Q. 84.3( 15 Votes )

Find the values of k for which the quadratic equation (3k + 1) x2 + 2(k + 1)x + 1 = 0 has real and equal roots.

Answer :

Given equation is (3k + 1) x2 + 2(k + 1)x + 1 = 0

Comparing with standard quadratic equation ax2 + bx + c = 0


a = (3k + 1) b = 2(k + 1) c = 1


Given that the roots of equation are real and equal


Thus D = 0


Discriminant D = b2 – 4ac = 0


(2k + 2)2 – 4.(3k + 1).1 = 0 using (a + b)2 = a2 + 2ab + b2


4k2 + 8k + 4 – 12k – 4 = 0


4k2 – 4k = 0


4k(k – 1) = 0


k = 0 (k – 1) = 0


k = 0 k = 1


The values of k are 0, 1 for which roots of the quadratic equation are real and equal.


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 Equations51 mins
Getting Familiar with Nature of Roots of Quadratic Equations51 mins
Quadratic Equation: Previous Year NTSE Questions32 mins
Understand The Concept of Quadratic Equation45 mins
Champ Quiz | Quadratic Equation33 mins
Quiz | Lets Solve Imp. Qs of Quadratic Equation43 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