Difference between revisions of "2002 AMC 12A Problems/Problem 1"

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== Problem ==
 
== Problem ==
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Compute the sum of all the roots of
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<math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math>
  
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<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 ==
{{solution}}
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That is equal to <math>(2x+3)(2x-10)=0</math>
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We can divide out by 4 to get
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<math>(x+\frac{3}{2})(x-5)=0</math>
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The sum of the roots is therefore
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<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 14:26, 17 October 2007

Problem

Compute the sum of all the roots of $(2x+3)(x-4)+(2x+3)(x-6)=0$

$\mathrm{(A) \ } \frac{7}{2}\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 13$

Solution

That is equal to $(2x+3)(2x-10)=0$

We can divide out by 4 to get

$(x+\frac{3}{2})(x-5)=0$

The sum of the roots is therefore

$5-\dfrac{3}{2}=\dfrac{7}{2} \Rightarrow \mathrm {(B)}$

See also

2002 AMC 12A (ProblemsAnswer KeyResources)
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