Problem: If c is a real number and the negative of one of the solutions of x2−3x+c=0 is a solution of x2+3x−c=0, then the solutions of x2−3x+c=0 are
Answer Choices:
A. 1,2
B. −1,−2
C. 0,3
D. 0,−3
E. 23​,23​
Solution:
Let r be a solution of x2−3x+c=0 such that −r is a solution of x2+3x−c =0. Then
r2−3r+cr2−3r−c​=0=0.​
which implies 2c=0. The solutions of x2−3x=0 are 0 and 3.