3rd degree equation asked by Asif Nebiyev in the Department of mathematics group

Exercise:

Solve in $\mathbb{R},$ ${{x}^{3}}+2\sqrt{3}\,{{x}^{2}}+3x+\sqrt{3}-1=0$

Solution :Let $p\left( x \right)={{x}^{3}}+2\sqrt{3}\,{{x}^{2}}+3x+\sqrt{3}-1\in \mathbb{R}\left[ x \right]$

So $p\left( x \right)=x\left( {{x}^{2}}+2\sqrt{3}x+3 \right)+\sqrt{3}-1=x{{\left( x+\sqrt{3} \right)}^{2}}+\sqrt{3}-1$

Take $u=x+\sqrt{3}\Leftrightarrow p\left( u \right)=\left( u-\sqrt{3} \right){{u}^{2}}+\sqrt{3}-1$

$\Leftrightarrow {{u}^{3}}-\sqrt{3}{{u}^{2}}+\sqrt{3}-1=0$ $\Leftrightarrow \left( {{u}^{3}}-1 \right)-\sqrt{3}\left( {{u}^{2}}-1 \right)=0$

But ${{u}^{3}}-1=\left( u-1 \right)\left( {{u}^{2}}+u+1 \right)$

So $p\left( u \right)=\left( u-1 \right)\left( {{u}^{2}}+u+1 \right)-\sqrt{3}\left( u-1 \right)\left( u+1 \right)=0$

$\Leftrightarrow \left( u-1 \right)\left( {{u}^{2}}+u+1-\sqrt{3}\left( u+1 \right) \right)=0$

$\Leftrightarrow u=1\,\,\,or\,\,{{u}^{2}}+u\left( 1-\sqrt{3} \right)+1-\sqrt{3}=0$

So we have two cases to be considered .

$if\,\,\,u=1\Leftrightarrow 1=x+\sqrt{3}\Leftrightarrow x=1-\sqrt{3}$

Now in 2nd case we will consider $m=1-\sqrt{3}<0$

So ${{u}^{2}}+um+m=0\Leftrightarrow {{u}^{2}}+2\frac{m}{2}u+\frac{{{m}^{2}}}{4}-\frac{{{m}^{2}}}{4}+m=0\Leftrightarrow {{\left( u+\frac{m}{2} \right)}^{2}}-\frac{{{m}^{2}}-4m}{4}=0$

Hence $u+\frac{m}{2}=\pm \frac{\sqrt{{{m}^{2}}-4m}}{2}\Leftrightarrow u=\frac{-m\pm \sqrt{{{m}^{2}}-4m}}{2}$

Thus $x=u-\sqrt{3}=\frac{-m\pm \sqrt{{{m}^{2}}-4m}-2\sqrt{3}}{2}=\frac{-1+\sqrt{3}\pm \sqrt{-4+4\sqrt{3}+4-2\sqrt{3}}-2\sqrt{3}}{2}$

$x=\frac{-1-\sqrt{3}\pm \sqrt{2\sqrt{3}}}{2}$