solving equation

**noob mathematician** · April 15th 2009, 11:59 PM

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

What is the standard method to solve for the following equation:
$4x^4-x-9=0$
without using software.

Thanks!

**ADARSH** · April 16th 2009, 12:09 AM

Follow this.
http://en.wikipedia.org/wiki/Quartic_equation

**Twig** · April 16th 2009, 12:13 AM

Are you sure it´s not supposed to be $4x^{4} -x^{2} -9 = 0$
That would simplify things..

**noob mathematician** · April 16th 2009, 12:38 AM

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?

**Twig** · April 16th 2009, 01:51 AM

hi

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

In this case:

let $x^2 = t$

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

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

Solving gives: $t_{1} = \frac{1+\sqrt{145}}{8}$
$t_{2} = \frac{1-\sqrt{145}}{8}$

But we have $x^2 = t$ , so two solutions are:
$\sqrt{\frac{1+\sqrt{145}}{8}} \mbox{ and } -\sqrt{\frac{1+\sqrt{145}}{8}}$

You will also get two imaginary roots, but I don´t write them out here...

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

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