Question

Solve the following quadratic equation:

x² – 4ax – b² + 4a² = 0

Answer 1

We write, –4ax = –(b + 2a)x + (b – 2a)x as 

x² × (–b² + 4a²) = (–b² + 4a²)x² = –(b + 2a)x × (b – 2a)x 

∴ x² – 4ax – b² + 4a² = 0 

⇒ x² – (b + 2a)x(b – 2a)x – (b – 2a)(b + 2a) = 0 

⇒ x[x – (b + 2a)] + (b – 2a)[x – (b + 2a)] = 0 

⇒ [x – (b + 2a)] [x + (b – 2a)] = 0 

⇒ x – (b + 2a) = 0 or x + (b – 2a) = 0 

⇒ x = 2a + b or x = –(b – 2a)

⇒ x = 2a + b or x = 2a – b 

Hence, (2a + b) and (2a – b) are the roots of the given equation.