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Question

The root of the equation, (x2+1)2=x(3x2+4x+3), are given by

 


  1. 23

  2. (1+i3)2,i=1

  3. 2+3

  4. (1i3)2,i=1


Solution

The correct options are
A

23


B

(1+i3)2,i=1


C

2+3


D

(1i3)2,i=1


Given equation is 

(x2+1)2=x(3x2+4x+3)

x43x32x23x+1=0

x2(x23x23x+1x2)=0

x0

x2+1x23(x+1x)2=0

(x+1x)23(x+1x)4=0

(x+1x4)(x+1x+1)=0

Or (x24x+1)(x2+x+1)=0

x=2±3, 1±i32

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