I can't find the root of this equation $\frac{x+3}{\sqrt[3]{(x^{2}-1)^{2}}+2\sqrt[3]{x^{2}-1}+4}+1-\frac{x^{2}+3x+9}{\sqrt{x^{3}-2}+5}=0 (1)$

Question

Solve $\left\{\begin{matrix} (x-\sqrt{y})(x^{2}+x\sqrt{y}+y-2) =6\ln(\frac{\sqrt{y}+\sqrt{y+9}}{x+\sqrt{x^2+9}})(*) \\ \sqrt[3]{y-1}+x = \sqrt{xy-2} \end{matrix}\right.$ $(x,y\in \mathbb{R})$

Here is the way i solve it $(*)\Leftrightarrow x^3-2x+6\ln(x+\sqrt{x^{2}+9})=(\sqrt{y})^3-2\sqrt{y}+6\ln(\sqrt{y}+\sqrt{y+9})$

So that, we have $f(t)= t^{3}-2t+6\ln(t+\sqrt{t^{2}+9}), t\in \mathbb{R}$ and $f(t)$ increases and $f(t)$ is a continuous function on $\mathbb{R}$

Thus, $x=\sqrt{y}$

After proving that $x=\sqrt{y}$, I have $\sqrt[3]{x^{2}-1}+x=\sqrt{x^{3}-2} \Leftrightarrow (x-3)(\frac{x+3}{\sqrt[3]{(x^{2}-1)^{2}}+2\sqrt[3]{x^{2}-1}+4}+1-\frac{x^{2}+3x+9}{\sqrt{x^{3}-2}+5})=0$

After this step, I can't find the root of this equation. $\frac{x+3}{\sqrt[3]{(x^{2}-1)^{2}}+2\sqrt[3]{x^{2}-1}+4}+1-\frac{x^{2}+3x+9}{\sqrt{x^{3}-2}+5}=0 (1)$

I don't know $(1)$ have its roots or not.Please give me some ideas to solve $(1)$ by not using complicate methods that is taught in the unversities.

I am not sure that the way i solve the problem is right or not. If you have any solutions, please show me!

Thank you very much!

Answer 1

"Please give me some ideas to solve (1)by not using complicate methods that is taught in the unversities."

Ok, well the simple method taught in universities is to use a software package. Here is some output from Mathematica:

---

You ask for an elementary approach.

Show that

(1) $x-1<(x^2-1)^{2/3}+2(x^2-1)^{1/3}$ for $x>1$ [Hint: AM/GM]

(2) $x^2+3x+9>(x^2-2)^{1/3}+5$ for $x^2>2$ [Hint: trivial]

which establishes that the original equation has no real roots.

Of course, the hard part is realising that (1) and (2) are true; proving them is easy.