When can this equation involving algebraic fractions hold?

Question

In this question, $a$ and $b$ are distinct, non-zero real numbers, and $c$ is a real number.

Show that, if $a$ and $b$ are either both positive or both negative, then the equation $\frac{x}{x-a}+\frac{x}{x-b}=1$ has two distinct real solutions.
Show that the equation $\frac{x}{x-a}+\frac{x}{x-b}=1+c$ has exactly one real solution if $c^2=-\dfrac{4ab}{(a-b)^2}$ . Show that this equation can be written $c^2=1-\left(\dfrac{a+b}{a-b}\right)^2$ and deduce that it can only hold if $0 < c^2 \leq 1$ .

Suggestion

How can we handle equations which involve fractions? Does it make a difference if $x$ is in the denominator?

This interactive applet shows the graph of $y = \dfrac{x}{x-a}+\dfrac{x}{x-b}-1$ . You can change the values of $a$ and $b$ .

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/VAbYspuB/width/600/height/400/border/888888/rc/false/ai/false/sdz/true/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

The blue lines are the lines $y=c$ which intersect the red curve exactly once.

Solution

Clearly $x \neq a$ and $x \neq b$ . We multiply through by $(x-a)(x-b)$ to clear the denominators, and get the equation $x(x-b)+x(x-a)=(x-a)(x-b),$ which, after multiplying out and rearranging, becomes $x^2=ab.$ This equation will have two distinct real roots ( $\pm\sqrt{ab}$ ) as long as $ab>0$ , which happens exactly when $a$ and $b$ are either both positive or both negative.

We multiply through by $(x-a)(x-b)$ as in the first part, this time finding, for $x$ not equal to $a$ or $b$ , that $x(x-b)+x(x-a) = (1+c)(x-a)(x-b).$ Expanding this gives $x^2-bx+x^2-ax = x^2-ax-bx+ab +cx^2-acx-bcx+abc,$ which rearranges to the quadratic $(1-c)x^2 + c(a+b)x - abc - ab = 0.$

We know that a quadratic equation has one (repeated) real solution if and only if the discriminant is equal to zero.

So the quadratic has one real solution if and only if $c^2(a+b)^2+4(1-c)(abc+ab)=0.$ Expanding this gives $c^2(a^2+2ab+b^2)+4abc-4c^2ab+4ab-4abc=0,$ that is, $c^2(a^2-2ab+b^2)+4ab=0.$ Rearranging this yields the result we desire: the equation has exactly one real solution if and only if $c^2=-\frac{4ab}{(a-b)^2}.$

Looking at the equation we have and the equation we’d like to get to, it would be good if we could prove that $-4ab=(a-b)^2-(a+b)^2.$

Expanding, we see that $(a-b)^2-(a+b)^2=a^2-2ab+b^2-a^2-2ab-b^2=-4ab,$ and so $c^2 = -\frac{4ab}{(a-b)^2} = \frac{(a-b)^2 - (a+b)^2}{(a-b)^2} = 1 - \left(\frac{a+b}{a-b}\right)^2.$

Looking closely at the equation $c^2=1-\left(\frac{a-b}{a+b}\right)^2,$ we see that since $\left(\frac{a-b}{a+b}\right)^2 \geq 0,$ we must have $c^2 \leq 1$ .

Furthermore, since $c$ is real, we must have $c^2 \geq 0$ . However, $c^2=0$ if and only if $4ab=0$ (using our first expression for $c$ ). But we were told at the beginning that $a$ and $b$ were both non-zero, so this cannot happen.

So $0<c^2\le 1$ .

In fact, we were also told that $a$ and $b$ are distinct, so we cannot have $a-b=0$ , and hence we cannot have $c=1$ either. So $0<c^2<1$ .

A subtle point; the question is worded as “show that $A$ can only hold if $B$ ”, where $A$ and $B$ are statements.

(Here $A$ is the statement “ $c^2=-\dfrac{4ab}{(a-b)^2}$ ”, while $B$ is the statement “ $0 < c^2 \leq 1$ ”.)

The statement means “if $A$ is true, then $B$ must also be true”. This is different from saying “ $A$ if $B$ ”, which means “if $B$ is true, then $A$ must also be true”. The two statements are converses of each other. The converse of a true result is sometimes, but not always, true, and we must take care to ensure that we know which statement we are attempting to prove.