A) 4
B) \[-4\]
C) 8
D) \[-8\]
Correct Answer: A
Solution :
Given \[2x=3+5i\] \[\Rightarrow \,\,\,\,\,\,x=\frac{3+5i}{2}\] \[\operatorname{Consider} \,{{x}^{3}} =\frac{27+125{{i}^{3}}+225{{i}^{2}}+135i}{8}\] \[=\,\,\frac{27 - 125i - 225 +135i}{8}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{matrix} \because \,\,\,{{i}^{2}}=1 \\ {{i}^{3}}=-i \\ \end{matrix} \right)\] \[=\,\,\,\frac{-198+10i}{8}=\frac{-99+5i}{4}\] and \[{{x}^{2}}=\frac{9+25{{i}^{2}}+30i}{4}\] \[=\,\,\,\frac{9-25+30i}{4}\,\,=\,\,\frac{-8+15\,i}{2}\] Now, Consider \[2{{x}^{3}} + 2{{x}^{2}}- 7x + 72\] \[=\,\,\left( \frac{-99+5i}{2} \right)+(-8+15i)-\frac{7(3+5i)}{2}9+72\] \[=\,\,\,-\frac{99}{2}+\frac{5i}{2}-8+15i-\frac{21}{2}-\frac{35}{2}i+72\] \[=\,\,\left( -\frac{99}{2}-8-\frac{21}{2}+72 \right)+\left( \frac{5}{2}+15-\frac{35}{2} \right)i\] \[=\frac{-99-16-21+144}{2}=\frac{8}{2}=4\]
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