Answer :

Given: x2 – 2ax – (4b2 – a2) = 0

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


A = 1, B = – 2a, C = – (4b2 – a2)


Discriminant D = B2 – 4AC


= (– 2a)2 – 4.1. – (4b2 – a2)


= 4a2 – 4a2 + 16 b2 = 16b2 > 0


Hence the roots of equation are real.



Roots are given by




x = (a + 2b) or x = (a – 2b)


Hence the roots of equation are (a + 2b) or (a – 2b)


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