Question

Find the value of x with method.

Answer 1

Given Equation is 2x^2 + 4x + 5 = 0.

Here a = 2, b = 4, c = 5.

The solutions are:

(i)

$x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$

$= > \frac{(-4) + \sqrt{(4)^2 - 4(2)(5)}}{2(2)}$

$= > \frac{-4 + \sqrt{16 - 40}}{4}$

$= > \frac{-4 + \sqrt{24}}{4}$

$= > \frac{-4 + 2\sqrt{6}i}{4}$

$= > \frac{-2 + \sqrt{6}i}{2}$

$= > -1 + \frac{\sqrt{6}}{2} i$

(ii)

$= > x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$

$= > \frac{-4 - \sqrt{4^2 - 2(4)(5)}}{2(2)}$

$= > \frac{-4 - \sqrt{16 - 24}}{4}$

$= > -\frac{2 + \sqrt{6}i}{2}$

$= > -1 - \frac{\sqrt{6}}{2} i$

Therefore,

$= > \boxed{x = -1 + \frac{\sqrt{6}}{2}i, -1 - \frac{\sqrt{6}}{2}i}$

Hope it helps!