Question

Solve the following equations:
$x^4+\frac{8}{9}{x}^2+1=3x^3+3x$.

Answer 1

$x^4+\frac{8}{9}{x}^2+1={3x}^3+3x$

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

$\frac{1+\frac{8}{9}{x}^2+x^4}{x^4}=\frac{3+3x^2}{x^3}1+\frac{8}{9}{x}^2+x^4=x(3+3x^2)1+\frac{8}{9}{x}^2+x^4=3x+3x^3$

On replacing $x$ by $\frac{1}{x}$ we saw thatthe equation becomes the same, so the roots of the equation are of the form

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

$\alpha +\frac{1}{\alpha }+\beta +\frac{1}{\beta }=-\frac{b}{a}=-\frac{-3}{1}=3\alpha +\frac{1}{\alpha }+\beta +\frac{1}{\beta }=\mathrm{3........}\left(i\right)\alpha .\frac{1}{\alpha }+\alpha .\beta +\alpha .\frac{1}{\beta }+\frac{1}{\alpha }.\beta +\frac{1}{\alpha \beta }+\beta .\frac{1}{\beta }=\frac{c}{a}=\frac{8}{9}\alpha \beta +\frac{1}{\alpha \beta }+\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+2=\frac{8}{9}\alpha \beta +\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+\frac{1}{\alpha \beta }=\frac{8}{9}-2=-\frac{10}{9}\alpha (\beta +\frac{1}{\beta })+\frac{1}{\alpha }(\beta +\frac{1}{\beta })=-\frac{10}{9}(\alpha +\frac{1}{\alpha })(\beta +\frac{1}{\beta })=-\frac{10}{9}$

Using $\left(i\right)$

$(\alpha +\frac{1}{\alpha })(3-(\alpha +\frac{1}{\alpha }))=-\frac{10}{9}$

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

$t(3-t)=-\frac{10}{9}27t-9t^2=-109t^2-27t-10=09t^2+3t-30t-10=03t(3t+1)-10(3t+1)=0(3t-10)(3t+1)=0t=\frac{10}{3},-\frac{1}{3}$

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

For $t=\frac{10}{3}$

$\alpha +\frac{1}{\alpha }=\frac{10}{3}\frac{{\alpha }^2+1}{\alpha }=\frac{10}{3}3{\alpha }^2-10\alpha +3=03{\alpha }^2-9\alpha -\alpha +3=03\alpha (\alpha -3)-1(\alpha -3)=0(3\alpha -1)(\alpha -3)=0\Rightarrow \alpha =\frac{1}{3},3$

For $t=-\frac{1}{3}$

$\frac{{\alpha }^2+1}{\alpha }=-\frac{1}{3}3{\alpha }^2+\alpha +3=0\alpha =\frac{-1\pm \sqrt{1-4\left(3\right)\left(3\right)}}{2\left(3\right)}=\frac{-1\pm \sqrt{-35}}{6}$

So the roots of the equation are $3,\frac{1}{3},\frac{-1\pm \sqrt{-35}}{6}$