2003 AMC 10A Problems/Problem 5: Difference between revisions
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== Solution == | == Solution == | ||
===Solution 1=== | |||
Using factoring: | Using factoring: | ||
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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)= | Therefore the answer is <math>\left(-\frac{5}{2}-1\right)(1-1)=\left(-\frac{7}{2}\right)(0)=\boxed{\mathrm{(B)}\ 0}</math> | ||
===Solution 2=== | |||
We can use the sum and product of a quadratic: | |||
<math>(d-1)(e-1)=de-(d+e)+1 \ | <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> | ||
== See Also == | == See Also == | ||
Revision as of 17:06, 31 July 2011
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:
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 | ||
