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Show that if $c\neq1$, the equation $$ \frac{x}{x-a} + \frac{x}{x-b} = 1+c$$ has exactly one real solution if $$c^2 = - \frac {4ab}{(a-b)^2}$$.

I know that a quadratic of the form $ax^2 +bx +c$ has only one real solution if $\sqrt{b^2-4ac} =0$, but I am not sure how to use this to solve the problem that has been set. I have also tried adding the two fractions and rearranging the result, but I have been unable to get anything resembling $ax^2 +bx +c$ .

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  • $\begingroup$ Just a minor correction: when $\Delta = 0$ the equation does not have one solution, but two equal ones. You can verify that on, for example, $x^2 - 2x + 1 = 0$. You can verify that the solutions are $x_{1, 2} = 1$ through polynomial division. $\endgroup$
    – rubik
    Feb 15 '15 at 12:24
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If you multiply both sides by $(x-a)(x-b)$, you get

$$ x(x-b)+x(x-a)=(1+c)(x-a)(x-b), $$

or $2x^2-(a+b)x=(1+c)(x^2)-(1+c)(a+b)x+(1+c)ab$, or

$(1-c)x^2+c(a+b)x-(1+c)ab=0$. This means that

$$ Ax^2+Bx+C=0, $$

where

$$ A=1-c,B=c(a+b), C=-(1+c)ab. $$

This has exactly one real solution iff $B^2-4AC =0\ldots$ You should be able to continue from here.

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  • $\begingroup$ wasn't $(1+c)(x^2) - 2(x^2) = (1+c-2)(x^2) = (c-1) (x^2)$ ? $\endgroup$
    – wuiyang
    Feb 18 '15 at 8:30
  • $\begingroup$ @user2324360 no, my computations are correct. Note that I switched the LHS and RHS $\endgroup$ Feb 18 '15 at 15:00
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A first attempt might be to write $\frac{x}{x-a}=\frac{x-a}{x-a}+\frac{a}{x-a}=1+\frac{a}{x-a}$ and so the equation is equivalent to $\frac{a}{x-a}+\frac{b}{x-b}=c-1$. Now, you can multiply both sides with $(x-a)(x-b)$ and simplify to get the usual quadratic form.

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Since $x \neq a$ and $x \neq b$ you can multiply both sides of the equation by $(x - a)(x - b)$ arriving, after a bit of basic algebra, to $$(1 - c)x^2 + c(a + b)x - ab(1 + c) = 0$$

Setting the condition $\Delta = 0$ will yield the desired result.

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