# Proving that an equation has exactly two solutions in the reals

Let $$a_1,a_2,a_3,b_1,b_2,b_3\in \mathbb{R}$$, $$a_1,a_2,a_3>0$$, $$b_1.
Prove that $$\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?

• You can use IVT on on [b_1+r,b_2-r] for small r and the property that the function is locally monotonous. Commented Jan 16, 2021 at 11:51
• @TomTom314 Can I just say "Let $r\in \mathbb R$ be a very small positive number?
– Lilo
Commented Jan 16, 2021 at 12:01
• Yes, I should do that. Commented Jan 16, 2021 at 12:04
• Here is the SAME question asked and answered yesterday math.stackexchange.com/questions/3986096/… Commented Jan 16, 2021 at 13:05

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.

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.

• Aren't $f(b_1), f(b_2), f(b_3)$ undefined since $x\neq b_1,b_2,b_3$?
– Lilo
Commented Jan 16, 2021 at 11:59
• Please mote that my $f(x)$ is quadratic not the original one you may call it $g(x)$ since $b_1,b_2,b_3$ cannot be the roots of this equation. Hence roots of $f(x)=0$ and $g(x)=0$ are the same. Commented Jan 16, 2021 at 12:02
• Do I need to explain the relation between $g(x)$ and $f(x)$ or is it trivial enough?
– Lilo
Commented Jan 16, 2021 at 12:03