The sum of all the rational roots of the equa | vedclass

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(D) Given equation: $6x^6-25x^5+31x^4-31x^2+25x-6=0$ Grouping terms: $6(x^6-1)-25x(x^4-1)+31x^2(x^2-1)=0$ Factoring out $(x^2-1)$: $(x^2-1)[6(x^4+x^2+

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$2.5$

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(D) Given equation: $6x^6-25x^5+31x^4-31x^2+25x-6=0$ Grouping terms: $6(x^6-1)-25x(x^4-1)+31x^2(x^2-1)=0$ Factoring out $(x^2-1)$: $(x^2-1)[6(x^4+x^2+1)-25x(x^2+1)+31x^2]=0$ $(x^2-1)(6x^4-25x^3+37x^2-25x+6)=0$ Roots from $x^2-1=0$ are $x=1, -1$. For $6x^4-25x^3+37x^2-25x+6=0$,divide by $x^2$: $6(x^2+\frac{1}{x^2})-25(x+\frac{1}{x})+37=0$ Let $t = x+\frac{1}{x}$,then $6(t^2-2)-25t+37=0 \Rightarrow 6t^2-25t+25=0$ $(2t-5)(3t-5)=0 \Rightarrow t=\frac{5}{2}$ or $t=\frac{5}{3}$ Case $1$: $x+\frac{1}{x}=\frac{5}{2}$ $\Rightarrow 2x^2-5x+2=0$ $\Rightarrow (2x-1)(x-2)=0$ $\Rightarrow x=2, \frac{1}{2}$ Case $2$: $x+\frac{1}{x}=\frac{5}{3} \Rightarrow 3x^2-5x+3=0$. Discriminant $D = 25-36 = -11 The rational roots are $1, -1, 2, \frac{1}{2}$. Sum $= 1-1+2+\frac{1}{2} = 2.5$.

The sum of all the rational roots of the equation $6x^6-25x^5+31x^4-31x^2+25x-6=0$ is

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