# Solving -4 = x(x^2 + 1)

• Sep 10th 2009, 12:30 PM
crazydrclaw
Solving -4 = x(x^2 + 1)
Ok, so I have a bit of a quandry. I was trying to find the x-intercept(s) of y = x + [4/(x^2 + 1)]. After setting y = 0, I get the equation -4 = x(x^2 + 1). That's where the problem sets in. I suddenly realized that I have no idea how to solve this. Does anybody else have an idea? I can put the equation in a calculator and get a good approximation, but I'd like a conceptual way of getting at the solution if possible, as I'm now really interested in this problem.

Thanks so much!

James
• Sep 10th 2009, 02:07 PM
Matt Westwood
Looks like you got yourself a *cubic*.

There is a standard algorithmic way of solving this, but it's unwieldy and long-winded (although very pretty).

Your best way of tackling one of these is (assuming you're given one of these as an exercise at elementary level) is seeing whether there's a factor.

That is, given an equation in the form:
$x^3 + a x^2 + b x + c = 0$
(which is effectively what you've got when you rearrange your equation above)

... you want to try and get it in the form:
$(x + p)(x^2 + qx + r) = 0$

Any of this familiar? If not, you've got a way to go before you're up to this - you will need to struggle through solving quadratics (which is like cubics but not as bad).
• Sep 10th 2009, 03:14 PM
crazydrclaw
I'm up to the task of factoring. Unfortunately, I don't think this one does. One solution is x=0 (which I found by factoring out that first x.) The other solution (the calculator showed me there was only one other) is irrational.

I can employ "completing the square" when I have to find solutions to quadratics that don't factor; are there any similar methods for cubics?

Thanks so much for the help! :)
• Sep 10th 2009, 03:20 PM
crazydrclaw
Ok, I'm an idiot :-P I said x=0, but that's only if the equation is set to equal zero, which it's not here. Nevermind. So yeah, still stuck since the only solution the calculator gave me was irrational :-P

And yeah, there was only one solution, not two.

Thanks again!

James
• Sep 10th 2009, 06:11 PM
11rdc11
I was going to say use rational zero theorem and Descarte's rule of signs but your calculator already told you it irrational. Using Descarte's rule of signs you can see there is only 1 negative real zero.
• Sep 10th 2009, 09:39 PM
Matt Westwood
Quote:

Originally Posted by crazydrclaw
I'm up to the task of factoring. Unfortunately, I don't think this one does. One solution is x=0 (which I found by factoring out that first x.) The other solution (the calculator showed me there was only one other) is irrational.

I can employ "completing the square" when I have to find solutions to quadratics that don't factor; are there any similar methods for cubics?

Thanks so much for the help! :)

Yes there is. Hang onto your hat:

Cardan's Formula - ProofWiki

... and there's also one for quartics:

Ferrari's Method - ProofWiki

... but there is no such formula for higher order polynomials.
• Sep 11th 2009, 08:13 AM
crazydrclaw
Thanks!
• Sep 11th 2009, 08:56 AM
CaptainBlack
Quote:

Originally Posted by crazydrclaw
Ok, so I have a bit of a quandry. I was trying to find the x-intercept(s) of y = x + [4/(x^2 + 1)]. After setting y = 0, I get the equation -4 = x(x^2 + 1). That's where the problem sets in. I suddenly realized that I have no idea how to solve this. Does anybody else have an idea? I can put the equation in a calculator and get a good approximation, but I'd like a conceptual way of getting at the solution if possible, as I'm now really interested in this problem.

Thanks so much!

James

Descartes' rule of signs tells you that this has one negative root and zero positive roots.

Now you can find its approximate value numerically (it is easy enough to show the root is between -1 and -2, ..)

CB