Answer
a). The roots of a quadratic equation $ax^{2}+bx+c=0$ are given by :-
$x=\frac{-b+\sqrt (b^{2}-4ac)}{2a}$
So, x can only have real roots only if $\sqrt (b^{2}-4ac)$ is real i.e. $(b^{2}-4ac)\geq 0$. Hence, $b^{2}\geq4ac$
b). The two roots are identical if $b^{2}-4ac=0$ or, $b^{2}=4ac$
Work Step by Step
a). The roots of a quadratic equation $ax^{2}+bx+c=0$ are given by :-
$x=\frac{-b+\sqrt (b^{2}-4ac)}{2a}$
So, x can only have real roots only if $\sqrt (b^{2}-4ac)$ is real i.e. $(b^{2}-4ac)\geq 0$. Hence, $b^{2}\geq4ac$
b). The two roots are identical if $b^{2}-4ac=0$ or, $b^{2}=4ac$
