Question

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

• A. 3
• B. 3.5
• C. $\frac{25}{6}$
• D. 2.5

Answer 1

Answer: (D)

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