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

Vedant Priyam , 9 Years ago
Grade 11
anser 2 Answers
MITTAL DALAL
x = -2-\sqrt{3}i  \therefore x2 = 1 +2\sqrt{3}i ; x3 = 4 – 5\sqrt{3}i  ; x4 = -11 + 4\sqrt{3}i 
substitute this values in the equation 2x4 + 5x3 + 7x2 – x + 41 ;
we get: -22 + 8\sqrt{3}i +20- 25\sqrt{3}i +7 +14\sqrt{3}i + 2 +\sqrt{3}i + 41
= 48 – 2\sqrt{3}i
Last Activity: 9 Years ago
Rajul
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
 
Last Activity: 7 Years ago
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