Show that the equation $$ \dfrac{a}{x-1} + \dfrac{b}{x-2} + \dfrac{c}{x-3} + \dfrac{d}{x-4} = 1 $$ cannot have imaginary roots if $a, b, c, d$ are any four real numbers of the same sign. I tried putting $x = a$ and some other variables to show that the equation changes sign 3 times but failed.
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The resulting fourth degree equation has necessarily 4 real roots.
Here is a convincing reason, followed by a proof.
Have a look at the graphics below. It is conversion of the purely algebraic issue into a search for the number of intersection points between the curves of functions $f$ and $g$ with the following equations:
$$\begin{cases}y=f(x)=\frac{0.1}{x-1}+\frac{0.3}{x-2}+\frac{0.5}{x-3}+\frac{0.2}{x-4}\\y=g(x)=1\end{cases}$$
The rigorous proof comes from the fact that, assuming all coefficients to be positive, $f'(x)<0$, thus $f$ is decreasing on all intervals where it is defined, i.e.,
$$(-\infty,1),(1,2),(2,3),(3,4),(4,\infty).$$
A thorough inspection of the last four intervals show that there is an intersection point in all of them.
For example, the existence of a unique intersection point in interval $(1,2)$,is due to the fact that $f$ is decreasing and that
$$\cases{\lim_{x \rightarrow 1_+}f(x)=+\infty\\ \lim_{x \rightarrow 2_-}f(x)=-\infty}$$
For interval $(4,+\infty)$, it is because $f$ is decreasing and:
$$\cases{\lim_{x \rightarrow 4+}f(x)=+\infty\\ \lim_{x \rightarrow \infty}f(x)=0_+}$$
Jean Marie
- 81,803
Assume that the equation has a complex root, i.e. of the form $\alpha+i\beta$ where $\alpha,\beta\in\mathbb{R}$.
As complex roots occur in conjugate pairs for an equation with real coefficients, $\alpha-i\beta$ is also a root of the given equation. Put $x=\alpha+i\beta$ and $x=\alpha-i\beta$ successively in the equation to get two new equations.
Subtract one of these equations from the other and show that $\beta$ must equal $0$ to avoid any ambiguity.
– StubbornAtom Sep 26 '16 at 13:13