Question:
If b = 0, c < 0, is it true that the roots of x2 + bx + c = 0 are numerically equal and opposite in sign? Justify your answer.
Solution:
Given that, b = 0andc < 0and quadratic equation,
$x^{2}+b x+c=0$ $\ldots(i)$
Put $b=0$ in Eq. (i), we get
$x^{2}+0+c=0$
$\Rightarrow$ $x^{2}=-c$ $\left[\begin{array}{l}\text { here } c>0 \\ \therefore-c>0\end{array}\right]$
$\therefore$ $x=\pm \sqrt{-C}$
So, the roots of x2 + bx+c = O are numerically equal and opposite in sign.