Question

The imaginary roots of the equation (x2 + 2)2 + 8x2 = 6x(x2 + 2) are?

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

The given equation is:

$(x^{2}+2)^{2}+8x^{2}=6x(x^{2}+2)$

Let $x^{2}+2=t$, then the equation becomes

$t^{2}-6xt+8x^{2}$=0

$t^{2}-4xt-2xt+8x^{2}$=0

$t(t-4x)-2x(t-4x)$=0

$(t-2x)(t-4x)$=0

Now, putting the value of t in above equation, we get

$(x^{2}-4x+2)(x^{2}-2x+2)$=0

⇒$x^{2}-4x+2=0$

$x=2,2$ which are real roots.

And $x^{2}-2x+2=0$

x=1±i which are the imaginary roots of the equation.