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set $a_1$,$a_2$,$a_3>0$ and $λ_3>λ_2>λ_1$ on $ℝ$. show that there are exactly two $x$’s for

$a_1/(x-λ_1) + a_2/(x-λ_2) + a_3/(x-λ_3) = 0$

I tried use the intermediate value theorem but I got only one $x$, for $x∈(x_1,x_2)$, $x_1<λ_1$ and $x_2>λ_3$. $a_1$,$a_2$,$a_3$ can be anything to I didn't find another $x$ thanks ahead

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let $f(x) = \dfrac{a_1}{x - \lambda_1} + \dfrac{a_2}{x - \lambda_2} + \dfrac{a_3}{x - \lambda_3}.$

you can verify that $f$ is continuous everywhere except at $\lambda_1, \lambda_2$ and $\lambda_3.$ at these points $f$ has a vertical asymptote.

then $$\lim_{x \to \lambda_1 + }f(x) = \infty, \lim_{x \to \lambda_2 - }f(x) = -\infty.$$ therefore by intermediate value theorem, there is a root of $f$ between $\lambda_1$ and $\lambda_2.$

can you take it from here.

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