Question

Solve the following equations:
$x^5-6x^4-17x^3+17x^2+6x-1=0$.

Answer 1

Given equation, $x^5-6x^4-17x^3+17x^2+6x-1=0$

Consider f(x) $=x^5-6x^4-17x^3+17x^2+6x-1$

Notice that this is a reciprocal equation of odd degree which has the opposite signs of the first and last term.

$\therefore (x-1)$ is one factor of the given equation and the quotient is another reciprocal function which has same signs of the first and last term.

$\therefore f\left(x\right)=(x-1)(A{x}^4+B{x}^3+C{x}^2+Bx+A)$

Comparing the coefficient, we have $A=1,B=-5,C=-22$

$⟹f\left(x\right)=(x-1)(x^4-5x^3-22x^2-5x+1)$

Consider $g\left(x\right)=x^4-5x^3-22x^2-5x+1=(x^4+1)-5(x^3+x)–22x^2$

We need to find the roots of $g\left(x\right)=0$

$⟹(x^2+x^{-2})–5(x+x^{-1})–22=0\left[\text{dividing by}{x}^2\right]$

Substitute $x+x^{-1}=y$ in the above equation

$⟹(y^2-2)-5y-22=0⟹y^2-5y-24=0⟹(y-8)(y+3)=0$

$\therefore x+x^{-1}=8$ and $x+x^{-1}=3$

Solving the first quadratic equations we have, $x=4\pm \sqrt{15}$

Solving the second quadratic equations we have, $x=\frac{3\pm \sqrt{5}}{2}$

$\therefore$ roots of the given equation are $1,4\pm \sqrt{15},\frac{3\pm \sqrt{5}}{2}$