40687

In response to 7:
If c=0 then ax^2 + bx = 0 where a>0 and lets assume that b ≠ 0.
It then follows that ax^2 + bx = 0 can be factorized to give x(ax+b) = 0.
What we then see is that the curve has a root at the origin and at -b/a.
If we aree to use the discriminant to show the quadratic ax^2 + bx + c = 0 (remember c=0) has real roots we find:
b^2 - 4(a)(c) = b^2 - 4(a)(0) = b^2
From that we have seen the discriminant to always be positive (when b ≠ 0) and so the quadratic ax^2 + bx = 0 always has two real distinct roots.
However, if b where to = 0 then the quadratic becomes ax^2 = 0 and the discriminant would be :
b^2 - 4ac = (0)^2 - 4(a)(0) = 0.
as the discriminant would always be 0, the quadratic ax^2 = 0 will always have one repeated root.