NDA Math Miscellaneous

Simplifying the above question, we get
(x2+2)2+8x2=6x(x2+2)
x4+4x2+4+8x2=6x3+12x
x4−6x3+12x2−12x+4=0
Let
f(x)=x4−6x3+12x2−12x+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
=

−Coefficient of x3

Coefficient ofx3

=−(

−6

1

)
=6
Hence sum of roots is 6.