# How do you solve  x^2+18=-81?

Apr 2, 2016

${x}^{2} = - 99 \implies x = \pm \sqrt{- 99} = \pm 3 \sqrt{11} i$

Apr 2, 2016

$\textcolor{b l u e}{\implies x = \pm 3 i \sqrt{11}}$

#### Explanation:

Given:$\text{ } \textcolor{b r o w n}{{x}^{2} + 18 = - 81}$

If so chosen you could write this as $y = {x}^{2} + 99$

$\textcolor{b l u e}{\text{Solving for "x" where the given equation is true}}$

Subtract $\textcolor{b l u e}{18}$ from both sides

$\text{ } \textcolor{b r o w n}{{x}^{2} + 18 \textcolor{b l u e}{- 18} = - 81 \textcolor{b l u e}{- 18}}$

$\text{ } {x}^{2} + 0 = - 99$

know: $\textcolor{g r e e n}{3 \times 33 = 99} \text{ so "color(blue)(3xx3xx11=99)" so } \textcolor{red}{{3}^{2} \times 11 = 99}$

But we have negative 99 so we have $\left(- 1\right) \times {3}^{2} \times 11$

Thus our equation becomes

${x}^{2} = \left(- 1\right) \times {3}^{2} \times 11$

Take the square root of each side

$x = \pm \sqrt{\left(- 1\right) \times {3}^{2} \times 11}$

$\implies x = \pm 3 \times \sqrt{11} \sqrt{- 1}$

But $\sqrt{- 1} = i$

$\textcolor{b l u e}{\implies x = \pm 3 i \sqrt{11}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Notice that the graph does not cross the x-axis.

But by writing the equation as ${x}^{2} + 18 = - 81$

$y = {x}^{2} + 99$

So to solve for $x$ we have to write

$y = 0 = {x}^{2} + 99$ but for $y = 0$ implies that the graph crosses the x-axis when in fact it does not. So to be able to solve for $y =$ we introduce 'complex numbers'. Hence the $\sqrt{- 1}$ 