How do I solve this equation; $x^4+4x-1=0$?

Question

I know that I can easly solve this with the $4$th degree's equation, but isn't there a smarter way? It is an olympiad's problem so it shouldn't be a formula, but more find a formula... I have remarked that it is equal to $(x^2+1)\cdot(x+1)\cdot(x-1)+4x=0$ It isn't just solving the equation, I want to know if it can be solved otherwise than that... Could someone explain why this is equal to $(x^2+1)^2-(\sqrt{2}(x-1))^2$ please ?

Answer 1

You may write it as $$ x^4+4x-1=(x^2+1)^2-2(x-1)^2 = (x^2+1)^2-(\sqrt{2}(x-1))^2$$ and factorize.

Answer 2

If you don't want to rely on tricks (tricks are always useful though), a more systematic approach is as follows. Let $$x^4+4x-1 = (x^2+ax+b)(x^2+cx+d)$$ We then have \begin{align*} a+c & = 0\\ b+d+ac & = 0\\ bc + ad & = 4\\ bd & = -1 \end{align*} Eliminating $c$, we obtain \begin{align*} b+d & = a^2\\ d-b & = \dfrac4{a}\\ bd & = -1 \end{align*} This gives us $d = \dfrac{a^2}2 + \dfrac2a$ and $b = \dfrac{a^2}2 - \dfrac2a$. Since $bd=-1$, we obtain $$\dfrac{a^4}4 - \dfrac4{a^2} = -1 \implies a^6 - 16 = -4a^2$$ Setting $a^2=y$, we have $$y^3+4y - 16 = 0$$ By rational root theorem, we find that $y=2$ as one root. The other two roots are complex and hence we are not interested in them. This gives us $a=\pm \sqrt2$. Taking $a=\sqrt2$, we see $c=-\sqrt2$. This gives us $$d=1+\sqrt2 \text{ and }b = 1-\sqrt2$$ Hence, we obtain the factorization as $$x^4+4x-1 = (x^2+\sqrt2 x+1-\sqrt2)(x^2-\sqrt2 x+1+\sqrt2)$$

Answer 3

There is a tricky way since our polynomial can be easily written as a difference of two squares: $$ (x^2+1)^2-2(x-1)^2 = \left[x^2+x\sqrt{2}+(1-\sqrt{2})\right]\cdot \left[x^2-x\sqrt{2}+(1+\sqrt{2})\right]\tag{1}$$ hence by the quadratic formula the roots are given by $$ -\frac{1}{\sqrt{2}}\pm\sqrt{\sqrt{2}-\frac{1}{2}},\qquad \frac{1}{\sqrt{2}}\pm i\sqrt{\sqrt{2}+\frac{1}{2}}.\tag{2} $$