Solution The given equation is of course very difficult (actually impossible) to be solved exactly. But we do not need to find solutions. We only want to understand how changing \(c\) affects the number of solutions. It is useful to re-phrase the question in terms of the function \(f(x)=x^5+c-x\). The question then becomes: Find all values of \(c\) for which \(f(x)\) has exactly two \(x\)-intercepts. Since changing \(c\) amounts to shifting the graph of \(f(x)\) up or down, if we understand the shape of the graph of \(f(x)\) we will be done. So we use the familiar methods of Calculus in order to understand the shape of a graph: we calculate first and second derivative, find the local max or min, etc. The first derivative is \(f'(x)=5x^4 -1\). So the critical numbers are the solutions of \(5x^4-1=0\), that is, \[x=\pm \left(\frac{1}{5}\right)^{1/4}.\] Let's call these two numbers \(-x_0\) and \(x_0\), just to simplify the writing. The second derivative is \(f''(x)=20x^3\). This is of course positive if \(x\) positive and negative if \(x\) is negative. So \(f''(x_0) \gt 0\) and \(f''(-x_0) \lt 0\). By the second derivative test, \((x_0,f(x_0))\) is a local minimum and \((-x_0,f(-x_0))\) is a local maximum. Since \(-x_0\) and \(x_0\) are the only zeros of \(f'(x)\), and \(f'(x)\) is positive before \(-x_0\) and after \(x_0\), we see that the graph of \(f(x)\) is increasing on \((-\infty,-x_0)\), decreasing on \((-x_0,x_0)\), and increasing on \((x_0,\infty)\). So the shape will be as shown: \(\img{1268_30a.png}{0em}{}{12em}\) The number of \(x\)- intercepts will depend on where we need to draw the \(x\)-axis. If we draw the \(x\)-axis below the point of local minimum, or above the point of local maximum, then clearly there will be only one \(x\)-intercept. If we draw it strictly between the minimum and the maximum, there will be exactly three \(x\)-intercepts. To get two \(x\)-intercepts, we need to draw the \(x\)-axis either to go through the local minimum, or through the local maximum. In the first case, it means that \(f(x_0)=0\), that is: \[x_0^5+c-x_0=0.\] Solving this for \(c\) we find \[c=x_0-x_0^5=x_0(1-x_0^4)=x_0\left(1-\frac{1}{5}\right)=\frac{4x_0}{5}=\frac{4}{5^{5/4}}.\] The other case (the \(x\)-axis through the local maximum) gives us, in a similar way, \[c=-\frac{4}{5^{5/4}}.\]