Question 629643
Using the Quadratic Formula on your equation we get:
{{{x = (-(-a) +- sqrt((-a)^2-4(1)(b)))/2(1)}}}
which simplifies as follows:
{{{x = (-(-a) +- sqrt(a^2-4(1)(b)))/2(1)}}}
{{{x = (-(-a) +- sqrt(a^2-4b))/2(1)}}}
{{{x = (a +- sqrt(a^2-4b))/2}}}
which is short for:
{{{x = (a + sqrt(a^2-4b))/2}}} or {{{x = (a - sqrt(a^2-4b))/2}}}<br>
The difference of these is:
{{{(a + sqrt(a^2-4b))/2 - (a - sqrt(a^2-4b))/2}}}
{{{(2sqrt(a^2-4b))/2}}}
{{{sqrt(a^2-4b)}}}<br>
We are told that this difference is less than c, and as is true of <i>any</i> square root, it is not negative. So
{{{sqrt(a^2-4b) < c}}} and {{{sqrt(a^2-4b) >= 0}}}
Squaring both sides of both we get:
{{{a^2-4b < c^2}}} and {{{a^2-4b >= 0}}}
Now we solve both for b:
{{{a^2-c^2 < 4b}}} and {{{a^2 >= 4b}}}
{{{(1/4)(a^2-c^2) < b}}} and {{{(1/4)a^2 >= b}}}<br>
P.S. To me, "between p and q" means "between p and q  but not equal to p or q". But I don't think there is any way to eliminate the "or equal to" part of {{{(1/4)a^2 >= b}}}. So the largest possible b would be equal to {{{(1/4)a^2}}} not between that and {{{(1/4)(a^2-c^2)}}}.