If the roots of the equations ax^2 + 2bx + c = 0 and bx^2 - 2√acx + b = 0 are simultaneously real, then prove that b^2 = ac.

Question

If the roots of the equations ax² + 2bx + c = 0 and \(b\text{x}^2-2\sqrt{ac\text{x}}+b=0\) are simultaneously real, then prove that b² = ac.

Answer 1

For a quadratic equation, ax² + bx + c = 0,

D = b² – 4ac

If D ≥ 0, roots are real

ax² + bx + c = 0,

⇒ b² ≥ ac ------ (1)

\(b\text{x}^2-2\sqrt{acx}+b=0\)

⇒ 4ac – 4b² ≥ 0 

⇒ b² ≤ ac ----- (2)

For both (1) and (2) to be true

⇒ b² = ac