Hello Stuck Man
It looks as if you don't know about this important result:
- If $\displaystyle \alpha$ and $\displaystyle \beta$ are the roots of the quadratic equation $\displaystyle ax^2+bx+c=0$, then:
The sum of its roots, $\displaystyle \alpha + \beta = -\frac{b}{a}$
andThe product of its roots, $\displaystyle \alpha\beta = \frac{c}{a}$
(This is very easy to prove: expand $\displaystyle (x-\alpha)(x-\beta)$ and compare it to $\displaystyle x^2 + \frac{b}{a}x + \frac{c}{a}$.)
Well then, if $\displaystyle \alpha$ and $\displaystyle \beta$ are the roots of $\displaystyle 3x^2 +x + 2 = 0$, we get:$\displaystyle \alpha + \beta = -\frac{1}{3}$
and$\displaystyle \alpha\beta = \frac{2}{3}$
So for part (i):$\displaystyle \frac{1}{\alpha^2}+\frac{1}{\beta^2} =\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}$$\displaystyle = \frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}$
Now substitute in the values of $\displaystyle \alpha + \beta$ and $\displaystyle \alpha\beta$, simplify and the job's done. (I get the answer $\displaystyle -\frac{11}{4}$. Do you agree?)
****************
Part (ii)
We know the value of $\displaystyle \frac{1}{\alpha^2}+\frac{1}{\beta^2}$ and clearly $\displaystyle \frac{1}{\alpha^2}\times\frac{1}{\beta^2}= \frac{1}{(\alpha\beta)^2}= \frac94$. So we can use the result in reverse to write down the new equation, knowing the sum and product of its roots; i.e.$\displaystyle x^2 - \text{(sum of roots)}x+ \text{(product of roots)}=0$
which (if my arithmetic is correct) is:$\displaystyle x^2 +\frac{11}{4}x+\frac94=0$
i.e. $\displaystyle 4x^2+11x+9=0$
****************
Part (iii)
$\displaystyle \alpha$ is a root of $\displaystyle 3x^2+x+2=0$. So:$\displaystyle 3\alpha^2+\alpha+2=0$
$\displaystyle \Rightarrow 3\alpha^2 = -\alpha-2$ (1)
$\displaystyle \Rightarrow 9\alpha^4 =(-\alpha-2)^2$$\displaystyle =\alpha^2 +4\alpha+4$
$\displaystyle \Rightarrow 27\alpha^4 = 3\alpha^2 +12\alpha+12$$\displaystyle =-\alpha - 2 +12 \alpha +12$ from (1)
$\displaystyle = 11\alpha+10$
Grandad