Number of irrational roots in polynomial equation

Question

Number of irrational roots of the equation $y^3-3y=\sqrt{y+2}$

Trial: put $y=2\cos \theta$ So we have $4\cos^3\theta-3\cos\theta =\bigg|\sin^2\frac{\theta}{2}\bigg|$

$\cos 3\theta = \bigg|\sin^2\frac{\theta}{2}\bigg|$

Could some help me to calculae Irrational roots, thanks

Answer 1

After your substitution we obtain $$\cos3\theta=\left|\cos\frac{\theta}{2}\right|$$ 1. $\cos3\theta=\cos\frac{\theta}{2}$ gives $$3\theta=\pm\frac{\theta}{2}+360^{\circ}k,$$ where $k\in\mathbb Z$, which gives $x=144^{\circ}k$ or $x=\frac{720^{\circ}}{7}k.$

Now, since $\cos3\theta\geq0$, we obtain here: $$x=2\cos144^{\circ}$$ or $$x=2\cos\frac{720^{\circ}}{7}.$$ The case $\cos3\theta=-\cos\frac{\theta}{2}$ for you.

Answer 2

$(y^3- 3y)^2 = y+2$ is a six degree polynomial with a potential of the $6$ roots.

$y^6 - 6y^4 + 9y^2 - y -2 = 0$.

If it has rational roots they will be by rational roots theorem $\pm 1, \pm 2$.

$1 - 6 + 9 \pm 1 - 2 = 1,3$ and $64 - 6*16 + 36 \pm 2 -1 = 1,5$ so there are no rational roots.

But it might have imaginary roots, and we must discard any irrational roots that are less than negative $-2$.

If $|y| > y$ then $|y|^3 = 4|y|$ so $|y^2 - 3y|> |y|$ and $(y^2 - 3y)^2 > y^2 > 2|y|> |y| + 2 > y+2$.

So all the solutions are within $-2 < y < 2$ and so $\sqrt{y+2}$ is always defined. So all six roots are in the range... if they are real.

Doing your trig substition there

We have $\cos \theta = \sin^2 \frac {\theta}2$ (no need for absolute value signs; $\sin^2 $ is non negative.

If $\theta_1$ is a solution then $-\theta_1$ is also a solution. (And since $y = 2\cos \theta$ is irrational they are distinct solutions.)

In the first quadrant $\cos$ is decreasing while $\sin$ is increasing so there will be exactly one solution when $0 < \theta < 30$ and $3\theta$ is in the first quadrant. There aren't any when $3\theta$ is in the second quadrant (as $\cos$ will be negative. Any solutions in the third or fourth quadrants are just the negatives of solutions in the first or second so we don't worry about them.

And when $360 < 3\theta < 450$, when $3\theta$ is back in the first quadrant and $120 < \theta < 150$, there is exactly one solution. But it's the exact same solution as $\cos 3\theta = \cos 3(120 + \theta)$.

So there is one solution for $\theta_1 < \theta < 30$ and one for $\theta_2$.

So there are two irrational solutions and $4$ complex solutions.