Question

Solve the following equations:
$4x^4-20x^3+33x^2-20x+4=0$.

Answer 1

Given equation, $4x^4-20x^3+33x^2-20x+4=0$

$⟹4(x^2+x^{-2})–20(x+x^{-1})+33=0\left[\text{dividing by}{x}^2\right]$

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

$⟹4(y^2-2)-20y+33=0⟹4y^2-20y+25=0⟹(2y-5{)}^2=0$

We have equal roots for $y=\frac{5}{2}$

$⟹x+x^{-1}=\frac{5}{2}⟹2x^2-5x+2=0⟹(2x-1)(x-2)=0$

$\therefore$ roots of the given equation are $\frac{1}{2},\frac{1}{2},2,2$