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

The sum of all the rational roots of the equation $6 x^6-25 x^5+31 x^4-31 x^2+25 x-6=0$ is
  1. 3
  2. 3.5
  3. $\frac{25}{6}$
  4. 2.5

Solution

Given, equation $ \begin{aligned} & 6 x^6-25 x^5+31 x^4-31 x^2+25 x-6=0 \\ & \Rightarrow 6\left(x^6-1\right)-25 x\left(x^4-1\right)+31 x^2\left(x^2-1\right)=0 \\ & \Rightarrow\left(x^2-1\right)\left[6\left(x^4+x^2+1\right)-25 x\left(x^2+1\right)+31 x^2\right]=0 \\ & \Rightarrow\left(x^2-1\right)\left[6 x^4-25 x^3+37 x^2-25 x+6\right]=0 \end{aligned} $ Either $x^2-1=0$ $ \begin{aligned} & \Rightarrow x=+1,-1 \\ & \text { or } 6 x^4-25 x^3+37 x^2-25 x+6=0 \\ & \Rightarrow 6\left(x^4+1\right)-25 x\left(x^2+1\right)+37 x^2=0 \\ & \Rightarrow 6\left(x^2+\frac{1}{x^2}\right)-25\left(x+\frac{1}{x}\right)+37=0 \\ & \Rightarrow 6\left(x+\frac{1}{x}\right)^2-25\left(x+\frac{1}{x}\right)+25=0 \\ & \Rightarrow 6\left(x+\frac{1}{x}\right)^2-15\left(x+\frac{1}{x}\right)-10\left(x+\frac{1}{x}\right)+25=0 \\ & \Rightarrow 3\left(x+\frac{1}{x}\right)\left[2\left(x+\frac{1}{x}\right)-5\right]-5\left[2\left(x+\frac{1}{x}\right)-5\right]=0 \\ & \left.\Rightarrow 2\left(x+\frac{1}{x}\right)-5\right]\left[3\left(x+\frac{1}{x}\right)-5\right]=0 \\ & \Rightarrow \text { Either } 2\left(x+\frac{1}{x}\right)=5 \text { or } 3\left(x+\frac{1}{x}\right)=5 \\ & \Rightarrow 2 x^2-5 x+2=0 \text { or } 3 x^2-5 x+3=0 \\ & \Rightarrow 2 x^2-4 x-x+2=0 \\ & \because 3 x^2-5 x+3=0 \text { has no solution because } D < 0 . \\ & \Rightarrow 2 x(x-2)-1(x-2)=0 \\ & \Rightarrow x=2, \frac{1}{2} \end{aligned} $ So, sum of all the rational roots $ \lambda=1-1+2+\frac{1}{2}=2.5 $ Hence option (d) is correct

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

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