$x^4-4x^3+a{x}^2+bx+1=0$ has $4$ positive roots, then $a+b=$

Let the roots be $p,q,r,s$

$\because$Sum of roots = -(coefficient of $x^3$)/coefficient of $x^4$

So $p+q+r+s=4..........\left(1\right)$

And $\frac{p+q+r+s}{4}=1=$arithmetic mean$..........\left(2\right)$

$\because$Product of roots=constant term/coefficient of $x^4$

$\therefore$$pqrs=1$

So geometric mean ${\left(pqrs\right)}^{1/4}=1.........\left(3\right)$

Using eq(2) and eq(3)

$\left(\frac{p+q+r+s}{4}\right)={\left(pqrs\right)}^{1/4}$

$\Rightarrow$Arithmetic mean=geometric mean

So all roots are same

$\Rightarrow p=q=r=s$

Put in eq(1)

 $$\begin{gathered} 4p=4\Rightarrow p=1 \\ And q=r=s=1 \end{gathered}$$

Now we have all four roots so equation is ${(x-1)}^4=0$

$$\begin{gathered} \Rightarrow {(x-1)}^2{(x-1)}^2=0 (x^2-2x+1)(x^2-2x+1)=0 \\ \therefore x^4-4x^3+6x^2-4x+1=0 \end{gathered}$$

Comparing above equation with $x^4-4x^3+a{x}^2+bx+1=0$

We get $a=6, b=-4$

So $a+b=2$