Find a Quadratic Equation with roots of... I was given this problem at school to look at home as a challenge, after spending a good 2 hours on this I can't seem to get further than the last part of the equation. I'd love to see the way to get through 2) before tomorrow's lesson as a head start.
So the problem is as follows:
1) Quadratic Equation $$2x^2 + 8x + 1 = 0$$ 
i. Find roots $$\alpha + \beta$$
ii. Find roots $$\alpha\beta$$
2) Find an Equation with integer coefficients who's roots are:
$$2\alpha^4+\frac{1}{\beta^2}$$$$2\beta^4+\frac{1}{\alpha^2}$$
I'm completely puzzled on the second part of the question and I've tried following the method I was taught. Sorry if formatting is a bit off, first time posting here :) 
Thanks in advance for any help!
 A: Given $\displaystyle \alpha,\beta$ are the roots of $2x^2+8x+4=0.$ 
So$\displaystyle \alpha+\beta = -\frac{8}{2}=-4$ and $\displaystyle \alpha\cdot \beta = \frac{1}{2}.$
Now for Second part, Using $\bullet\; \bf{x^2-(sum \; of \; roots)x+(product\; of \; roots) =0}$
So here $\displaystyle \bf{sum\; of \; roots } = 2\alpha^4+\frac{1}{\beta^2}+2\beta^4+\frac{1}{\alpha^2} = 2\left[\alpha^4+\beta^4\right]+\frac{1}{\alpha^2}+\frac{1}{\beta^2}$
So we get $\displaystyle = 2\left[(\alpha^2+\beta^2)^2-2(\alpha\cdot \beta)^2\right]+\frac{(\alpha+\beta)^2-2\alpha\cdot \beta}{(\alpha\cdot \beta)^2}$
$\displaystyle = 2\left[\left\{(\alpha+\beta)^2-2\alpha\cdot \beta\right\}^2-2(\alpha\cdot \beta)^2\right]+\frac{(\alpha+\beta)^2-2\alpha\cdot \beta}{(\alpha\cdot \beta)^2} = $
and $\displaystyle \bf{product\; of roots} = \left(2\alpha^4+\frac{1}{\beta^2}\right)\times \left(2\beta^4+\frac{1}{\alpha^2}\right)$
$\displaystyle  = 4(\alpha\cdot \beta)^4+2\left[(\alpha+\beta)^2-2\alpha\cdot \beta\right]+\frac{1}{(\alpha\cdot \beta)^2}=$
A: 1) Quadratic Equation $$2x^2 + 8x + 1 = 0$$ 
i.  $$\alpha + \beta=\frac{-b}{a}=\frac{-8}{2}=-4$$
ii.  $$\alpha\beta=\frac{c}{a}=\frac{1}{2}$$
For remain just find roots and use this fact that if $x_1+x_2=s$ and $x_1 x_2=p$ equation will be $x^2 -sx+p=0$.  
A: Hint:
You must calculate $S=\alpha^4+\dfrac1{\beta^2}+\beta^4+\dfrac1{\alpha^2}$ and $P=\Bigl(\alpha^4+\dfrac1{\beta^2}\Bigr)\Bigl(\beta^4+\dfrac1{\alpha^2}\Bigr)$. They will be the roots of the quadratic equation $\;x^2-Sx+P=0$.
Now any symmetric rational function of $\alpha$  and $\beta$, by a theorem of Newton, can be expressed as a rational function of the elementary symmetric functions $s=\alpha+\beta$ and $p=\alpha\beta$:
\begin{align*}
\alpha^2+\beta^2&=s^2-2p,\enspace\text{hence}\quad\frac1{\alpha^2}+\frac1{\beta^2}=\frac{s^2-2p}{p^2},\\
\alpha^4+\beta^4&=\bigl(\alpha^2+\beta^2)^2-2\alpha^2\beta^2=(s^2-2p)^2-2p^2=s^2-4ps^2+2p^2
\end{align*}
whence $S$.
Similarly:
$$P=\alpha^4\beta^4+\alpha^2+\beta^2+\frac1{\alpha^2\beta^2}=p^4+s^2-2p+\frac1{p^2}.$$
