A) both positive
B) both negative
C) one positive and one negative
D) both equal
Correct Answer: B
Solution :
| [b] Let \[p(x)={{x}^{2}}+99+127\] |
| Let \[\alpha \] and \[\beta \] be the zeroes of \[p(x)\]. |
| Then, sum of roots |
| \[=\alpha +\beta =-\frac{b}{a}=-99\] (1) |
| product of roots \[=\alpha \beta =\frac{c}{a}=127\] ...(2) |
| Now, \[{{(\alpha -\beta )}^{2}}={{(\alpha +\beta )}^{2}}-4\alpha \beta \] |
| \[={{(-99)}^{2}}-4\times 127=9293\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha -\beta =\sqrt{9293}=\,\,\pm 96.4\] |
| When \[\alpha -\beta =96.4\] ...(3) |
| Solving eqs. (1) and (2), we get |
| \[\alpha =-13,\,\,\beta =-97.7\] |
| \[\therefore \] Both zeroes are negative. |
| Similarly, for \[\alpha -\beta =-96.4,\]we get both the zeroes as negative. |
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