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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