My friend asked me for math help for her pre-calc class and I can't for the life of me do the problem--- I'm out of practice.

Solve for x:

x^4-3x^2+2=0.

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- Aug 30th 2009, 08:00 PMjelloishSolving a simple equation
My friend asked me for math help for her pre-calc class and I can't for the life of me do the problem--- I'm out of practice.

Solve for x:

x^4-3x^2+2=0. - Aug 30th 2009, 08:08 PMWilmer
- Aug 30th 2009, 08:10 PMpickslides
Hi

What is needed here is some simple substitution

you have

$\displaystyle x^4-3x^2+2 = 0$

Replace $\displaystyle x^2$ with a dummy variable, lets call it $\displaystyle a$

Now $\displaystyle x^4-3x^2+2 = 0$ becomes $\displaystyle a^2-3a+2 = 0$

This is a simple quadratic

$\displaystyle a^2-3a+2 = 0$

$\displaystyle (a-2)(a-1)= 0$

By the null factor law

$\displaystyle a = 1,2$

but we are after $\displaystyle x$ so

$\displaystyle x^2 = 1,2 \Rightarrow x = \pm 1,\pm \sqrt{2}$ - Aug 30th 2009, 08:14 PMjelloish
Okay so when solving it that way I got K being -2 and -1, which then makes x= 2 and 1. However when put back into the equation it doesn't solve it correctly.

Then again I'm doing this all in my head and without pen+paper. - Aug 30th 2009, 08:18 PMpickslides
It shouldn't, that is the solution for k.

My post gives a detailed answer, try the solutions $\displaystyle x = \pm 1,\pm \sqrt{2}$

these will work. - Aug 30th 2009, 08:23 PMjelloish
I still tried those and they aren't working.

When substituting 2 (for all intents and purposes it works for the -2 too because of the squaring...) the equation ends up being

2^4 - 3(2^2) +2 which is..

16-12+2

which is 6.

The 1, -1 do work, though. My brains are frieeeeed. - Aug 30th 2009, 08:23 PMjelloish
Oh boy... I forgot that it was a root two.

Thanks. hahaha :)