Evaluate the mentioned below-v2-4-5-3-7-2-41-when -2-3 -

Given that 𝑥=−2−√(3) 𝑖 +2=−√(3) 𝑖 ⇒(𝑥+2)^2=(−√(3) 𝑖)^2 nbsp; squaring on both side ^2+4𝑥+7=0 nbsp; nbsp; …(1) ⇒2𝑥^4+5𝑥^3+7𝑥^2−𝑥+41 =2𝑥^2 (𝑥^2+4𝑥+7)−3𝑥^3− ...

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Standard XI

Mathematics

Strictly Increasing Functions

Evaluate the ...

Question

Evaluate the mentioned below:

(v) $2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41, w h e n x = - 2 - \sqrt{3} i$

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Solution

Given that-
$x = - 2 - \sqrt{3} i$
$\Rightarrow x + 2 = - \sqrt{3} i$
$\Rightarrow (x + 2)^{2} = (- \sqrt{3} i)^{2} s q u a r i n g o n b o t h s i d e$
$\Rightarrow x^{2} + 4 x + 7 = 0 \dots (1)$
$\Rightarrow 2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41$
$= 2 x^{2} (x^{2} + 4 x + 7) - 3 x^{3} - 7 x^{2} - x + 41$
$= 2 x^{2} (x^{2} + 4 x + 7) - 3 x (x^{2} + 4 x + 7) + 5 x^{2} + 20 x + 41$
$= 2 x^{2} (x^{2} + 4 x + 7) - 3 x (x^{2} + 4 x + 7) + 5 (x^{2} + 4 x + 7) + 6$
$= 2 x^{2} \times 0 - 3 x \times 0 + 5 \times 0 + 6 f r o m e q n (1)$
=6
Therefore, $2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41 = 6$

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Strictly Increasing Functions

Standard XI Mathematics

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Evaluate the mentioned below: (v)2x4+5x3+7x2−x+41,whenx=−2−√3i

Evaluate the mentioned below:

(v) $2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41, w h e n x = - 2 - \sqrt{3} i$