How do you find the zeros, real and imaginary, of  y=1/4(x+4)^2+4  using the quadratic formula?

Apr 4, 2017

$\textcolor{red}{x = - 4 + 4 i , - 4 - 4 i}$

Explanation:

Finding the zeros/roots of an equation of the form $y = f \left(x\right)$ means finding solutions of the equation $f \left(x\right) = 0$; $i . e .$ finding those values of $x$ for which $f \left(x\right) = 0$

$\therefore$ zeros/roots of $\textcolor{red}{y = \frac{1}{4} {\left(x + 4\right)}^{2} + 4}$ are basically solutions of $\textcolor{red}{\frac{1}{4} {\left(x + 4\right)}^{2} + 4 = 0}$

$\frac{1}{4} {\left(x + 4\right)}^{2} + 4 = 0$
$\implies \frac{{\left(x + 4\right)}^{2} + 16}{4} = 0$

$\implies \left({x}^{2} + 8 x + 16\right) + 16 = 0$
$\implies {x}^{2} + 8 x + 32 = 0$

Using the quadratic formula;

$x = \frac{- 8 \pm \sqrt{{8}^{2} - 4 \cdot 1 \cdot 32}}{2 \cdot 1}$

$x = \frac{- 8 \pm \sqrt{- 64}}{2} = \frac{- 8 \pm 8 i}{2} = - 4 \pm 4 i$

$\therefore \textcolor{red}{x = - 4 + 4 i , - 4 - 4 i}$ are the two imaginary zeros/roots of the given equation.