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 - 2 \(\sqrt{ac}\) x + b = 0
d = b2 – 4ac ≥ 0
\(=(2\sqrt{ac})^2-4\) b xb ≥ 0
= 4ac – 4 b2 ≥ \(0\)
From 1 and 2 we get only one case where b2 = ac