Help with Inverse function question. Thank you :D

Question

wcrmelissa2001 · Answer

$$y=\frac{ x }{ x ^{2} -2}+C$$

alexishope47 · Answer

You want to find the inverse of that equation?

wcrmelissa2001 · Answer

yeah. basically find x

alexishope47 · Answer

Okay let me see if I can figure it out.

wcrmelissa2001 · Answer

thanks ;D

alexishope47 · Answer

Well when you simply solve for x you get this but i'm not sure if this is what you are looking for. I couldn't figure it out so I put it into my online calculator and it got this.  Sorry I couldn't be more help.

wcrmelissa2001 · Answer

sokay :D the answer's wrong though (I have the answer with me but idk how to do) THANKS ANYWAY!

wcrmelissa2001 · Answer

@usercode3rror

alexishope47 · Answer

Yeah sorry. I'm usually better with math haha

wcrmelissa2001 · Answer

btw shoulda put this earlier but the answer is $$x=\frac{ 1\pm \sqrt{8(y-C)^2+1} }{ 2(y-C) }$$

sirm3d · Answer

transpose C, $$y-C=\frac{x}{x^2-2}$$ then solve the quadratic equation in $x$

wcrmelissa2001 · Answer

but how..because you have x2 down there and x so if you cross multiply then it gets weird

sirm3d · Answer

$$\frac{y-C}{1}=\frac{x}{x^2-2}$$ cross multiply $$(x^2-2)(y-C)=x$$

wcrmelissa2001 · Answer

yeah got till there too but then couldn't isolate x properly

sirm3d · Answer

$$(y-C)x^2-x-2(y-C)=0$$
compare it with $$ax^2+bx+c=0$$ you can see that the coefficient of $x^2$ is $(y-c)$, the coefficient of $x$ is $-1$, and the constant in the equation is $-2(y-C)$

wcrmelissa2001 · Answer

ohhhhhhhhhh wow. I actually get it. THANK YOU :D

sirm3d · Answer

you're welcome

wcrmelissa2001 · Answer

:D