1. ## 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!

2. 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 ...

$\displaystyle x^4 - 2x^2 + 1 = (x^2 - 1)^2$

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

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

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

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

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

which is an easy quadratic to solve. Once you have $\displaystyle \displaystyle X$ you can find $\displaystyle \displaystyle x$.
I still dont get how you went from my equation to what you have up there ^.
Thanks

5. 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?

6. 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:

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

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

This means that your original equation can be rewritten as

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

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

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

This equation can be solved for $\displaystyle X$ with the quadratic formula, as it's a quadratic equation with $\displaystyle X$ as an unknown, and as we know that $\displaystyle X$ stands for $\displaystyle x^2$, this means that If we know $\displaystyle X$ we can get that $\displaystyle x=\pm \sqrt{X}$, and then solve for $\displaystyle x$ when we have figured out $\displaystyle X$.

7. 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!