Can You Solve This Cubic and Square Root Equation?

[anemone]

Here is this week's POTW:

-----

Solve the equation $x^3-3x=\sqrt{x+2}$.

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Olinguito
2. castor28
3. Cbarker1
4. Opalg

Solution from Opalg:

Spoiler

Any solution to the equation $x^3 - 3x = \sqrt{x+2}$ must satisfy $-2\leqslant x\leqslant 2$, because if $x<-2$ then the right side is not defined, and if $x>2$ then the left side is greater than the right.

Therefore all solutions must be of the form $x = 2\cos\theta$ for some $\theta$. Substitute that into the equation to get $$8\cos^3\theta - 6\cos\theta = \sqrt{2\cos\theta + 2},$$ $$2(4\cos^3\theta - 3\cos\theta) = \sqrt{2(\cos\theta + 1)},$$ $$2\cos(3\theta) = \sqrt{2(\cos\theta + 1)}.\qquad(*)$$ Now square both sides: $$4\cos^2(3\theta) = 2(\cos\theta+1),$$ $$2\cos^2(3\theta) - 1 = \cos\theta,$$ $$\cos(6\theta) = \cos\theta.$$ It follows that $6\theta = 2k\pi \pm\theta$ (for some integer $k$), so that either $\theta = \frac{2k\pi}5$ or $\theta = \frac{2k\pi}7$. But that includes several extraneous solutions arising from when the equation was squared. In fact, it follows from equation $(*)$ that we must have $\cos(3\theta)\geqslant0$. The only remaining values of $\theta$ to satisfy that are $\theta = 0$, $\frac{4\pi}5$ and $\frac{4\pi}7$. So the solutions of the original equation are $$ x = 2\cos 0 = 2,$$ $$x = 2\cos\tfrac{4\pi}5 = -\tfrac12(\sqrt5 + 1),$$ $$x = 2\cos\tfrac{4\pi}7.$$