# Quod Erat Demonstrandum

## 2012/09/25

### 答家軒

Filed under: Pure Mathematics — johnmayhk @ 7:07 下午

Question

For positive numbers $a,b$, prove that

$\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2}=0$

has at least one root in ($-1,1$).

Solution

Let

$f(x)=\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2}=\frac{a}{(x+1)(x^2+x-1)}+\frac{b}{(x-1)(x^2+x+2)}$

For $x\rightarrow 1^{-}$,

$(x-1)(x^2+x+2) \rightarrow 0^{-}$

$\frac{b}{(x-1)(x^2+x+2)} \rightarrow -\infty$

also,

$\frac{a}{(x+1)(x^2+x-1)} \rightarrow \frac{a}{2}$;

hence $f(x) \rightarrow -\infty$ as $x \rightarrow 1^{-}$.

For $x\rightarrow (\frac{-1+\sqrt{5}}{2})^{+}$,

$(x+1)(x^2+x-1) \rightarrow 0^{+}$

$\frac{a}{(x+1)(x^2+x-1)} \rightarrow +\infty$

also,

$\frac{b}{(x-1)(x^2+x+2)}$ is a finite negative number;

hence $f(x) \rightarrow +\infty$ as $x \rightarrow (\frac{-1+\sqrt{5}}{2})^{+}$.

And, $f(x)$ is continuous on ($\frac{-1+\sqrt{5}}{2},1$);

thus, $f(x)=0$ has at least one root in ($\frac{-1+\sqrt{5}}{2},1$) $\subset$ ($-1,1$).