Can you figure out my approach?

We want $f(a)=f'(a)=0$ or $a^5+ka+4=5a^4+k=0$ or $a^5-5a^5+4=0$ , so $a=1$ and $k=\boxed{-5}$

At a multiple root the derivative is zero. Thus $5x^4 + k = 0\ \ \therefore\ \ 5x^5 + kx = 0,$ and subtracting this from the original equation $-4x^5 + 4 = 0\ \ \therefore\ \ x^5 = 1\ \ \therefore\ \ x = 1;$ substitute in derivative gives $5 + k = 0\ \ \therefore\ \ k = -5.$

I did not use the condition $x > 0$ . Am I overlooking something?

Since the polynomial only crosses the x-axis once and is tangent to the x-axis, we conclude that $x^5+kx+4 \geq 0$ for all positive x, which simplifies to $x^5+4 \geq -kx$ .

By AM-GM Inequality,

$\frac{x^5+1+1+1+1}{5} \geq \sqrt[5]{x^5} = x$

So, $x^5+4 \geq 5x$ for all positive $x$ .

So, we get $-k = 5$ , $k = \boxed{-5}$

Assume polynomial to be $(x-a)^{2} \times (x^{3}+bx^{2}+cx+d)$ . Make coefficients of $x^{3},x^{2},x^{4}=0$ and constant =4 to get a=1,b=2,c=3,d=4. Substitute these values in coefficient of x to get it -5, which is k also!