# How do you find the roots, real and imaginary, of y= 2(x+1)^2+(-x-2)^2  using the quadratic formula?

Jul 20, 2016

Zeros are complex conjugate numbers $- \frac{4}{3} - i \frac{\sqrt{2}}{3}$ and $- \frac{4}{3} + i \frac{\sqrt{2}}{3}$

#### Explanation:

Let us first expand RHS using identity ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

$y = 2 {\left(x + 1\right)}^{2} + {\left(- x - 2\right)}^{2}$

= 2(x^2+2×x×1+1^2)+(-x)^2+2×(-x)×(-2)+(-2)^2

= $2 {x}^{2} + 4 x + 2 + {x}^{2} + 4 x + 4$

= $3 {x}^{2} + 8 x + 6$

Now zeros of a quadratic function are those values of variable, here $x$, who make the value of function $0$ i.e. $3 {x}^{2} + 8 x + 6 = 0$. Quadratic formula gives the solution of $a {x}^{2} + b x + c = 0$ as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

Hence, zeros of $3 {x}^{2} + 8 x + 6$ are x=(-8+-sqrt(8^2-4×3×6))/(2×3) or

$x = \frac{- 8 \pm \sqrt{64 - 72}}{6}$

= $\frac{- 8 \pm \sqrt{- 8}}{6}$

= $- \frac{8}{6} \pm i \frac{2 \sqrt{2}}{6}$

= $- \frac{4}{3} \pm i \frac{\sqrt{2}}{3}$.

Hence zeros are complex conjugate numbers $- \frac{4}{3} - i \frac{\sqrt{2}}{3}$ and $- \frac{4}{3} + i \frac{\sqrt{2}}{3}$