115. An difficult equation with one radical.

by Virgil Nicula, Sep 12, 2010, 8:37 PM

Quote:
Solve the equation $x^3-3x=\sqrt {x+2}$ , where $x\in\mathbb R$ .

Proof. $\left\{\begin{array}{c}
x+2\ge 0\\\
x(x^2-3)\ge 0\end{array}\right|\ \implies\ x\in \left[-\sqrt 3,0\right]\cup\left[\sqrt 3,\infty\right)\ (*)$ . Observe that for $x>2$ we have

$x^3-3x=(x-2)^3+6x^2-15x+8>6x^2-15x+8=6(x-2)^2+9x-16>9x-16$

and $9x-16>\sqrt {x+2}$ $\iff$ $(9x-16)^2>x+2$ $\iff$ $(81x-127)(x-2)>0$ . Therefore, $x>2\implies$

$x^3-3x>\sqrt {x+2}$ , absurd. In conclusion, $x\le 2$ and using the relation $(*)$ obtain that $\boxed{x\in \left[-\sqrt 3,0\right]\cup\left[\sqrt 3,2\right]}$ .

Using the substitution $x=2\cos \phi$ where $\phi\in \left[0,\frac {\pi}{6}\right]\cup\left[\frac {\pi}{2},\frac {5\pi}{6}\right]\subset\left[0,\pi\right)$ , the given equation becomes

$4\cos^3\phi -3\cos\phi =\cos\frac {\phi}{2}\iff \cos 3\phi =\cos\frac {\phi}{2}$ from where obtain that $\phi\in\left\{0,\frac {4\pi}{7},\frac {4\pi}{5}\right\}$ $\iff$ $x\in\left\{\ 2\ ,\ 2\cos\frac {4\pi}{7}\ ,\ 2\cos \frac {4\pi}{5}\ \right\}$ .



Another similar proposed problem. Find all solutions of the equation $x^3-3x=\sqrt{4-x^2}$ , where $x\in\mathbb R$ .

Proof 1. Is clearly that for any $x\in [-2,2]$ exists uniquely $\theta\in [0,\pi ]$ so that $x=2cos\theta$ . Thus, our equation becomes $\cos 3\theta=\sin\theta$ a.s.o.

Proof 2. Squaring and factorizing gives us $(x^2-2)(x^4-4x^2+2)=0$ a.s.o.
This post has been edited 19 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:49 AM

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