Question

Solve the following quadratic equation:

x² + 6x – (a² + 2a – 8) = 0

Answer 1

We write, 6x = (a + 4)x – (a – 2)x as 

x² × [–(a² + 2a – 8)] = –(a² + 2a – 8)x² = (a + 4)x × [–(a – 2)x] 

∴ x² + 6x – (a² + 2a – 8) = 0 

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

⇒ x[x + (a + 4)] – (a – 2)[x + (a + 4)] = 0 

⇒ [x + (a + 4)][x – (a – 2)] = 0 

⇒ x + (a + 4) = 0 or x – (a – 2) = 0 

⇒ x = –(a + 4) or x = a – 2 

Hence, –(a + 4) and (a – 2) are the roots of the given equation.