Proving that an equation has exactly two solutions in the reals

Question

Let $a_1,a_2,a_3,b_1,b_2,b_3\in \mathbb{R}$, $a_1,a_2,a_3>0$, $b_1<b_2<b_3$.
Prove that $\frac{a_1}{x-b_1}+\frac{a_2}{x-b_2}+\frac{a_3}{x-b_3} = 0$ has exactly two solutions in $\mathbb{R}$.

First, I showed that the solutions must be bounded in $[b_1,b_3]$.
Since there are 3 points that the function is undefined at, I couldn't use any theorem that requires the function to be continuous. I know that it's continuous between $(b_1,b_2)$ and $(b_2,b_3)$, so I was trying to use IVT, but it also didn't work.

Could you please give me a hint?

Answer 1

You need to use IVT in the two intervals $(b_1,b_2)$ and $(b_2, b_3)$ near the endpoints.

At $b_1^+$, $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3} \to \infty$.

That is because $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3}> \frac{{a_1}}{x-b_1}+\frac{{a_2}}{b-b_2}+\frac{{a_3}}{b-b_3} \to \infty$, where $b = \frac {b_1+b_2}2$.

Similarly at $b_2^-$, $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3} \to -\infty$.

By IVT, this shows that $\frac{{a_1}}{x-b_1}+\frac{{a_2}}{x-b_2}+\frac{{a_3}}{x-b_3} =0$ for some $x \in (b_1,b_2)$. The result for $(b_2, b_3)$ is similar.

Answer 2

Obviously $x\ne b_1,b_2mb_3$ a simplification gives: $$f(x)=a_1(x-b_2)(x-b_3)+a_2(x-b_1)(x-b_3)+a_3(x-b_1)(x-b_2)$$ So we get $$f(b_1)=a_3(b_1-b_2)(b_1-b_3)>0$$ $$f(b_2)=a_2(b_2-b_1)(b_1-b_3)<0$$ $$f(b_3)=a_3(b_3-b_1)(b_2-b_2)>0$$ By IVT $f(x)=0$ has two real roots one in $(b_1,b_2)$ and one in $(b_2,b_3)$. Effectively, the given eq. is a quadratic which has both rotts real.