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

The given equations are

ax² + 2bx + c = 0             ............ (1)

`bx^2-2sqrt(ac)x+b = 0` ............. (2)

Roots are simultaneously real

Then prove that b² = ac

Let D₁ and D₂ be the discriminants of equation (1) and (2) respectively,

Then,

D₁ = (2b)² - 4ac

= 4b² - 4ac

And

`D_2=(-2sqrt(ac))^2-4xxbxxb`

= 4ac - 4b²

Both the given equation will have real roots, if D₁ ≥ 0 and D₂ ≥ 0

4b² - 4ac ≥ 0

4b² ≥ 4ac

b² ≥ ac                ............... (3)

4ac - 4b² ≥ 0

4ac ≥ 4b²

ac ≥ b²                          ................... (4)

From equations (3) and (4) we get

b² = ac

Hence, b² = ac.