(I've written this topic specifically for students taking MEI FP1.)
Observe that any quadratic equation can be written in the following form
$$ (x-\alpha)(x-\beta) = 0 $$
Where $\alpha$ and $\beta$ are the quadratic's roots.
Multiply out the left hand side and you get
$$ x^{2} - (\alpha+\beta)x+\alpha\beta = 0 $$
Now observe the general quadratic equation with real coefficients $a$, $b$, and $c$
$$ ax^{2}+bx+c = 0$$
Dividing through by $a$ you get
$$ x^{2} + \frac{b}{a}x+\frac{c}{a} = 0$$
Comparing coefficients with the equation formed by the roots
$$ \begin{align} \alpha+\beta &= -\frac{b}{a} \\ \alpha\beta &= \frac{c}{a} \end{align} $$
We can see that there is a clear relationship between the coefficients of a quadratic and its roots
$$ x^{2} -(\textrm{sum of roots})x + (\textrm{product of roots}) = 0 $$
Q) Given that $\alpha$ and $\beta$ are the roots of $x^{2}+x-12$, form a new quadratic with roots $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$.
A) From the given quadratic we see that
$$ \begin{align} \alpha+\beta &= -1 \\ \alpha\beta &= -12 \end{align} $$
So by summing the new roots we get
$$ \begin{align} \alpha+\frac{1}{\beta}+\beta+\frac{1}{\alpha} &= \alpha+\beta+\frac{\alpha+\beta}{\alpha\beta} \\ &= -1 + \frac{-1}{-12} \\ &= -\frac{11}{12} \end{align} $$
And by multiplying the new roots together we get
$$ \begin{align} \left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right) &= \alpha\beta+2+\frac{1}{\alpha\beta} \\ &= -12 +2 + \frac{1}{-12} \\ &= -\frac{121}{12} \end{align} $$
Therefore the new quadratic is $12x^{2}+11x-121$.
Q) Given that $\alpha$ and $\beta$ are the roots of $10x^{2}+x-3$, form a new quadratic with roots $2\alpha-1$ and $2\beta-1$.
A) From the given quadratic we see that
$$ \begin{align} \alpha+\beta &= -\frac{1}{10} \\ \alpha\beta &= -\frac{3}{10} \end{align} $$
So by summing the new roots we get
$$ \begin{align} 2\alpha-1+2\beta-1 &= 2(\alpha+\beta)-2 \\ &= 2\left(-\frac{1}{10}\right)-2 \\ &= -\frac{11}{5} \end{align} $$
And by multiplying the new roots together we get
$$ \begin{align} (2\alpha -1)(2\beta -1) &= 4\alpha\beta-2(\alpha+\beta)+1 \\ &= 4\left(-\frac{3}{10} \right)-2\left(-\frac{1}{10} \right)+1 \\ &= 0 \end{align} $$
Therefore the new quadratic is $5x^{2}+11x$.