A) \[2\pm \sqrt{3}\]
B) \[3\pm \sqrt{2}\]
C) \[2\pm \sqrt{5}\]
D) \[5\pm \sqrt{2}\]
Correct Answer: C
Solution :
Let \[\alpha \] and \[{{\alpha }^{2}}\] be the roots of\[{{x}^{2}}-x-k=0\]. Then \[\alpha +{{\alpha }^{2}}=1\]and\[{{\alpha }^{3}}=-k\]. \[\therefore \,\,\,\,\,{{(-k)}^{1/3}}+{{(-k)}^{2/3}}=1\,\,\,\Rightarrow \,\,\,-{{k}^{1/3}}+{{k}^{2/3}}=1\] Þ \[{{({{k}^{2/3}}-{{k}^{1/3}})}^{3}}=1\,\,\,\,\Rightarrow {{k}^{2}}-k-3k({{k}^{2/3}}-{{k}^{1/3}})=1\] Þ\[{{k}^{2}}-k-3k(1)=1\]Þ \[{{k}^{2}}-4k-1=0\,\,\,\Rightarrow k=2\pm \sqrt{5}\].
You need to login to perform this action.
You will be redirected in
3 sec
