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Solve the following equation $$x^3-3x=\sqrt{x+2}$$ without squaring of the both sides.

The domain is $x\geq-2$ but what is the rest?

For example, after squaring of the both sides we obtain: $$(x^3-3x)^2=x+2$$ with $x(x-\sqrt3)(x+\sqrt3)\geq0$, which gives $$x^6-6x^4+9x^2-x-2=0.$$ Since $$2^6-6\cdot2^4+9\cdot2^2-2-2=0,$$ we can get a factor $x-2$, but the rest is not so easy.

Id est, it's better to find a solution without squaring.

Thank you!

Michael Rozenberg
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  • By inspection, I would note that $x=2$ solves the equation (the $\sqrt{ }$ denotes positive square root, I presume). – Sarvesh Ravichandran Iyer Mar 26 '17 at 04:36
  • You could let $x+2=y^2$ and try to factor the resulting sextic. That will still leave a not-so-nice cubic to solve, with roots in range. – dxiv Mar 26 '17 at 04:37

2 Answers2

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Hint:

Set $x=2\cos2y,0\le2y\le\pi\ \ \ \ (1)$

$$\cos6y=\cos y$$

$\implies6y=2m\pi\pm y$ where $m$ is any integer.

Choose $m$ such that $(1)$ is satisfied

lab bhattacharjee
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Idea 1: $\Leftrightarrow (x-2)(x^2+x-1)(x^3+x^2-2x-1) = 0$

Idea 2:$\Leftrightarrow (x^3-3x-2)+(2-\sqrt{x+2})=0$

$\Leftrightarrow (x-2)(x^2+2x+1)- \frac{x-2}{2+\sqrt{x+2}}=0$

$\Leftrightarrow (x-2)(x^2-2x+1-\frac{1}{2+\sqrt{x+2}}) = 0$