Question

Solve:
$6x^6-25x^5+31x^4-31x^2+25x-6=0$

Answer 1

$6x^6-25x^5+31x^4-31x^2+25x-6=0$

This is a reciprocal equation of even degree with opposite signs.

Therefore,

$x=1,-1$ are the roots.

Now, dividing the equation by $(x-1)(x+1)$, we have

$6x^6-25x^5+31x^4-31x^2+25x-6÷(x+1)(x+1)=6x^4-25x^3+37x^2-25x+6$

Thus the equation will be-

$6x^4-25x^3+37x^2-25x+6=0$

$6x^2-25x+37-\frac{25}{x}+\frac{6}{x^2}=0$

$6(x^2+\frac{1}{x^2})-25(x+\frac{1}{x})+37=0$

Let $x+\frac{1}{x}=t$

$\Rightarrow x^2+\frac{1}{x^2}=t^2-2$

Therefore, the equation becomes-

$6(t^2-2)-25t+37=0$

$\Rightarrow 6t^2-25t+25=0$

$\Rightarrow (3t-5)(2t-5)=0$

$\Rightarrow t=\frac{5}{3},\frac{5}{2}$

Case I:- $t=\frac{5}{3}$

$x+\frac{1}{x}=\frac{5}{3}$

$3x^2-5x+3=0$

By quadratic formula,

$x=\frac{-(-5)\pm \sqrt{{(-5)}^2-4\times 3\times 3}}{2\times 3}$

$\Rightarrow x=\frac{5\pm \sqrt{11}i}{6}$

Case II:- $t=\frac{5}{2}$

$x+\frac{1}{x}=\frac{5}{2}$

$2x^2-5x+2=0$

$(2x-1)(x-2)=0$

$x=2,\frac{1}{2}$

Hence the roots of the equation $6x^6-25x^5+31x^4-31x^2+25x-6=0$ are $-1,1,\frac{1}{2},2$ and $\frac{5\pm \sqrt{11}i}{6}$