How many roots does this have

[Xylitol]

2 right?

 

[Xylitol]

my bad it's x^2 not x - aha sorry

 

[ConstipatedVigilante]

Isn't 5? The xy^5 makes it 5th degree, right?

 

[Ticky]

...And you couldn't just use a numeric solver to answer this?


2 real roots.


Edit: Assuming you want roots where y=0.

 

[frostedflakes]

I'm not seeing how you can determine anything from a single equation that has more than one variable. Wouldn't you need a system of equations to solve for the zeros?

 

[Aluvus]

Originally posted by: frostedflakes
I'm not seeing how you can determine anything from a single equation that has more than one variable. Wouldn't you need a system of equations to solve for the zeros?

If you set y=0, you are left with -2x = x^2.

 

[Random Variable]

Originally posted by: Aluvus

Originally posted by: frostedflakes
I'm not seeing how you can determine anything from a single equation that has more than one variable. Wouldn't you need a system of equations to solve for the zeros?

If you set y=0, you are left with -2x = x^2.

Why would you do that? You need to solve f(x,y)=0, where f(x,y)=xy^5-2x-3y-x^2.

 

[Ticky]

Originally posted by: Random Variable

Originally posted by: Aluvus

Originally posted by: frostedflakes
I'm not seeing how you can determine anything from a single equation that has more than one variable. Wouldn't you need a system of equations to solve for the zeros?

If you set y=0, you are left with -2x = x^2.

Why would you do that? You need to solve f(x,y)=0, where f(x,y)=xy^5-2x-3y-x^2.

Wouldn't that have infitely many solutions? Isn't that then the equation of some sort of surface, which, I would bet, intersects the xy-plane on at the very least a line. Also, why are you introducing f(x,y) into this? It's a two dimensional problem, that can't easily be solved for y. He just wanted to know how many roots it had, and I assume he wanted the case where y=0.

 

[Random Variable]

Originally posted by: Ticky

Originally posted by: Random Variable

Originally posted by: Aluvus

Originally posted by: frostedflakes
I'm not seeing how you can determine anything from a single equation that has more than one variable. Wouldn't you need a system of equations to solve for the zeros?

If you set y=0, you are left with -2x = x^2.

Why would you do that? You need to solve f(x,y)=0, where f(x,y)=xy^5-2x-3y-x^2.

Wouldn't that have infitely many solutions? Isn't that then the equation of some sort of surface, which, I would bet, intersects the xy-plane on at the very least a line. Also, why are you introducing f(x,y) into this? It's a two dimensional problem, that can't easily be solved for y. He just wanted to know how many roots it had, and I assume he wanted the case where y=0.

Yes, it would most likely have infinitely many solutions. I guess you could try to find the lines along which it intersects the xy-plane (but even that might be infinite).

If it's not a three dimensional problem, then ignore what I said.