#1**+5 **

Well...maybe we can.....

y=(-6/y)²-1 simpify

y = 36/y^2 - 1 multiply through by y^2

y^3 = 36 - y^2

And using the onsite solver, we have one real solution and two non-real solutions...

$${{\mathtt{y}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{36}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{2}}\\

{\mathtt{y}} = {{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{2}}\\

{\mathtt{y}} = {\mathtt{3}}\\

\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{2.828\: \!427\: \!124\: \!746\: \!190\: \!1}}{i}\\

{\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.828\: \!427\: \!124\: \!746\: \!190\: \!1}}{i}\\

{\mathtt{y}} = {\mathtt{3}}\\

\end{array} \right\}$$

CPhill
Feb 13, 2015

#2**+5 **

y=(-6/y)²-1

$$\\y=\left(\frac{-6}{y}\right)^2-1\\\\

y=\frac{36}{y^2}-1\\\\

y^3=36-y^2\qquad y\ne 0\\\\

y^3+y^2-36=0\\\\$$

now I am going to use reaminder theorum to look for a factor

If y=3 then 27+9-36=0 great (y-3) is a factor and 3 is a root

$$\\(y^3+y^2-36)\div (y-3)=y^2+4y+12 \qquad $I used polynomial division to get this$\\\\

so\\

y^3+y^2-36=(y-3)(y^2+4y+12)\\\\\\

$Now find roots of $y^2+4y+12\\\\

\triangle = 16-48<0\qquad $therefore there are no real roots to this$\\\\

so\\

$The only real solution is $y=3$$

check

Melody
Feb 13, 2015