Question

Determine the zeros of the function y= 2x² + 12x + 22

Answer 1

zeros of the function y= $2x^2 + 12x + 22$ is $x = -3+\frac{ \sqrt{8}i }{2}$ &  $x = -3-\frac{ \sqrt{8}i }{2}$.

Step-by-step explanation:

Here we have , y=$2x^2 + 12x + 22$ in order to find zeros of this quadratic function we get: $2x^2 + 12x + 22 = 0$

⇒ $2x^2 + 12x + 22 = 0$

⇒ $x^2 + 6x + 11 = 0$

⇒ $x = \frac{-b \pm \sqrt{b^2-4a(c)} }{2a}$

⇒ $x = \frac{-(6) \pm \sqrt{6^2-4(1)(11)} }{2(1)}$

⇒ $x = \frac{-6 \pm \sqrt{36-44)} }{2}$

⇒ $x = \frac{-6 \pm \sqrt{8}i }{2}$

⇒ $x = -3 \pm \frac{ \sqrt{8}i }{2}$

Since, we have two root one with positive & other with negative sign so :

⇒ $x = -3-\frac{ \sqrt{8}i }{2}$

Therefore, zeros of the function y= $2x^2 + 12x + 22$ is $x = -3+\frac{ \sqrt{8}i }{2}$ &  $x = -3-\frac{ \sqrt{8}i }{2}$.