If $b=0$ and $c < 0$,is it true that the roots of $x^{2}+bx+c=0$ are numerically equal and opposite in sign? Justify.

(A) Given the quadratic equation: $x^{2}+bx+c=0$ ....$(i)$
Substitute $b=0$ into equation $(i)$:
$x^{2}+0(x)+c=0$
$x^{2}+c=0$
$x^{2}=-c$
Since it is given that $c < 0$,let $c = -k$ where $k > 0$. Then:
$x^{2} = -(-k) = k$
$x = \pm \sqrt{k}$
Thus,the roots are $\sqrt{k}$ and $-\sqrt{k}$.
These roots are numerically equal (both have magnitude $\sqrt{k}$) and opposite in sign (one is positive and one is negative). Therefore,the statement is true.