Difference between revisions of "2003 AMC 10A Problems/Problem 5"
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So <math>d</math> and <math>e</math> are <math>-\frac{5}{2}</math> and <math>1</math>. | So <math>d</math> and <math>e</math> are <math>-\frac{5}{2}</math> and <math>1</math>. | ||
| − | Therefore the answer is <math>(-\frac{5}{2}-1)(1-1)=(-\frac{7}{2})(0)=0 \Rightarrow B</math> | + | Therefore the answer is <math>(-\frac{5}{2}-1)(1-1)=(-\frac{7}{2})(0)=0 \Rightarrow B</math> |
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| + | OR we can use sum and product. | ||
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| + | <math>(d-1)(e-1)=de-(d+e)+1 \Rightarrow</math> | ||
| + | <math>product-sum+1 \Rightarrow</math> | ||
| + | <math>c/a-(-b/a)+1 \Rightarrow</math> | ||
| + | <math>(b+c)/a+1 \Rightarrow</math> | ||
| + | <math>0 \Rightarrow B</math> | ||
== See Also == | == See Also == | ||
Revision as of 17:47, 3 June 2008
Problem
Let 
and 
denote the solutions of 
. What is the value of 
?
Solution
Using factoring:
or 
So and are 
and 
.
Therefore the answer is 
OR we can use sum and product.
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 | ||
