Q. 414.6( 11 Votes )

Solve each of the following quadratic equations:

4x2 - 4a2x + (a4 - b4) = 0

Answer :

4x2 - 4a2x + (a4 - b4) = 0

Using the splitting middle term - the middle term of the general equation is divided in two such values that:


Product = a.c


For the given equation a = 4 ;b = - 4a2 ; c = (a4 - b4)


= 4. (a4 - b4)


= 4a4 - 4b4


And either of their sum or difference = b


= - 4a2


Thus the two terms are - 2(a2 + b2) and - 2(a2 - b2)


Difference = - 2(a2 + b2) - 2(a2 - b2)


= - 2a2 - 2b2 - 2a2 + 2b2


= - 4a2


Product = - 2(a2 + b2). - 2(a2 - b2)


= 4(a2 + b2)(a2 - b2)


= 4. (a4 - b4)


( using a2 - b2 = (a + b) (a - b))


4x2 - 4a2x + (a4 - b4) = 0


4x2 - 4a2x + ((a2)2 – (b2)2) = 0


( using a2 - b2 = (a + b) (a - b))


4x2 - 2(a2 + b2) x - 2(a2 - b2) x + (a2 + b2) (a2 - b2) = 0


2x [2x - (a2 + b2)] - (a2 - b2) [2x - (a2 + b2)] = 0


[2x - (a2 + b2)] [2x - (a2 - b2)] = 0


[2x - (a2 + b2)] = 0 or [2x - (a2 - b2)] = 0



Hence the roots of given equation are


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
Champ Quiz | Quadratic Equation33 mins
Balance the Chemical Equations49 mins
Understand The Concept of Quadratic Equation45 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