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 bx² - 2\(\sqrt{acx}\)+ b = 0 are simultaneously real then prove that b² = ac

Answer 1

It is given that the roots of the equations ax² + 2bx + c = 0 are real.

∴ D₁ = (2b)² - 4 x a x c ≥ 0

⇒ 4(b² - ac) ≥ 0

⇒ b² - ac ≥ 0.........(i)

Also, the roots of the equation bx² - 2\(\sqrt{acx}\) + b = 0 are equal

∴ D₁ = (-2\(\sqrt{ac}\))² - 4 x b x b ≥ 0

⇒ 4(ac - b²) ≥ 0

⇒ -4(b² - ac) ≥ 0

⇒ b² - ac ≥ 0........(2)

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if

b² - ac = 0

⇒ b² = ac