Question

# which among the following is a root of the equation x6−2x3+2=0 is :

A
21/6eπi/12
B
21/2eπi/12
C
6eπi/12
D
eπi/12

Solution

## The correct option is A 21/6eπi/12Let x3=z Hence the equation we get z2−2z+2=0 z=x3=1+i or z=x3=1−i By using formula, we get x3=√2(1+i√2) x=216ei(2kπ+π4)3,   k∈ 0, 1, 2 Hence roots of the x3=1+i is 216eiπ12, 216ei9π12,216ei17π12 Similarly for x3=1−i  Here θ=−π4 and  roots are 216e−iπ12, 216ei7π12,216ei15π12 Hence 6 roots of the equation are  216eiπ12, 216ei9π12,216ei17π12,216e−iπ12, 216ei7π12,216ei15π12  Mathematics

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