Quadratic Equations

We start with the equation: $6 x^4-5 x^3+13 x^2-5 x+6=0$.Divide both sides by $x^2$ (where $x \neq 0$ ) to get:$6 x^2-5 x+13-\frac{5}{x}+\frac{6}{x^2}=0$Rewrite the expression by grouping terms: $6 \left(x^2+\frac{1}{x^2}\right) -5 \left(x+\frac{1}{x}\right) +13=0$Let $t=x+\frac{1}{x}$. Then $x^2+\frac{1}{x^2}=t^2-2$.Substitute these into the equation: $6 (t^2-2) -5 t+13=0$Simplify: $6 t^2-12-5 t+13=06 t^2-5 t+1=0$This is a quadratic equation in $t$. The discriminant $(D)$ is:$D=(-5)^2-4 \cdot 6 \cdot 1=25-24=1$ (which is greater than 0 ).This means there are two real solutions for $t$ :$t=\frac{5 \pm 1}{12}=\frac{6}{12} \text{ or } \frac{4}{12} t=\frac{1}{2} \text{ or } \frac{1}{3}$ So, $x+\frac{1}{x}=\frac{1}{2}$ or $x+\frac{1}{x}=\frac{1}{3}$.Now solve for $x$ in each case:If $x+\frac{1}{x}=\frac{1}{2}$, multiply both sides by $x: x^2+1=\frac{1}{2} x$. Rearranged:$2 x^2-x+2=0$If $x+\frac{1}{x}=\frac{1}{3}$, multiply both sides by $x: x^2+1=\frac{1}{3} x$. Rearranged:$3 x^2-x+3=0$For both equations, the discriminant is less than 0 . This means there are no real solutions for $x$.Therefore, all the solutions for $x$ are complex numbers.