If the roots of the equations ax2 + 2bx + c = 0 and bx2 - 2\(\sqrt{acx

1.	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» 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

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»

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