Answer :

Given the roots of both the equations are real

For first equation ax2 + 2bx + c = 0


Its discriminant; d ≥ 0


D = b2 – 4ac


D = (2b)2 – 4 × a × c ≥ 0


4b2


b2 ac …1


For second equation


bx2 - 2x + b = 0


d = b2 – 4ac 0


= (2)2 – 4 b xb 0


= 4ac – 4 b2


ac b2 …2


From 1 and 2 we get only one case where b2 = ac


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