# How do you solve by completing the square for y=-x^2+6?

Jul 7, 2015

If $- {x}^{2} + 6 = 0$ then $x = \pm \sqrt{6}$

#### Explanation:

As written there is an infinite number of "solutions" in the form of $\left(x , y\right)$pairs for which the equation is valid, including: $\left(0 , 6\right) , \left(1 , 5\right) , \left(- 2 , 2\right)$, etc.

I will assume you wanted the solution for the x intercept, which is equivalent to the solution to
$\textcolor{w h i t e}{\text{XXXX}}$$0 = - {x}^{2} + 6$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(if this is not the case, re-post your question with clarification)

If $- {x}^{2} + 6 = 0$
then, equivalently,
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 6 = 0$

$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} = 6$

$\textcolor{w h i t e}{\text{XXXX}}$${\left(x + 0\right)}^{2} = 6$$\textcolor{w h i t e}{\text{XXXX}}$would technically "complete the square... but to what point?

$\textcolor{w h i t e}{\text{XXXX}}$$\left(x + 0\right) = \pm \sqrt{6}$$\textcolor{w h i t e}{\text{XXXX}}$carrying on as if it were necessary

$\textcolor{w h i t e}{\text{XXXX}}$$x = \pm \sqrt{6}$