We are given two numbers $a>0$ and $b>0$ and the equation given is
$$ \frac{a}{x^3 + 2x^2 – 1} + \frac{b}{x^3 + x – 2} = 0 $$
Prove that this equation has at least one solution in the interval $(-1,1)$. Now I can let
$$ f(x) = \frac{a}{x^3 + 2x^2 – 1} + \frac{b}{x^3 + x – 2} $$
And then I can think of using intermediate value theorem (IVT) on the interval $[-1,1]$. But the problem is that $ x^3 + 2x^2 – 1 $ has zeroes at $-1$ and $0.618$ and $x^3 + x – 2$ has a real zero at $1$. So, that means that $f(x)$ is not continuous on $[-1,1]$. So, how can IVT be used here ?
Thanks
Best Answer
The two cubics do not have a common root.That can be exploited as follows:
Let $g(x)=a(x^{3}+x-2)+b(x^{3}+2x^{2}-1)$. Then $g(1)=2b>0$ and $g(-1)=-4a<0$. Hence, there exists $x \in (-1,1)$ such that $g(x)=0$. At this point $x$ neither $x^{3}+x-2$ nor $x^{3}+2x^{2}-1$ can be $0$: If $x^{3}+x-2=0$ we get $g(x)=b(x^{3}+2x^{2}-1)\neq 0$ and if $x^{3}+2x^{2}-1=0$ then $g(x)=a(x^{3}+x-2) \neq 0$. Hence, we can divide the equation $g(x)=0$ by $(x^{3}+x-2) (x^{3}+2x^{2}-1)$.
EDIT: A simple proof of the fact that the two cubics do not have a common root is posted below by user2661923