Solve the equation x^3 −3x=(√(x+2))


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Question Number 152543 by liberty last updated on 29/Aug/21

$Solve the equation$ $x^{3} - 3 x = \sqrt{x + 2}$
	Answered by EDWIN88 last updated on 29/Aug/21
	$it is clear for x≥−2 case(1) −2≤x≤2 ,let x=2cos y ,0≤y≤π 8cos^3 y−6cos y=(√(2(cos y+1))) 2cos 3y=(√(4cos^2 (y/2))) cos 3y=cos (y/2) ; 3y=±(y/2)+2kπ x=2cos 0=2, x=2cos ((4π)/5) , x=2cos ((4π)/7) the other case x>2 have no solution . Therefore the solution is ={2,2cos ((4π)/5),2cos ((4π)/7)}$
	$i t i s c l e a r f o r x ⩾ - 2$ $c a s e (1) - 2 ⩽ x ⩽ 2, l e t x = 2 \cos y, 0 ⩽ y ⩽ π$ $8 \cos^{3} y - 6 \cos y = \sqrt{2 (\cos y + 1)}$ $2 \cos 3 y = \sqrt{4 \cos^{2} \frac{y}{2}}$ $\cos 3 y = \cos \frac{y}{2}; 3 y = \pm \frac{y}{2} + 2 k π$ $x = 2 \cos 0 = 2, x = 2 \cos \frac{4 π}{5}, x = 2 \cos \frac{4 π}{7}$ $t h e o t h e r c a s e x > 2$ $h a v e n o s o l u t i o n . T h e r e f o r e$ $t h e s o l u t i o n i s = {2, 2 \cos \frac{4 π}{5}, 2 \cos \frac{4 π}{7}}$
	Answered by MJS_new last updated on 29/Aug/21

	$squaring (beware of false solutions!) and$ $transforming \Rightarrow$ $x^{6} - 6 x^{4} + 9 x^{2} - x + 2 = 0$ $(x - 2) (x^{2} + x - 1) (x^{3} + x^{2} - 2 x - 1) = 0$ $\Rightarrow$ $x_{1} = 2 [true] ★$ $x_{2} = - \frac{1}{2} - \frac{\sqrt{5}}{2} [true] ★$ $x_{3} = - \frac{1}{2} + \frac{\sqrt{5}}{2} [false]$ $x^{3} + x^{2} - 2 x - 1 = 0$ $\Rightarrow$ $x_{4} = - \frac{1}{3} - \frac{2 \sqrt{7}}{3} \cos (\frac{π}{6} + \frac{1}{3} \arcsin \frac{\sqrt{7}}{14}) [false]$ $x_{5} = - \frac{1}{3} - \frac{2 \sqrt{7}}{3} \sin (\frac{1}{3} \arcsin \frac{\sqrt{7}}{14}) [true] ★$ $x_{6} = - \frac{1}{3} + \frac{2 \sqrt{7}}{3} \sin (\frac{π}{3} + \frac{1}{3} \arcsin \frac{\sqrt{7}}{14}) [false]$
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