Quartics are fun!

Try to factor the equation $x^4-4x-1=0$ into the form of the product of two quadratics of the form $(x^2+ax+b)(x^2-ax+c)=0$ . Expand:

$x^4+(b+c-a^2)x^2+a(c-b)x+bc=0$

And compare the coefficients:

$b+c=a^2$

$b-c=\dfrac{4}{a}$

$bc=-1$

Solve for $b$ and $c$ with the first two equations:

$b=\dfrac{a^2+\dfrac{4}{a}}{2}$

$c=\dfrac{a^2-\dfrac{4}{a}}{2}$

Replace it in the third equation and simplify:

$a^6+4a^2-16=0$

By the Rational Root Test, we find that $a^2-2$ is a factor. So,

$(a^2-2)(a^4+2a^2+8)=0$

With this we get that $a=\sqrt{2}$ . Now, obtain $b$ and $c$ :

$b=1+\sqrt{2}$

$c=1-\sqrt{2}$

Hence, $x^4-4x-1=(x^2+\sqrt{2}x+1+\sqrt{2})(x^2-\sqrt{2}x+1-\sqrt{2})$

If we solve each factor with the quadratic formula, we get our four solutions:

$x_{1,2}=\dfrac{-\sqrt{2} \pm i\sqrt{2+4\sqrt{2}}}{2}$

$x_{3,4}=\dfrac{\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}$

Only the last two are real, hence our final answer is $x_{3}+x_{4}=\sqrt{2} \approx 1.4142$

polynomial- x^4+0 x^3+0 x^2-4*x-1=0 MATLAB CODE: roots([1 0 0 -4 -1]) % a bulit-in function, roots(coefficients of the polynomial in descending order of the power of x) this returns- ans =

1.6633
-0.7071 + 1.3836i -0.7071 - 1.3836i -0.2490

taking the real solutions 1.6633 & -0.2490 and adding them results in--> 1.6633-0.2490=1.4143