Difference between revisions of "2003 AMC 10A Problems/Problem 5"

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(Solution)
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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 \Rightarrow B</math>  
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Therefore the answer is <math>(-\frac{5}{2}-1)(1-1)=(-\frac{7}{2})(0)=0 \Rightarrow B</math>
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OR we can use sum and product.
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<math>(d-1)(e-1)=de-(d+e)+1 \Rightarrow</math>
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<math>product-sum+1 \Rightarrow</math>
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<math>c/a-(-b/a)+1 \Rightarrow</math>
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<math>(b+c)/a+1 \Rightarrow</math>
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<math>0 \Rightarrow B</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 17:47, 3 June 2008

Problem

Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$. What is the value of $(d-1)(e-1)$?

$\mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$

Solution

Using factoring:

$2x^{2}+3x-5=0$

$(2x+5)(x-1)=0$

$x = -\frac{5}{2}$ or $x=1$

So $d$ and $e$ are $-\frac{5}{2}$ and $1$.

Therefore the answer is $(-\frac{5}{2}-1)(1-1)=(-\frac{7}{2})(0)=0 \Rightarrow B$

OR we can use sum and product.

$(d-1)(e-1)=de-(d+e)+1 \Rightarrow$ $product-sum+1 \Rightarrow$ $c/a-(-b/a)+1 \Rightarrow$ $(b+c)/a+1 \Rightarrow$ $0 \Rightarrow B$

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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