Question 408299
Interesting thing about this problem is that the roots and the value of p are not uniquely determined. You'll see why:


Suppose the two roots are {{{r}}} and {{{2r}}}. The product of the roots is 8 by Vieta's formulas, so {{{2r^2 = 8}}} --> {{{r^2 = 4}}} --> r = +/- 2. Hence, the roots are {2, 4} or {-2, -4}. Also, the sum of the roots is -p, so p = -6 or 6, respectively.


Hence, the quadratic {{{x^2 - 6x + 8}}} has roots {2, 4} and {{{x^2 + 6x + 8}}} has roots {-2, -4}.