Question:
Given $\lambda_1<\lambda_2<\lambda_3 \in \Bbb R, a_1,a_2,a_3>0 $
Show that this equation has exactly 2 solutions:
$\frac {a_1}{x-\lambda_1}+\frac {a_2}{x-\lambda_2}+\frac {a_3}{x-\lambda_3}=0$
What I did:
Try 1: Since we are studying continuity, Weirstrass theorem and Mean value theorem at the moment, I think one of these theorems should somehow be involved in the proof…
Assuming that x is not equal to any of the lambdas I multiplied by the denumerators and got a quadratic equation- From this I figure that there must be at most 2 solutions. I thought about using the discriminant to prove that that there are exactly 2, but the calculation is very long, and I don't think this is what the lecturer intended for us to do, considering what we are studying now.
Try 2: Thought of using the Intermediate value theorem. So I defined a function $f(x)=\frac {a_1}{x-\lambda_1}+\frac {a_2}{x-\lambda_2}+\frac {a_3}{x-\lambda_3}$ and I want to find when it equals 0.
but Intermediate value theorem only involves two lambdas, and also I don't know if the lambdas are negative or positive (some/all of them)- Should I divide to cases?
Hints are welcome at first 🙂
Best Answer
Yes, try to divide into cases.
More complete answer :
For $x \in ]- \infty, \lambda_1[$, all your terms are negative -> No solutions
For $x \in ]\lambda_1, \lambda_2[$ -> use the continuity of the function as you said on the try 2
For $x \in ]\lambda_2, \lambda_3[$ -> use the continuity of the function as you said on the try 2
For $x \in ]- \infty, \lambda_3[$, all your terms are positive -> No solutions