Question

Solve the following quadratic equation:

x² – 2ax – (4b² – a²) = 0

Answer 1

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

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

∴ x² – 2ax – (4b² – a²) = 0 

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

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

⇒ [x + (2b – a)][x – (2b + a)] = 0 

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

x = –(2b – a) or x = 2b + a 

⇒ x = a – 2b or x = a + 2b 

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