How to solve a quadratic equation with a negative value for the variable?

[peterpanhandle]

Homework Statement

Solve 4y-y²=x for y.

The Attempt at a Solution

First I tried using the quadratic equation with a=-1 b=4 c=-x

y=(-b±(4²-4(-1)(-x))^(1/2))/2(-1)

That got me this far: y=2±(8-2x)^(1/2)

Then I checked wolfram and now I am confused

Spoiler

y=±2-sqrt(4-x

. Did I not approach this corectly?

 

[vela]

Homework Statement

Solve 4y-y²=x for y.

The Attempt at a Solution

First I tried using the quadratic equation with a=-1 b=4 c=-x

y=(-b±(4²-4(-1)(-x))^(1/2))/2(-1)

That got me this far: y=2±(8-2x)^(1/2)

Then I checked wolfram and now I am confused

Spoiler

y=±2-sqrt(4-x

. Did I not approach this corectly?

Your approach is fine. You just made an algebra mistake while simplifying.

You had $y = \frac{-4 \pm \sqrt{16 - 4x}}{-2}$. Note that you can't simply divide the stuff inside the radical by 2 when you bring the 2 from the denominator inside.

 

[peterpanhandle]

You just made an algebra mistake while simplifying.

Thank you for the quick reply and sorry for my slow thanks. How very typical of me that it's some algebraic error Thanks again!

 

[SammyS]

...

Then I checked wolfram and now I am confused [ SPOILER]y=±2-sqrt(4-x)[/SPOILER]. Did I not approach this correctly?

As the originator of the thread, you don't really need to use a Spoiler.