A) \[{{x}^{2}}-16\]
B) \[{{x}^{2}}+16\]
C) \[{{x}^{2}}+4\]
D) \[{{x}^{2}}-4\]
Correct Answer: A
Solution :
| [a] Let \[\alpha \] and \[\beta \] be the zeroes of the quadratic polynomial |
| Then, \[\alpha +\beta =0\] [Given] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\alpha +4=0\] [ one zero is 4 (Given)] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\alpha =-4\] |
| \[\therefore \] Zeroes of the polynomial are 4 and \[-4\] |
| Product of the zeroes \[=4\times (-4)=-16\] |
| \[\therefore \]The required polynomial |
| \[={{x}^{2}}-(\text{Sum of zeroes})x+\text{Product of the zeroes}\] \[={{x}^{2}}-0x+(-16)={{x}^{2}}-16\] |
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