | <math>(d-1)(e-1)=de-(d+e)+1 \implies\text{product}-\text{sum}+1 \implies \dfrac{c}{a}-\left(-\dfrac{b}{a}\right)+1 \implies \dfrac{b+c}{a}+1= \dfrac{5}{-5}+1=\boxed{\mathrm{(B)}\ 0}</math> | ||
Revision as of 22:12, 8 May 2020
Problem
Let 
and 
denote the solutions of 
. What is the value of 
?
Solution
Solution 1
Using factoring:
or 
So and are 
and 
.
Therefore the answer is 
Solution 2
We can use the sum and product of a quadratic (a.k.a Vieta):
See Also
| 2003 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
