A variant of the quadratic formula is
x = (2
c)/(-
b +/- sqrt[
b^2-4
ac])
From this it is easy to see that for a variable a that one root is always greater than -c/b and the other is always less than -c/b.
Also note that from this form of the quadratic equation, lim {a -> 0} (2c)/(-b +/- sqrt[b^2-4ac]) is equal to -c/b or diverges to infinity, depending on which root is examined.
Let p1 and p2 be the roots of ax^2+ bx + c = 0.
Let q1 and q2 be the roots of -ax^2+ bx + c = 0.
Let r1 and r2 be the roots of (a/2)x^2+ bx + c = 0.
Without loss of generality, p2 < -c/b < p1; q2 < -c/b < q1; and r2 < -c/b < r1.
From the limit earlier, exactly one of r1 and r2 approaches -c/b. If that root is r1 then q2 < -c/b < r1 < p1. Also, if that root is r2 then p2 < r2 < -c/b < q1.
Therefore (a/2)x^2+ bx + c = 0 has root between some p and q. More generally this is true for any k with abs(a)>abs(k)>0.
Edited on July 14, 2016, 1:56 pm
Solution
