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Let $ a_1,a_2,a_3 >0$

$\lambda_1\lt \lambda_2 \lt \lambda_3 \in \mathbb{R}$

prove that the following equation has exactly 2 solutions

$\dfrac{a_1}{x-\lambda_1} + \dfrac{a_2}{x-\lambda_2} + \dfrac{a_3}{x-\lambda_3} = 0$

this should be closely related the continuity and the intermediate value theorem, but i couldn't connect the dots on this one.

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Just examine the signs of the function on the left hand side (call it $f$) for all intervals where it is continuous.

For $x < \lambda_1$, $f$ is always negative. So there are no roots here.

For $\lambda_1 < x < \lambda_2$, $f(x) \to \infty$ as $x \to {\lambda_1}^+$ but $f(x) \to -\infty$ as $x \to {\lambda_2}^-$. So there is a root here. Also, $f$ is decreasing so there is exactly one root.

For $\lambda_2 < x < \lambda_3$, $f(x) \to \infty$ as $x \to {\lambda_2}^+$ but $f(x) \to -\infty$ as $x \to {\lambda_3}^-$. So there is another root here. Again, $f$ is decreasing so there is exactly one root.

For $\lambda_3 < x$, $f$ is always positive. So there are no roots here.

So there are exactly two roots in total.

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    $\begingroup$ what is the proof that $f$ is monotone decreasing when $\lambda_1 < x < \lambda_2$? $\endgroup$ – guynaa Dec 21 '13 at 13:21
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    $\begingroup$ @guynaa Each fraction is a monotonically decreasing function since $a_1, a_2, a_3 > 0$. So the sum is also decreasing. $\endgroup$ – Pratyush Sarkar Dec 21 '13 at 16:35
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Letting $f(x)$ be the left hand side of the equation, then imagine the graph of $y=f(x)$.

We can get the followings :

$$\lim_{x\to \pm\infty}f(x)=0, \lim_{x\to\lambda_i\pm0}f(x)=\pm\infty.$$

With the continuity of the graph except $x=\lambda_i$, this leads what you desire. Do you see the monotonicity of each part?

The answer $X$ will be in $\lambda_1\lt X\lt \lambda_2$ and $\lambda_2\lt X\lt \lambda_3$.

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  • $\begingroup$ i think i got it, thanks! $\endgroup$ – guynaa Dec 20 '13 at 16:49
  • $\begingroup$ My big pleasure! If you need to write an answer property, you know you need to use the intermediate value theorem, don't you? $\endgroup$ – mathlove Dec 20 '13 at 16:49
  • $\begingroup$ yeah, this question comes from homework after the continuity and interm value theorem lecture $\endgroup$ – guynaa Dec 20 '13 at 16:54

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