Question

z² + 6z – 8 = 0

Answer 1

Given: z² + 6z – 8 = 0

$\Rightarrow$z² + 6z = 8 … (1)

We make L.H.S. a perfect square by using

Third term = .${\left(\frac{1}{2}\times \mathrm{cofficient}\mathrm{of}\mathrm{z}\right)}^2={\left(\frac{1}{2}\times 6\right)}^2=3^2=9$

On adding 9 to both sides of equation (1), we get:

z² + 6z + 9 = 8 + 9

$\Rightarrow$z² + 2 × z × 3 + (3)² = 17

$\Rightarrow$(z + 3)² = ${\left(\sqrt{17}\right)}^2$

On taking square root of both sides, we get:

z + 3 = $\pm \sqrt{17}$

$\Rightarrow$z = $-3\pm \sqrt{17}$