Difference between revisions of "2003 AMC 10A Problems/Problem 5"
(→Solution) |
(→Solution 4) |
||
| Line 30: | Line 30: | ||
===Solution 4=== | ===Solution 4=== | ||
| − | The form <math>(d-1)(e-1)</math> resembles the factored form for the quadratic, namely <math>(x-d)(x-e)</math>. Note putting in 1 for <math>x</math> in the that quadratic immediately yields the desired expression. Thus, <math>2(1)^2+3(1)-5 = \boxed{\mathrm{(B)}\ 0}</math> | + | The form <math>(d-1)(e-1)</math> resembles the factored form for the quadratic, namely <math>(x-d)(x-e)</math> based on the information given. Note putting in 1 for <math>x</math> in the that quadratic immediately yields the desired expression. Thus, <math>2(1)^2+3(1)-5 = \boxed{\mathrm{(B)}\ 0}</math> |
== See Also == | == See Also == | ||
Revision as of 20:49, 25 April 2022
Contents
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):
Solution 3
By inspection, we quickly note that is a solution to the equation, therefore the answer is
Solution 4
The form resembles the factored form for the quadratic, namely 
based on the information given. Note putting in 1 for 
in the that quadratic immediately yields the desired expression. Thus, 
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. 
