Question

Using quadratic formula find the value of x.

p²x² + (p² – q²)x – q² = 0

Answer 1

We have p²x² + (p² – q²)x – q² = 0

Comparing this equation to ax² + bx + c = 0, we have

a = p², b = p² – q² and c = – q²

∴ D = b² – 4ac

= (p² – q²)² – 4 × p² + (– q)²

= (p² – q²)² + 4p²q²

= (p² + q²)² > 0

So, the given equation has real roots given by

`alpha = (-b + sqrt(D))/(2a)`

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

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

= `q^2/p^2`

And `beta = (-b - sqrt(D))/(2a)`

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

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

= –1

Hence roots are `q^2/p^2` and –1