Problem:
Let d and e denote the solutions of 2x2+3x−5=0. What is the value of (d−1)(e−1)?
Answer Choices:
A. −25
B. 0
C. 3
D. 5
E. 6
Solution:
Since
0=2x2+3x−5=(2x+5)(x−1) we have d=−25 and e=1.
So (d−1)(e−1)=0.
OR
If x=d and x=e are the roots of the quadratic equation ax2+bx+c=0, then
de=ac and d+e=−ab.
For our equation this implies that
(d−1)(e−1)=de−(d+e)+1=−25−(−23)+1=0.
Answer: B.
The problems on this page are the property of the MAA's American Mathematics Competitions