Quadratic Equations, SPM Practice (Paper 2)

Question 4:
It is given α and β are the roots of the quadratic equation x (x – 3) = 2k – 4, where k is a constant.
$\begin{array}{l}\text{(a) Find the range of values of }k\text{ if }\alpha \ne \beta \text{.}\\ \text{(b) Given }\frac{\alpha }{2}\text{ and }\frac{\beta }{2}\text{ are the roots of another quadratic equation}\\ 2x^2+tx-4=0,\text{ where }t\text{ is a constant, find the value of }t\text{ and of }k\text{.}\end{array}$

Solution:
$\begin{array}{l}\text{(a)}\\ x\left(x-3\right)=2k-4\\ x^2-3x+4-2k=0\\ a=1,b=-3,c=4-2k\\ b^2-4ac>0\\ {\left(-3\right)}^2-4\left(1\right)\left(4-2k\right)>0\\ 9-16+8k>0\\ 8k>7\\ k>\frac{7}{8}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{From the equation }{x}^2-3x+4-2k=0,\\ \alpha +\beta =-\frac{b}{a}\\ =-\frac{-3}{1}\\ =\mathrm{3.............}\left(1\right)\\ \alpha \beta =\frac{c}{a}\\ =\frac{4-2k}{1}\\ =4-2k\mathrm{.............}\left(2\right)\\ \\ \text{From the equation 2}{x}^2+tx-4=0,\\ \frac{\alpha }{2}+\frac{\beta }{2}=-\frac{t}{2}\\ \alpha +\beta =-t\mathrm{.............}\left(3\right)\\ \frac{\alpha }{2}\times \frac{\beta }{2}=-\frac{4}{2}\\ \alpha \beta =-\mathrm{8.............}\left(4\right)\\ \\ \text{Substitute }(1)=(3),\\ 3=-t\\ t=-3\\ \text{Substitute }(2)=(4),\\ 4-2k=-8\\ 4+8=2k\\ k=6\end{array}$