Please teach me how to do part b(iii).
note that $\displaystyle (\alpha - \beta)^2 = \alpha^2 - 2 \alpha \beta + \beta^2$. with this, you can find $\displaystyle \alpha - \beta$ by square rooting both sides.
but how to find $\displaystyle \alpha^2 - 2 \alpha \beta + \beta^2$?
note that $\displaystyle (\alpha + \beta)^2 = \alpha^2 + 2 \alpha \beta + \beta^2$
$\displaystyle \Rightarrow (\alpha + \beta)^2 - 4 \alpha \beta = \color{red}\alpha^2 - 2 \alpha \beta + \beta^2$
and you know $\displaystyle \alpha + \beta$ as well as $\displaystyle \alpha \beta$ from part b(i)
lets assume we are going to do this without using the quadratic formula.
$\displaystyle (\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha \beta=(\alpha+\beta)^2-4\alpha \beta $
so:
$\displaystyle \alpha-\beta=\sqrt{(\alpha+\beta)^2-4\alpha \beta} $
and we know that $\displaystyle \alpha+\beta=k=7$ and that $\displaystyle \alpha \beta=2k-3=11$
CB