Efficient Solutions for Fractional Polynomial Equations

[ACLerok]

Hi, I was just reading up on some material for one of my exams and came across something. I realize this may seem like a trivial, obvious question but:
What is the easiest way to find the solutions of the polynomial (for example):
(x^2) - (1/6)x - 1/6 = 0
Other than guessing and testing for the correct values of x, what is the quickest and most efficient method of solving for x when the coefficients of each term is a fraction rather than a whole number? Thanks in advance.

 

[Gib Z]

well you just let your fraction, 1/6, equate to b, as the equation was a quadratic of the form ax^2 + bx + c = 0. Then substitute your values for a, b and c into the quadratic formula, x=\frac {-b(+-)(b^2 - 4ac)^0.5}{2a} If you mean in general, you it is quite hard to solve equations of degree 3 and 4 analytically, and after 5 it can only be approximated using methods such as Newtons Method.

 

[Edgardo]

Given the equation

x^2 + px + q = 0

the solutions are:

x_{1} = - \frac{p}{2} - \sqrt{ \left( \frac{p}{2} \right)^2-q }

and

x_{2} = - \frac{p}{2} + \sqrt{ \left( \frac{p}{2} \right)^2-q }

So in your example:

p=-1/6 and q=-1/6

 

[ACLerok]

Thanks a lot!

 

[ACLerok]

one more thing: if I am trying to find the solutions if the polynomial is of order 3, for example:
x^3 + 0.5x^2 - 0.25x - 0.125 = 0

Do I add 0.125 to both sides then factor an x or do must I do something else? Thanks

 

[murshid_islam]

see cardano's method here:
http://en.wikipedia.org/wiki/Cubic_equation
http://mathworld.wolfram.com/CubicFormula.html

 

[ACLerok]

Wow what a long and involved process
There is no other process of finding them? oh well
Thanks

 

[nworm]

You can try to find (adjust) one solution (x=0.5). After that you can divide x^3 + 0.5x^2 - 0.25x - 0.125 by x-0.5.
The result is x^2+x+0.25.
So x^3 + 0.5x^2 - 0.25x - 0.125=(x-0.5)(x^2+x+0.25)=0
You already can solve it.

 

[arildno]

Wow what a long and involved process
There is no other process of finding them? oh well
Thanks

Not all problems easily stated are easily solved.

Most happen to be insoluble, I'm afraid.

 

[nworm]

arildno
I don't think that ACLerock solve very difficult problems.
It is possible to apply more easy methods when solving home and exam tasks.
(for example the method from my post).