Question

Solve the following quadratic equation:

4x² + 4bx – (a² – b²) = 0

Answer 1

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

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

∴ 4x² + 4bx – (a² – b²) = 0 

⇒ 4x² + 2(a + b)x – 2(a – b)x – (a – b)(a + b) = 0 

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

⇒ [2x + (a + b)][2x – (a – b)] = 0 

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

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

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