Solution assignment 07 Quadratic functions and graphs

For which value of $p$ does the parabola:

$y=x^2-(p+1)x+p$

have two intersection points with the $X$ -axis having a mutual distance of $4$ ?

According to the $abc$ -formula, the intersection points with the $X$ -axis are:

$x_{1,2}=\displaystyle\frac{p+1\pm\sqrt{(p+1)^2-4p}}{2}=\displaystyle\frac{p+1\pm(p-1)}{2}$

So:

$x_1=1$ or $x_2=p$

The distance between both intersection points is $4$ and thus we have to calculate $p$ from (see also Absolute value |x|):

$|p-1|=4$

so:

$p-1=4$ of $p-1=-4$

so:

$p=5$ or $p=-3$