Question

Solve the following equations:
$x^4+1-3(x^3+x)=2x^2$.

Answer 1

$x^4+1-3(x^3+x)={2x}^2{x}^4-3x^3-2x^2-3x+1=0$

Replace $x$ by $\frac{1}{x}$

$\frac{1}{x^4}-\frac{3}{x^3}-\frac{2}{x^2}-\frac{3}{x}+1=0\frac{1-3x-2x^2-3x^3+x^4=0}{x^4}1-3x-2x^2-3x^3+x^4=0\Rightarrow x^4-3x^3-2x^2-3x+1=0$

We saw that the equation become same by replacing $x$ by $\frac{1}{x}$ . So the roots of the equation are of the form

$\alpha ,\frac{1}{\alpha },\beta ,\frac{1}{\beta }$

$\alpha +\frac{1}{\alpha }+\beta +\frac{1}{\beta }=-\frac{b}{a}=-\frac{-3}{1}\alpha +\frac{1}{\alpha }+\beta +\frac{1}{\beta }=\mathrm{3...........}\left(i\right)\alpha .\frac{1}{\alpha }+\alpha .\beta +\alpha .\frac{1}{\beta }+\frac{1}{\alpha }.\beta +\frac{1}{\alpha \beta }+\beta .\frac{1}{\beta }=\frac{c}{a}=\frac{-2}{1}\alpha \beta +\frac{1}{\alpha \beta }+\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+2=-2\alpha \beta +\frac{1}{\alpha \beta }+\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=-4(\alpha +\frac{1}{\alpha })(\beta +\frac{1}{\beta })=-4$

Using $\left(i\right)$
$(\alpha +\frac{1}{\alpha })(3-(\alpha +\frac{1}{\alpha }))=-4$

Put $\alpha +\frac{1}{\alpha }=t$

$t(3-t)=-4t^2-3t-4=0t^2-4t+t-4=0t(t-4)+1(t-4)=0(t+1)(t-4)=0t=-1,4$

Now $\alpha +\frac{1}{\alpha }=t$

For $t=-1$

$\alpha +\frac{1}{\alpha }=-1\frac{{\alpha }^2+1}{\alpha }=-1{\alpha }^2+1+\alpha =0\alpha =\frac{-1\pm \sqrt{1-4\left(1\right)\left(1\right)}}{2}=\frac{-1\pm \sqrt{-3}}{2}$

For $t=4$

$\alpha +\frac{1}{\alpha }=4\alpha +\frac{1}{\alpha }=4{\alpha }^2+1-4\alpha =0\alpha =\frac{4\pm \sqrt{16-4\left(1\right)}}{2}=\frac{4\pm 2\sqrt{3}}{2}=2\pm \sqrt{3}$

So the values of $x$ are $\frac{-1\pm \sqrt{-3}}{2},2\pm \sqrt{3}$