The imaginary roots of the equation `(x^2+2)^2+8x^2=6x(x^2+2)` are

Options

A. ` 1 pm i `
B. ` 2 pm i`
C. ` 1 pm sqrt(2)`
D. `2 pm I sqrt(2)`

Answer

A

Explanation

Given `eq^(n) "is " (x^(2)+2)^(2) + 8x^(2) = 6x (x^(2) + 2)`
`rArr x^(4) + 4x^(2) + 4 + 8x^(2) = 6x^(3) + 12x`
`rArr x^(4) - 6x^(3) + 12x^(2) - 12x + 4 = 0 `
`(x^(2) + 2)^(2) + 8x^(2) = 6x (x^(2) + 2)`
`rArr (x^(2) + 2)^(2) + 8x^(2) = 6x(x^2+2)`
`rArr x^(4) + 4x^(2) + 4 + 8x^(2)= 6x^(3) + 12x`
`rArr x^(4) - 6x^(3) + 12x^(2) - 12x + 4 = 0 ` ,
This can be factorised into `(x^(2)-4x+2)(x^(2)-2x+2)=0`
Consider , `x^(2) - 2x + 2 =0`
Roots are `(-(-2)pm sqrt((-2)^(2)-4(2)(1)))/(2(1))=(2pmsqrt(4-8))/2`
` (2pm sqrt(-4))/2 = (2pm 2i)/2= 1 pm i`
Also , this equation is satisfied by `1 pm i`. Hence , required roots are `1 pm i.`