2002 AMC 12A Problems/Problem 1: Difference between revisions
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== Problem == | == Problem == | ||
Compute the sum of all the roots of | |||
<math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math> | |||
<math> \mathrm{(A) \ } \frac{7}{2}\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 13 </math> | |||
== Solution == | == Solution == | ||
{{ | That is equal to <math>(2x+3)(2x-10)=0</math> | ||
We can divide out by 4 to get | |||
<math>(x+\frac{3}{2})(x-5)=0</math> | |||
The sum of the roots is therefore | |||
<math>5-\dfrac{3}{2}=\dfrac{7}{2} \Rightarrow \mathrm {(B)}</math> | |||
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=A|before=First Question|num-a=2}} | {{AMC12 box|year=2002|ab=A|before=First Question|num-a=2}} | ||
Revision as of 13:26, 17 October 2007
Problem
Compute the sum of all the roots of
Solution
That is equal to 
We can divide out by 4 to get
The sum of the roots is therefore
See also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First Question |
Followed by Problem 2 |
| 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 12 Problems and Solutions | |
