Q. 3 A4.0( 5 Votes )

Find k, if the roots of x2 — (3k — 2)x + 2k = 0 are equal and real.

Answer :

Comparing equation x2 — (3k — 2)x + 2k = 0 with ax2 + bx + c = 0 we get


a = 1, b = – (3k – 2) and c = 2k


Discriminant (D) = b2 – 4ac


As roots are real and equal D = 0


b2 – 4ac = 0


[ – (3k – 2)]2 – 4(1)(2k) = 0


(3k – 2)2 – 8k = 0


Expand using (a – b)2 = a2 – 2ab + b2


9k2 – 12k + 4 – 8k = 0


9k2 – 18k – 2k + 4 = 0


taking 9k common from first two terms and – 2 common from next two


9k(k – 2) – 2(k – 2) = 0


(9k – 2)(k – 2) = 0


9k – 2 = 0 or k – 2 = 0


9k = 2 or k = 2


Therefore k = and k = 2


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Get To Know About Quadratic Formula42 mins
Quiz | Knowing the Nature of Roots44 mins
Take a Dip Into Quadratic graphs32 mins
Foundation | Practice Important Questions for Foundation54 mins
Nature of Roots of Quadratic EquationsFREE Class
Getting Familiar with Nature of Roots of Quadratic Equations51 mins
Quadratic Equation: Previous Year NTSE Questions32 mins
Champ Quiz | Quadratic Equation33 mins
Balance the Chemical Equations49 mins
Champ Quiz | Quadratic Equation48 mins
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
view all courses