# How do you find the zeros, real and imaginary, of  y=(x-3)^2+41  using the quadratic formula?

Dec 13, 2017

$x = 3 + \sqrt{41} i ,$$3 - \sqrt{41} i$

#### Explanation:

Given:

$y = {\left(x - 3\right)}^{2} + 41$

Expand ${\left(x - 3\right)}^{2}$.

$y = {x}^{2} - 6 x + 9 + 41$

Simplify.

$y = {x}^{2} - 6 x + 50$ is a quadratic equation in standard form:

$y = a {x}^{2} + b x + c$,

where:

$a = 1$, $b = - 6$, and $c = 50$

The zeroes are the x-intercepts, which are the values of $x$ when $y = 0$.

Substitute $0$ for $y$.

$0 = {x}^{2} - 6 x + 50$

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

Plug in the known values.

$x = \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{6} - 4 \cdot 1 \cdot 50}}{2 \cdot 1}$

Simplify.

$x = \frac{6 \pm \sqrt{- 164}}{2}$

Prime factorize $164$.

$x = \frac{6 \pm \sqrt{2 \times 2 \times 41}}{2}$

Simplify.

$x = \frac{6 \pm 2 \sqrt{41} i}{2}$

Reduce.

$x = \frac{{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}}^{3} \pm {\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}^{1} \sqrt{41} i}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} ^ 1$

Simplify.

$x = 3 \pm \sqrt{41} i$

Values for $x$.

$x = 3 + \sqrt{41} i ,$$3 - \sqrt{41} i$