Question

Solve the equation:
$x^5-5x^4+9x^3-9x^2+5x-1=0$.

Answer 1

Given equation, $x^5-5x^4+9x^3-9x^2+5x-1=0$

Consider f(x) = x^5-5x^4+9x^3-9x^2+5x-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 $A=1,B=-4,C=5$

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

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

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

$⟹(x^2+x^{-2})–4(x+x^{-1})+5=0\left[\text{dividing by}{x}^2\right]$

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

$⟹(y^2-2)-4y+5=0⟹y^2-4y+3=0⟹(y-3)(y-1)=0$

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

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

Solving the second quadratic equations we have, $x=\frac{1\pm \sqrt{3}i}{2}$

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