Problem: Consider x2+px+q=0 where p and q are positive numbers. If the roots of this equation differ by 1, then p equals
Answer Choices:
A. 4q+1
B. q−1
C. −4q+1
D. q+1
E. 4q−1
Solution:
The roots are 21(−p+p2−4q) and 21(−p−p2−4q).
The difference is p2−4q=1. Hence p2=4q+1 and p=4q+1.