MHT CET · Maths · Complex Number
If \(x=-2-\sqrt{3} i\), where \(i=\sqrt{-1}\), then the value of \(2 x^4+5 x^3+7 x^2-x+41\) is
- A 6
- B -6
- C 75
- D -76
Answer & Solution
Correct Answer
(A) 6
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & x=-2+\sqrt{3} i \\ & \Rightarrow x+2=-\sqrt{3} i \\ & \Rightarrow(x+2)^2=(-\sqrt{3} i)^2 \\ & \Rightarrow x^2+4 x+4=-3 \\ & \Rightarrow x^2+4 x+7=0 \\ & \text { Now } 2 x^4+5 x^3+7 x^2-x+41 \\ & =\left(2 x^2-3 x+5\right)\left(x^2+4 x+7\right)+6 \\ & =\left(2 x^2-3 x+5\right) \times 0+6 \\ & =6\end{aligned}\)
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