Question

If the roots of the equation $x^2-ax+b=0$ are real and differ by a quantity which is less than c(c > 0), then b lies between $\frac{a^2-c^2}{4}$ and $\frac{a^2}{4}.$

Answer 1

Given $a^2-4b>0\Rightarrow b<a^2/4\cdots \left(1\right)$
Again $\alpha$ and $\beta$ differ by a quantity less than $c(c>0)$
$\therefore |\alpha +\beta |<c$
$\Rightarrow {|\alpha +\beta |}^2-4\alpha \beta <c^2$
$\Rightarrow a^2-4b<c^{2}or\frac{a^2-c^2}{4}<b\cdots \left(2\right)$
$\therefore \frac{a^2-c^2}{4}<b<\frac{a^2}{4}$ by (1) and (2)