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

Given:

\( b=0, c<0 \)

To do:

We have to find whether the roots of \( x^{2}+b x+c=0 \) are numerically equal and opposite in sign.

Solution:

$x^2+bx+c=0$

Substituting $b=0$, we get,

$x^2+0x+c=0$

$x^2=-c$

$x=\pm \sqrt{-c}$ [$c<0 \Rightarrow -c>0$]

Therefore,

The roots of \( x^{2}+b x+c=0 \) are numerically equal and opposite in sign.

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