Question

It is given that p²x² + (p² – q²)x – q² = 0; (p ≠ 0).

1. Show that the discriminant (D) of the above equation is a perfect square.
2. Find the roots of the equation.

Answer 1

p²x² + (p² – q²)x – q² = 0; (p ≠ 0)

Comparing with ax² + bx + c = 0

Then a = p², b = (p² – q²) and c = – q²

(i) Now, D = b² – 4ас

= (p² – q²)² – 4(p²) (– q²)

= p⁴ + q⁴ – 2p²q² + 4p²q²

= p⁴ + q⁴ + 2p²q²

= (p² + q²)²

Therefore, the discriminant (D) is the perfect square.

Hence Proved.

(ii) To find the roots of the equation.

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`

= `(-(p^2 - q^2) +- sqrt((p^2 + q^2)^2))/(2p^2)`   ...[∵ b² – 4ac = (p² + q²)²]

= `(-p^2 + q^2 +- (p^2 + q^2))/(2p^2)`

= `(-p^2 + q^2 + p^2 + q^2)/(2p^2)` or `(-p^2 + q^2 - p^2 - q^2)/(2p^2)`

= `(2q^2)/(2p^2)` or `(-2p^2)/(2p^2)`

= `q^2/p^2` or (–1)

Therefore, the roots of the equation are `q^2/p^2` and (–1).