Question

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

Answer 1

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

∴ D₁ = (2b)² – 4 × a × c ≥ 0

⇒ 4(b² – ac) ≥ 0 

⇒ –4(b² – ac) ≥ 0 

⇒ b² – ac ≥ 0   ...(1) 

Also, the roots of the equation `bx^2 - 2sqrt(ac)x + b = 0` are real. 

∴ `D_2 = (-2sqrt(ac))^2 - 4 xx b xx 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