2003 AMC 10A Problems/Problem 5

Problem

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

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

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.

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

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