(Solved): Q2. (12 marks) Solve the equation \[ y^{-4}+y^{-2}+2 y^{-1}+6=0 \] for \( y \), by using the follo ...

Q2. (12 marks) Solve the equation \[ y^{-4}+y^{-2}+2 y^{-1}+6=0 \] for \( y \), by using the following approach: Letting \( x=y^{-1} \), we write the equation as \[ \left(x^{2}+\frac{1}{2}\right)^{2}=x^{4}+x^{2}+\frac{1}{4}=-2 x-6+\frac{1}{4}=-2 x-\frac{23}{4} . \] Then, let us introduce an unspecified additive term \( z \), so that the above equation can also be written as \[ \left(x^{2}+\frac{1}{2}+z\right)^{2}=\left(x^{2}+\frac{1}{2}\right)^{2}+2 z\left(x^{2}+\frac{1}{2}\right)+z^{2}=-2 x-\frac{23}{4}+2 z x^{2}+z+z^{2} . \] We look for a specific value of \( z \) that makes the right hand side of the equation just obtained a perfect square; that is, \[ 2 z x^{2}-2 x+\left(-\frac{23}{4}+z+z^{2}\right)=\left(\sqrt{2 z} x-\frac{1}{\sqrt{2 z}}\right)^{2} . \] This analytical step requires that \[ -\frac{23}{4}+z+z^{2}=\left(\frac{1}{\sqrt{2 z}}\right)^{2}=\frac{1}{2 z} \text { or, equivalently, } z^{3}+z^{2}-\frac{23}{4} z-\frac{1}{2}=0 . \] It is known that the three roots of the above cubic equation are \[ z=2 \text { and } z=-\frac{3}{2} \pm \sqrt{2} \] Find the four roots of \( y \), by using the above information. If any of the roots of \( y \) is a complex number, express it in the form of \( p+q i \), where \( p \) and \( q \) are real numbers and \( i \) is \( \sqrt{-1} \). (Notice that it is unacceptable to use any online equation solver directly, without using any of the above information. Notice also that results without analytical justification will be assigned zero marks, even if they are correct.)