$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.
Best Answer
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.