Review question

When can this equation involving algebraic fractions hold? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8407

Solution

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

1. 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.

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.

1. Show that the equation $\frac{x}{x-a}+\frac{x}{x-b}=1+c$ has exactly one real solution if $c^2=-\dfrac{4ab}{(a-b)^2}$.

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 $c^2=-\frac{4ab}{(a-b)^2}.$

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 $0 < c^2 \leq 1$.

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.