Hello.. I'm not sure whether this should be posted under here:

What is the standard method to solve for the following equation:

$\displaystyle 4x^4-x-9=0$

without using software.

Thanks!

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- Apr 15th 2009, 11:59 PMnoob mathematiciansolving equation
Hello.. I'm not sure whether this should be posted under here:

What is the standard method to solve for the following equation:

$\displaystyle 4x^4-x-9=0$

without using software.

Thanks! - Apr 16th 2009, 12:09 AMADARSH
Follow this.

http://en.wikipedia.org/wiki/Quartic_equation - Apr 16th 2009, 12:13 AMTwig
Are you sure itīs not supposed to be $\displaystyle 4x^{4} -x^{2} -9 = 0 $

That would simplify things.. - Apr 16th 2009, 12:38 AMnoob mathematician
That's the equation.. Thanks for the link too.. but it's so difficult.. wondering anyone with the software can help me with the 4 roots?

- Apr 16th 2009, 01:51 AMTwig
hi

Did u mean its supposed to be an $\displaystyle x^2 $ term?

In this case:

let $\displaystyle x^2 = t $

Giving us: $\displaystyle 4t^2 -t -9 = 0 $

Which can be written: $\displaystyle (t-\frac{1}{8})^{2} = \frac{145}{8} $

Solving gives: $\displaystyle t_{1} = \frac{1+\sqrt{145}}{8} $

$\displaystyle t_{2} = \frac{1-\sqrt{145}}{8} $

But we have $\displaystyle x^2 = t $, so two solutions are:

$\displaystyle \sqrt{\frac{1+\sqrt{145}}{8}} \mbox{ and } -\sqrt{\frac{1+\sqrt{145}}{8}} $

You will also get two imaginary roots, but I donīt write them out here...

Btw: If you DID mean $\displaystyle 4x^4-x-9 = 0$, Mathematica gives me a HORRIBLE answer...