NDA Math Miscellaneous

Simplifying the above question, we get $(x^2+2)^2+8x^2=6x(x^2+2)$ $x^4+4x^2+4+8x^2=6x^3+12x$$x^4-6x^3+12x^2-12x+4=0$Let$f(x)=x^4-6x^3+12x^2-12x+4$Then we get$f(0)=4$$f(1)=-1$We can see that the function changes sign from positive to negative between 0 and -1. Hence it cuts the x-axis at least at one point between the above intervals.Therefore it has at least one real root in the above interval.Now sum of the roots will be equal to$= \frac{-\text{Coefficient of }x^3}{\text{Coefficient of}x^3}$$= -\left(\frac{-6}{1}\right)$$=6$Hence sum of roots is 6.