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2003 AMC 10A Problems/Problem 5

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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 (ProblemsResources)
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
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