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# Find the value of 2x^4+5x^3+7x^2-x+41, when x=-2-√3¡

MITTAL DALAL
11 Points
4 years ago
x = -2-$\dpi{80} \sqrt{3}i$  $\dpi{80} \therefore$ x2 = 1 +2$\dpi{80} \sqrt{3}i$ ; x3 = 4 – 5$\dpi{80} \sqrt{3}i$  ; x4 = -11 + 4$\dpi{80} \sqrt{3}i$
substitute this values in the equation 2x4 + 5x3 + 7x2 – x + 41 ;
we get: -22 + 8$\dpi{80} \sqrt{3}i$ +20- 25$\dpi{80} \sqrt{3}i$ +7 +14$\dpi{80} \sqrt{3}i$ + 2 +$\dpi{80} \sqrt{3}i$ + 41
= 48 – 2$\dpi{80} \sqrt{3}i$
Rajul
13 Points
3 years ago
x=-2-√3i, x2=1+4√3i, x3=9√3i-10, x^4=8√3i-47
2(8√3i-47)+5(9√3i-10)+7(1+4√3i)+2+√3i+41
=6
So this is the correct answer we get after calculating the question