prove that the following equation has exactly 2 solutions 
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.
 A: 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.
A: 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 $\lambda_2\lt X\lt \lambda_3$.
