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

1 Answer

Jul 7, 2015

If $-x^2+6=0$ then $x=+-sqrt(6)$

Explanation:

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

I will assume you wanted the solution for the x intercept, which is equivalent to the solution to
$color(white)("XXXX")$ $0 = -x^2+6$
$color(white)("XXXX")$ $color(white)("XXXX")$ (if this is not the case, re-post your question with clarification)

If $-x^2+6 = 0$
then, equivalently,
$color(white)("XXXX")$ $x^2-6 = 0$

$color(white)("XXXX")$ $x^2 = 6$

$color(white)("XXXX")$ $(x+0)^2 = 6$ $color(white)("XXXX")$ would technically "complete the square... but to what point?

$color(white)("XXXX")$ $(x+0) = +-sqrt(6)$ $color(white)("XXXX")$ carrying on as if it were necessary

$color(white)("XXXX")$ $x = +-sqrt6$

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