Question

how do I issolate 'y' in this equasion. . . .
xy-2y^2=-4

Answer 1

\[\large xy-2y^2=-4\]


\[\large xy-2y^2+4 = 0\]


\[\large -2y^2+xy+4 = 0\]


The last equation is in the form ay^2 + by + c = 0 where

a = -2

b = x

c = 4


From here, you plug all this into the quadratic formula given below, and simplify, to solve for y.


\[\large y=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]


I'll let you take over from here.

Answer 2

thank you

Answer 3

you're welcome

Answer 4

this is what i got \[y=\frac{ x \pm \sqrt{x^2+32} }{ -4 }\]
what do i do about the plus or minus?

Answer 5

Here's how I would do it

\[\large y=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

\[\large y=\frac{-x \pm \sqrt{(-x)^2-4(-2)(4)}}{2(-2)}\]

\[\large y=\frac{-x \pm \sqrt{x^2+32}}{-4}\]

\[\large y=\frac{-x + \sqrt{x^2+32}}{-4} \ \text{or} \ \frac{-x - \sqrt{x^2+32}}{-4}\]

\[\large y=\frac{x - \sqrt{x^2+32}}{4} \ \text{or} \ \frac{x + \sqrt{x^2+32}}{4}\]

\[\large y=\frac{x + \sqrt{x^2+32}}{4} \ \text{or} \ \frac{x - \sqrt{x^2+32}}{4}\]

So you just needed to break up the +- into two different equations

Answer 6

ok. that makes sense

Answer 7

ok great, glad it does