Show that the equation $x^3+3x+1=0$ has exactly one real solution

Question

Show that $x^3+3x+1=0$ has exactly one real solution.

Answer 1

Hint: use Calculus. Note that derivative always takes positive value.

Answer 2

The discriminant is

$$\Delta = - 4 \cdot 3^3 - 27 = - 5 \cdot 27 < 0$$

From the negativity of $\Delta$, we conclude that the cubic polynomial has only one real root.

Answer 3

A general method would be to check value of polynomial at Maxima and minima (Wiggles). If both values have same sign then only one real root. If no Wiggles then too single root.

Answer 4

Let $f$ defined by $f(x)=x^3+3x+1$. we know there is at least one real root.

assume there are $2$ real roots $a$ and $b$.

the polynomial function $f$ is continuous at $[a,b]$ and has a derivative in $(a,b)$.

Role's Theorem allows us to say there exists $c\in(a,b)$ such that

$f'(c)=3c^2+3=0$ which is not possible.

so there is only one real root.