Finding parameter of equation $$x^4 + 1 = kx; k>0$$
The question is at what k the equation has 1 solution.
I understand that it could rephrased as follows: at what k $f(x) = x^4 - kx + 1$ has 1 $x:f(x) = 0$. but I've no idea how to solve quadratic equation. 
Also, the problem may be seen as finding point of intersection of parabola-like $x^4$ and the line $kx - 1$. In this case, the line is just a tangent for $x^4$. Thus, we can find its equation via derivative, i.e. $f'(x) = 4 x^3 \rightarrow y-1 = 4 x^3 (x-1) \rightarrow y = 4 x^4 - 4 x^3 + 1$. Now, we can compare two forms of the same line: $4 x^4 - 4 x^3 + 1 = kx - 1 $. I feel I've made a mistake somewhere, because if both equations describe the same line, they should give equal results for any x, for example: $x = 0 \rightarrow 0 - 0 + 1 \neq 0 - 1$
Could you help, please?
 A: If we count multiplicity, a quartic polynomial can't have exactly $1$ real root, so the real root must be a double root. 
A double root is also a root of the derivative, which is $4x^3-k$.
Thus the root must be $\sqrt[3]{k/4}$.
Substitute in the equation. We get $(k/4)^{4/3}+1=k(k/4)^{1/3}$. This is easy to solve for $k$ if we rewrite the right-hand side as $4(k/4)^{4/3}$. 
A: Let line $y=kx-1$ is tangent to $y=x^4$ at point $(\alpha,\alpha^4)$, but it lies on line also , therefore, $\alpha^4=k\alpha-1\tag{1}$
Also, for tangent slope of tangent at $(\alpha,\alpha^4)$ must be equal to slope of the line $=k\implies 4\alpha^3=k\implies \alpha=\left(\frac{k}{4}\right)^{1/3}$
Putting it in $(1)$ gives,
$\left(\frac{k}{4}\right)^{4/3}=k\left(\frac{k}{4}\right)^{1/3}-1\implies \frac{{k}^{4/3}}{4^{4/3}}=\frac{k^{4/3}}{4^{1/3}}-1$
Now, it's easy to solve from here on. 
A: Your approach with the tangent line is a good one.  When you wrote $y-1=4x^3(x-1)$ you assumed the tangent point is $(1,1)$ but there is nothing to tell you that.  Let the point of tangency be $(a,a^4)$.  Then the slope is $4a^3=-k$, so the tangent point requires $1-ka=a^4$ or $k=\frac {1-a^4}a$ giving you two equations in two unknowns.
