Given equation is (x2+2)2+8x2=6x(x2+2) ....(1) Put x2+2=t Equation (1) becomes, ⇒t2+8x2=6tx ⇒t2−6tx+8x2=0 ⇒t2−4tx−2tx+8x2=0 ⇒t(t−4x)−2x(t−4x)=0 ⇒(t−4x)(t−2x)=0 Substitute the value of t, we get ⇒(x2+2−4x)(x2+2−2x)=0 Consider, (x2−2x+2)=0 ⇒x=
2±√22−4(2)(1)
2(1)
⇒x=
2±√22−4(2)(1)
2(1)
⇒x=
2±2i
2
⇒x=1±i Hence, the two roots of the equation (x2+2)2+8x2=6x(x2+2) are 1±i.