Problem: Let the roots of ax2+bx+c=0 be r and s. The equation with roots ar+b and as+b is:
Answer Choices:
A. x2−bx−ac=0
B. x2−bx+ac=0
C. x2+3 bx+ca+2 b2=0
D. x2+3bx−ca+2 b2=0
E. x2+bx(2−a)+a2c+b2(a+1)=0
Solution:
We have r+s=−ab and rs=ac. The equation with roots ar+b and as+b is (x−(ar+b))(x−(as+b))=0
∴x3−(a(r+s)+2b)x+a2rs+ab(r+s)+b2=0
∴x2−(a(−ab)+2b)x+a2(ac)+ab(−ab)+b2=0
∴x2−bx+ac=0