Solve $3x^2+2=0$ for $ x$?

Question

Here is the full question:

$ f(x) = x^3 + 2λx $ where $\lambda $ is a real parameter.

Find its stationary point(s) and discuss the nature of the stationary point(s), in the case where $\lambda > 0$, $\lambda = 0$, and $\lambda < 0$.

I tried to solve $3x^2 + 2 = 0$ for $x$, but I can only get $\sqrt{-\frac 23}$ as a root.

So I don't know how to find stationary points of $f$.

Answer 1

we get $$x^2=-\frac{2}{3}$$ and this is $$x_{1,2}=\pm\sqrt{\frac{2}{3}}i$$ with $$i^2=-1$$

Answer 2

You would get complex numbers: $$x=\pm\sqrt{\frac{2}{3}}i$$

Answer 3

There are no solutions in the $\mathbb{R}$.

The solutions are in $\mathbb{C}$

$ x=\pm i\sqrt{\frac{2}{3}} $

Answer 4

Using the quadratic formula, where $a=3$, $b=0$, and $c=2$.

$$x=\dfrac{\pm\sqrt{(-0)^2-4(3)(2)}}{6}=\dfrac{\pm\sqrt{-24}}{6}=\dfrac{\pm2\sqrt{-6}}{6}=\pm\dfrac{\sqrt{-6}}{3}$$

Therefore, there are no solutions in the real world.

Here, $\pm\dfrac{\sqrt{-6}}{3}=\pm\dfrac{\sqrt{6}\sqrt{-1}}{3}=\pm\dfrac{\sqrt{6}i}{3}$

Where $i=\sqrt{-1}$ and $i^2=-1$. When introducing $i$, you get into the imaginary/complex world.

Answer 5

Have you tried moving things around? $$3x^2 + 2 = 0$$ Subtract $2$ from both sides: $$3x^2 = -2$$ Divide both sides by $3$: $$x^2 = -\frac{2}{3}$$ Then, taking the square root of both sides: $$x = \sqrt{-\frac{2}{3}}$$ Uh oh... have you learned about imaginary numbers yet? Like $i = \sqrt{-1}$?