Solving for x in terms of y

the problem is y=x/[sqrt(x+1)]

I feel like an idiot because I can't figure this out. everytime i try, i get stuck with xy+y=x^2 which seems like i could get nowhere with singling out an x by itself >.< I appreciate any help. just trying to review my algebra for calculus but i seem to have forgotten.

You're on the right track, almost done --- you need to solve the quadratic equation!

First of all, squaring your equation
y=x/[sqrt(x+1)]
gives
y^2=x^2/(x+1)
so
y^2*x+y^2=x^2.
You miss the square on the y in your equation.

Now, solve the quadratic equation in x, treating y as a fixed parameter:
x^2 - y^2*x - y^2 = 0
The two solutions you should get are:
x1=(1/2)*y^2 + (1/2)*sqrt(y^4+4*y^2)
simplifies to
x1=(1/2)*y^2 + (1/2)*abs(y)*sqrt(y^2+4)
and
x2=(1/2)*y^2 - (1/2)*sqrt(y^4+4*y^2)
simplifies to
x2=(1/2)*y^2 - (1/2)*abs(y)*sqrt(y^2+4)

The sqrt(y^2+4) is always real because the y^2+4 is always greater than 0, so x1 and x2 are always real numbers, and both are valid.

y = x / √(x+1)
y√(x + 1) = x
y² (x + 1) = x²
y²x + y² = x²
x² - y²x - y² = 0

This is a quadratic equation. Solve it.
Discriminant D = (b² - 4ac) = (y^4 + 4y²) = y²(y² + 4)

x = [-b ± √D] / 2a
= [y² ± √y²(y² + 4)] / 2
= [y² ± y√(y² + 4)] / 2
= (y/2)[y ± √(y² + 4)]