Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 41 of Exercise 10A
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Solve each of the following quadratic equations:
4x2 - 4a2x + (a4 - b4) = 0
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 
