Question

Solve the following quadratic equation:

4x² – 4a²x + (a⁴ – b⁴) = 0

Answer 1

We write, –4a²x = –2(a² + b²)x – 2(a² – b²)x as 

4x² × (a⁴ – b⁴) = 4(a⁴ – b⁴)x² = [–2(a² + b²)]x × [–2(a² – b²)]x 

∴ 4x² – 4a²x + (a⁴ – b⁴) = 0 

⇒ 4x² – 2(a² + b²)x – 2(a² – b²)x + (a² – b²)(a² + b²) = 0 

⇒ 2x[2x – (a² + b²)] – (a² – b²)[2x – (a² + b²)] = 0 

⇒ [2x – (a² + b²)][2x – (a² – b²)] = 0 

⇒ 2x – (a² + b²) = 0 or 2x – (a² – b²) = 0 

⇒ `x = (a^2 + b^2)/2` or `x = (a^2 - b^2)/2` 

Hence, `(a^2 + b^2)/2` and `(a^2 - b^2)/2` are the roots of the given equation.