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

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

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Answer 1 · 2015-07-07T22:42:00

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

Explanation:

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

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

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

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

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

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

$XXXX$ $color(white)("XXXX")$ $x = \pm \sqrt{6}$ $x = +-sqrt6$

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

1 Answer

Explanation:

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How do you solve by completing the square for y=-x^2+6y=−x2+6y=-x^2+6?

1 Answer

Explanation:

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