# How do you solve x^2+16=0 using the quadratic formula?

Feb 22, 2018

If ${x}^{2} + 16 = 0$

$x = \pm \sqrt{- 16}$

#### Explanation:

If $a {x}^{2} + b x + c = 0 ,$
then
$x = \frac{1}{2 a} \left(- b \pm \sqrt{{b}^{2} - 4 a c}\right)$

For ${x}^{2} + 16 = 0$

$a = 1 , b = 0 , c = 16$
Substituting

$x = \frac{1}{2 \times 1} \left(- 0 \pm \sqrt{{0}^{2} - 4 \times 1 \times 16}\right)$

$= \frac{1}{2} \times \pm \sqrt{- 64}$

$= \pm \sqrt{- 16}$

Feb 22, 2018

$\pm 4 i$

#### Explanation:

The quadratic formula is given by:

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The equation in standard form is

${x}^{2} + 0 x + 16$

and $a = 1 , b = 0 , c = 16$

Plug in:

$\frac{- 0 \pm \sqrt{{0}^{2} - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

$\frac{0 \pm \sqrt{- 64}}{2}$

The number under the square root is negative, so there are no real solutions.

Finding the imaginary solutions ($i = \sqrt{- 1}$):

$\frac{\pm \sqrt{64 \cdot - 1}}{2}$

$\frac{\pm \sqrt{64} \cdot \sqrt{- 1}}{2}$

$\frac{\pm 8 i}{2}$

$\pm 4 i$

Feb 22, 2018

$x = \pm 4 i$

#### Explanation:

$\text{An alternative approach}$

${x}^{2} + 16 = 0$

$\Rightarrow {x}^{2} = - 16$

$\text{this has no real solutions but will have 2 complex}$
$\text{solutions}$

$\textcolor{b l u e}{\text{take square root of both sides}}$

$\Rightarrow x = \pm \sqrt{- 16} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x = \pm \sqrt{16 \times - 1} = \pm 4 i$

Feb 23, 2018

x=(0±sqrt(−64))/2
x=(−b±sqrt(b^2−4ac))/(2a)
x=(−(0)±sqrt((0)^2−4(1)(16)))/(2(1))
x=(0±sqrt(−64))/2