If the roots of the equations ax2 + 2bx + c = 0 and \(b\text{x}^2-2\s

1.	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» 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

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»

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