Show quadratic equation has two distinct real roots. $x^2 - (5-k) x + (k+2) = 0  $ has two distinct real roots.
So, in the markscheme of this question, they take the discriminant ($-b^2 + 4ac$) and say it is greater than 0. That is, $( (-(5-k)^2 - 4(1)(k+2) > 0)$. 
Then, they take its discriminant and say it is less than 0.
$36 - (4 * 32) < 0 $
And ,then they say it is true. I didn't understand how taking the discriminant of the discriminant helped prove.
 A: The first quadratic is in $x$ where you want the solutions to be real. Therefore the discriminant must be non-negative. Furthermore you want distinct real roots, therefore the discriminant must be positive.
When you compute the discriminant for the quadratic equation in $x$, you end up getting a quadratic expression in $k$. Then for it to be positive, you ended up with a quadratic inequality
$$(5-k)^2-4(k+2)>0$$
This is equivalent to 
$$k^2-14k+17 >0$$
Think of this as a parabola in $k$ and you want this parabola to be always above the $k-$axis. That means NO real solutions. For this the discriminant of this quadratic expression in $k$ must be negative.
PS: However please note that the last inequality is not true for all values of $k$. In fact
it is only true when 
$$k \in (-\infty, 7-4\sqrt{2}) \cup (7+4\sqrt{2}, \infty).$$
So for these values of $k$ the quadratic equation will have real roots.
A: Hint: Showing that the discriminant is less than zero implies that there are no real solutions. So, if the discriminant has no real solutions, then it is greater than $0$. (That is $(5−k)^2 −4(1)(k+2)>0$ for all values of $k$.)
A: Taking the discriminant of the discriminant is not clarified in the accepted solution.
The discriminant of the quadratic equation $x^2-(5-k)x+(k+2)=0$ is $\Delta=k^2-14k+17$. We want $\Delta$ to be always positive, then the given equation will always have two distinct real roots.
For this to happen, the discriminant of $\Delta$ must be negative. But, the discriminant of $\Delta$ is $\Delta'=(-14)^2-4.17=128>0$ is positive. Therefore, the given quadratic equation has not always two distinct real roots but only if $k\in(-\infty,7-4\sqrt2)\cup(7+4\sqrt2,\infty)$ as found in the accepted answer.
