A) \[-\frac{b}{d}\]
B) \[-\frac{c}{b}\]
C) \[\frac{b}{d}\]
D) \[\frac{c}{b}\]
Correct Answer: B
Solution :
[b] Let \[\alpha ,\beta \] be the zeroes of \[b{{x}^{2}}+cx+d.\] |
\[\therefore \] Sum of zeroes \[=-\frac{c}{b}\] |
Now, one zero = 0 [Given] |
\[\therefore \] Other zero\[=-\frac{c}{b}\] |
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