Answer :

Given: x2 – (2b – 1)x + (b2 – b – 20) = 0

Comparing with standard quadratic equation Ax2 + Bx + C = 0


A = 1, B = – (2b – 1), C = (b2 – b – 20)


Discriminant D = B2 – 4AC


= [ – (2b – 1)2] – 4.1. (b2 – b – 20) Using a2 – 2ab + b2 = (a – b)2


= 4b2 – 4b + 1 – 4b2 + 4b + 80 = 81 > 0


Hence the roots of equation are real.



Roots are given by




x = (b + 4) or x = (b – 5)


Hence the roots of equation are (b + 4) or (b – 5)


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses