# Solving a Quartic

• June 2nd 2011, 04:56 AM
theleontan
Solving a Quartic
Solve the equation x^4 - 2x^2 +1 = 0 for x, where x is real. What on earth do i do? How do i start!
• June 2nd 2011, 05:06 AM
skeeter
Quote:

Originally Posted by theleontan
Solve the equation x^4 - 2x^2 +1 = 0 for x, where x is real. What on earth do i do? How do i start!

note that the quartic will factor ...

$x^4 - 2x^2 + 1 = (x^2 - 1)^2$
• June 2nd 2011, 05:23 AM
Prove It
If you were not able to recognise what Skeeter posted straight away, notice that if you let $\displaystyle X = x^2$, then your equation becomes

$\displaystyle X^2 - 2X + 1 = 0$

which is an easy quadratic to solve. Once you have $\displaystyle X$ you can find $\displaystyle x$.
• June 2nd 2011, 05:05 PM
theleontan
Quote:

Originally Posted by Prove It
If you were not able to recognise what Skeeter posted straight away, notice that if you let $\displaystyle X = x^2$, then your equation becomes

$\displaystyle X^2 - 2X + 1 = 0$

which is an easy quadratic to solve. Once you have $\displaystyle X$ you can find $\displaystyle x$.

I still dont get how you went from my equation to what you have up there ^.
Thanks
• June 2nd 2011, 05:37 PM
Wilmer
Quote:

Originally Posted by theleontan
Solve the equation x^4 - 2x^2 +1 = 0 for x, where x is real.

Let's try again...
let u = x^2 ; then your equation becomes: u^2 - 2u + 1 = 0
Are you saying you can't follow that?
• June 2nd 2011, 05:37 PM
scounged
Quote:

Originally Posted by theleontan
I still dont get how you went from my equation to what you have up there ^.
Thanks

Let's look at your original equation:

$x^4 - 2x^2 +1 = 0$

note that $x^4=(x^2)^2$

This means that your original equation can be rewritten as

$(x^2)^2- 2(x^2) +1 = 0$

What Prove It did was to substitute $x^2$ for $X$.
If you substitute $x^2$ for $X$ in the equation above,
you will get this:

$(X)^2- 2(X) +1 = 0$

This equation can be solved for $X$ with the quadratic formula, as it's a quadratic equation with $X$ as an unknown, and as we know that $X$ stands for $x^2$, this means that If we know $X$ we can get that $x=\pm \sqrt{X}$, and then solve for $x$ when we have figured out $X$.
• June 3rd 2011, 04:35 AM
jgv115
This is a longer method but the more ways you know how to do something the better! Alternatively, the factor theorem could be used.

Subbing x = 1 into the quartic gives you 0 so x-1 is a factor. Using long division we can get a cubic answer. Using the factor theorem again on that will get you a factorisable quadratic.

Then you will have 4 factors!