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Blog Post 51
EpicSkills32   1
N Feb 2, 2014 by briantix
$\ [\text{Blog Post 51}] $

So a few blog posts ago, I posted a problem.

Find all solutions to the following:
\[ x^4-4x=1 \]

Oh dear. We have something to the fourth power. What's even worse, there isn't anything cubed or squared. hm....

At first, we see the 1 and the even power, and we do something like:

\[ x^4-1=4x \]
\[ (x^2+1)(x^2-1)=4x \]
\[ (x^2+1)(x+1)(x-1)=4x \]

And then we get stuck.
Stumped?

Instead, let's try to force a nicer polynomial.
We add $\ 2x^2+1 $ to both sides.

\[ x^4+2x^2-4x+1=2x^2+2 \]
\[ x^4+2x^2+1=2x^2+4x+2 \]

Aha! Something that can be factored!

\[ (x^2+1)^2=2(x^2+2x+1) \]
\[ (x^2+1)^2=2(x+1)^2 \]

Oh but wait a second. The term on the left can be changed so that there is no coefficient in front.

\[ 2(x+1)^2 = 2x^2+4x+2 = (\sqrt{2}x+\sqrt{2})^2 \]

Now we can express our equation with two squared terms:

\[ (x^2+1)^2=(\sqrt{2}x+\sqrt{2})^2 \]

Since both sides are squared, we move both terms to one side to get a difference of squares:

\[ (x^2+1)^2-(\sqrt{2}x+\sqrt{2})^2 \]

Expanding:

\[ \left((x^2+1)+(\sqrt{2}x+\sqrt{2})\right)\left((x^2+1)-(\sqrt{2}x+\sqrt2})\right)= 0 \]

\[ (x^2+\sqrt{2}x+\sqrt{2}+1)(x^2-\sqrt{2}-\sqrt{2}+1) = 0 \]

We have two factors that we need to set equal to zero. Since they are just quadratics, we can do so with the quadratic formula:

\[ (x^2+\sqrt{2}x+\sqrt{2}+1)=0 \]

\[ a= 1 \] \[b=\sqrt{2}\] \[c=\sqrt{2}+1\]

\[ \therefore x= \dfrac{-\sqrt{2}\pm\sqrt{2-4(\sqrt{2}+1)}}{2} \]
\[ x= \dfrac{-\sqrt{2}\pm\sqrt{-2-4\sqrt{2}}}{2} \]

We have these ugly negatives under a terrible radical, so we change that:

\[ x= \dfrac{-\sqrt{2}\pm i\sqrt{2+4\sqrt{2}}}{2} \]

Again for the other factor:

\[ (x^2-\sqrt{2}-\sqrt{2}+1)=0 \]

\[ a=1 \] \[b=-\sqrt{2}\] \[c=-\sqrt{2}+1 \]

\[ \therefore x= \dfrac{\sqrt{2}\pm\sqrt{2-4(-\sqrt{2}+1)}}{2} \]
\[ x= \dfrac{-\sqrt{2}\pm\sqrt{-2+4\sqrt{2}}}{2} \]

In the end, our solutions for $\ x^4-4x=1 $ are the following values:

\[ x= \left(\dfrac{-\sqrt{2}\pm i\sqrt{2+4\sqrt{2}}}{2}\right),\left(\dfrac{-\sqrt{2}\pm\sqrt{-2+4\sqrt{2}}}{2}\right) \]

Yay. :)
1 reply
EpicSkills32
Jan 24, 2014
briantix
Feb 2, 2014
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