Ok so I have a quadratic equation in standard form. x^2 - 8x

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Ok so I have a quadratic equation in standard form. x^2 - 8x

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Ok so I have a quadratic equation in standard form.x^2 - 8x + 16I completed the square and got it down to (x-4)^2Now to inverse (x-4)^2 would I do it likey = (x-4)^2 (sqrt (y)) = x-4 4+/- (sqrt (y))Just want to double check make sure I've not made a silly mistake before handing it in.

Yes, you did a great job! Those are the exact steps to take.

The answer is 4 +/- sqrt(x) (Note I used x in the answer, not y. But you had the right steps. -- this is why I originally used f(x) and x when solving it. If you're going to use y, don't forget to swap it back out for x at the end.)

The domain here is x >= 0, so that the inside of the square root is >= 0.

Best,

Scott

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