Question

If the largest real root of the equation $x^4-4 x^3+5 x^2-4 x+1=0$ can be expressed as $\frac{a+\sqrt{b}}{c}$ (where $b$ does not contains any perfect square), then find the value of $(a+b+c)$.

Answer 1

Dividing the equation by $x^2$ (as coefficient of $x^3$ and $x$ are equal)
we get, $x^2-4\left(x+\frac{1}{x}\right)+\frac{1}{x^2}+5=0 \Rightarrow\left(x+\frac{1}{x}\right)^2-4\left(x+\frac{1}{x}\right)+3=0$
Put $t=x+\frac{1}{x} \Rightarrow t^2-4 t+3=0 \Rightarrow t=3$ or $t=1$ (not possible, think!)
$\therefore x+\frac{1}{x}=3 \Rightarrow x^2-3 x+1=0 \Rightarrow x=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3+\sqrt{5}}{2}$ or $\frac{3-\sqrt{5}}{2}$
$\therefore$ Largest real root is $\frac{3+\sqrt{5}}{2}$.
$\Rightarrow \frac{3+\sqrt{5}}{2} \equiv \frac{a+\sqrt{b}}{c} \Rightarrow(a+b+c)=3+5+2=10$