Given: (x2+2)2+8x2=6x(x2+2) Let x2+2=t⇒t2+8x2=6xt ⇒t2−6xt+8x2=0 ⇒t2−4xt−2xt+8x2=0 ⇒t(t−4x)−2x(t−4x)=0 ⇒(t−2x)(t−4x)=0 Now, putting the value of t in above equation, we get ⇒(x2+2−2x)(x2+2−4x)=0 Now, (x2−2x+2)=0 and (x2−4x+2)=0 ⇒x2−2x+2=0 x=
−(−2)±√(−2)2−4×1×2
2×1
=
2±√−4
2
=
2±2i
2
=1±i Roots are imaginary. Now, ⇒(x2−4x+2)=0 x=−(−4)±
√(−4)2−4×1×2
2×1
=
4±√8
2
=
4±2√2
2
=2±√2 Roots are real. We can say that not all roots of the equation are complex.So statement 1st is wrong. Now, Sum of all the roots =(1+i)+(1–i)+(2+√2)+(2−√2)=6 So statement 2nd is true.