Find the number of real solutions of $x^3-3x-\sqrt{5}=0$ Find the number of real solutions of the equation  $x^3-3x-\sqrt{5}=0$
I don't know that how start for find !
 A: Hint :
Let $f : x \mapsto x^3-3x-\sqrt{5}$. Then $f$ is differentiable and $f'(x)=3(x^2-1)$, so $f'$ vanishes for $x=-1$ and $x=1$, and $f$ has a local maximum at $x=-1$.
But $f(-1)=2-\sqrt{5} < 0$, so $f$ has only one real zero, i.e. the equation has only one solution.
A: You can be sure that the equation has atleast one real solution since $f(1)<0$ and $f(3)>0$ and by Intermediate value theorem for continuous functions(where, $f(x)=x^3-3x-\sqrt5$ )there exists an $x_0 \in (1,3)$ such that $x_0^3 - 3x_0 -\sqrt5=0$. Moreover, you can also be certain that this is the only real solution of the equation since the function increases on $(3, \infty)$ (can be proven use second derivative test) and is negative on $(-\infty,1)$.
Given $x^3-3x-\sqrt5 = 0$
$$x^3-3x=\sqrt5$$
Squaring on both sides(since $\sqrt5$ is irrational and to use Rational root theorem I want all coefficients to be integers),
$$x^6-6x^4+9x^2-5=0$$
Using Rational root theorem the only possible rational solutions of the above equation are $1,-1,\frac{1}{5},-\frac{1}{5}$ and checking manually none of them are roots of the above equation. So the given equation has certainly no rational solutions.
Therefore $x^3-3x-\sqrt5$ has only one real solution(in addition you can also be certain that it is not a rational number).
