Question

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

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

Answer 1

Answer: (D)

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