# Solving a simple equation

• Aug 30th 2009, 08:00 PM
jelloish
Solving 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 PM
Wilmer
Quote:

Originally Posted by jelloish
Solve for x:
x^4-3x^2+2=0.

Let k = x^2
k^2 - 3k + 2 = 0
Solve that for k, then for x
• Aug 30th 2009, 08:10 PM
pickslides
Hi

What is needed here is some simple substitution

you have

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

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

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

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

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

By the null factor law

$a = 1,2$

but we are after $x$ so

$x^2 = 1,2 \Rightarrow x = \pm 1,\pm \sqrt{2}$
• Aug 30th 2009, 08:14 PM
jelloish
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 PM
pickslides
It shouldn't, that is the solution for k.

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

these will work.
• Aug 30th 2009, 08:23 PM
jelloish
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 PM
jelloish
Oh boy... I forgot that it was a root two.

Thanks. hahaha :)