If frac{5}{2} is the sum of two roots of the | vedclass

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(C) The given equation is $6x^6-25x^5+31x^4-31x^2+25x-6=0$.
This is a reciprocal equation of the first kind.
By testing $x=1$ and $x=-1$,we find tha

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$\frac{25}{6}$

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(C) The given equation is $6x^6-25x^5+31x^4-31x^2+25x-6=0$.
This is a reciprocal equation of the first kind.
By testing $x=1$ and $x=-1$,we find that $x=1$ and $x=-1$ are roots of the equation.
Let the roots be $a_1, a_2, a_3, a_4, a_5, a_6$.
Given $a_1+a_2 = \frac{5}{2}$.
Since $1$ and $-1$ are roots,let $a_3=1$ and $a_4=-1$.
The sum of all roots is given by $-\frac{	ext{coefficient of } x^5}{	ext{coefficient of } x^6} = -\frac{-25}{6} = \frac{25}{6}$.
Thus,$a_1+a_2+a_3+a_4+a_5+a_6 = \frac{25}{6}$.
Substituting the known values: $\frac{5}{2} + 1 - 1 + a_5 + a_6 = \frac{25}{6}$.
$a_5+a_6 = \frac{25}{6} - \frac{5}{2} = \frac{25-15}{6} = \frac{10}{6} = \frac{5}{3}$.
Since the equation is a reciprocal equation of the first kind,the roots occur in pairs $(r, 1/r)$. The roots $1$ and $-1$ are real. The remaining roots $a_1, a_2, a_5, a_6$ are non-real.
The sum of all non-real roots is $(a_1+a_2) + (a_5+a_6) = \frac{5}{2} + \frac{5}{3} = \frac{15+10}{6} = \frac{25}{6}$.

If $\frac{5}{2}$ is the sum of two roots of the equation $6x^6-25x^5+31x^4-31x^2+25x-6=0$,then the sum of all non-real roots of the equation is

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