Difference between revisions of "2003 AMC 10A Problems/Problem 5"
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<math> \mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math> | <math> \mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math> | ||
| − | == Solution == | + | == Solutions == |
| + | ===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)=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 (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> | ||
| + | |||
| + | |||
| + | ===Solution 3=== | ||
| + | By inspection, we quickly note that <math>x=1</math> is a solution to the equation, therefore the answer is | ||
| + | |||
| + | <math>(d-1)(e-1)=(1-1)(e-1)=\boxed{\mathrm{(B)}\ 0}</math> | ||
| + | |||
| + | ===Solution 4=== | ||
| + | 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 == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:37, 19 August 2023
Contents
Problem
Let 
and 
denote the solutions of 
. What is the value of 
?
Solutions
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. 
