how do i solve this guy?

 

thanks

solve what?

 

i assume u mean finding the root of that function

 

it looks like u will need to find the roots of a quartic of a form ax^4 + bx^3 + c = 0

there should be a general formula for that, i dont know it. lemme just try to look for it on the net

well do it this way.

[math]x^{2}(x-1)-\frac{2}{x}=0[/math]

[math]x^{3}-x^{2}=\frac{2}{x}[/math]

[math]x^{4}-x^{3}-2=0[/math]

 

then -1 iby inspection is a root, therefore (x+1) is a factor of f(x). then you can reduce the polynomial into a quintic

 

information on solving quartic

http://mathworld.wolfram.com/QuarticEquation.html

there is also a link at the bottom to solutions to quintic

 

I banged the equation into Maple, and the solutions are horrible. the only nice solutions is -1, the others are massive strings of numbers.

That should have been fairly obvious, as you can see that x +- 2 can't be a factor, so you've straight ruled out all the other integrals.

jakiri i dont understand your last sentence, could u explain in more detail.

I think he meant intervals instead of integrals, but I could be wrong.

thanks.

i could see the -1 as being factor but i thought there might be a way to figure out the other one(s): relatively easily i've looked on the net but i don't seem to be coming up with any answer. :S

well u know there is 1 real root, so there must be another real root, and then two complex roots which are conjugates.

 

here is my screenshot from maple

I think he meant intervals instead of integrals, but I could be wrong.

 

All the other integral roots.

 

You have an equation which ends in a prime. To factorise it into an (x+a)(x+b)(x+....) form, a*b*c*whatever must equal the prime. To get integral values, one of those numbers must be the prime, and the rest must be one, by the definition of a prime number.

 

As it's fairly clear that 2 is not a factor, the solutions therefore can't be nice integrals.

Another useful test in factorising polynomials over rationals, or to test if it has any rational roots would be the Rational Root Test

thanks for your help