Question

Solve the following equation in $R$
${\left(x^2+2\right)}^2+8x^2=6x\left(x^2+2\right)$

Answer 1

let $x^2+2=t$

$t^2+8x^2=6xt$

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

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

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

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

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

$x^2+2-2x$ has imaginary roots.

$x^2-4x+2$ has real roots

$x=\frac{4\pm \sqrt{16-8}}{2}=\frac{4\pm 2\sqrt{2}}{2}=2\pm \sqrt{2}$