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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 $f$.

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    $\begingroup$ No solutions for real numbers ... $\endgroup$ – Bram28 Mar 15 '18 at 17:52
  • $\begingroup$ You need $\pm \sqrt{-\frac 23}$ i.e, $x = \pm \sqrt{\frac 23}i$.Can you please tell us how this relates to finding stationary points? $\endgroup$ – Namaste Mar 15 '18 at 18:05
  • $\begingroup$ How in the world is this calculus or linear algebra? $\endgroup$ – user535339 Mar 15 '18 at 18:06
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    $\begingroup$ Dave: I'd suggest you post the entire question. Do you have a function $f(x) = y = x^3 + 2x +C$ (where C is some constant)? for which you are trying to find the stationary points? Please double check your function and your derivative. In particular, please check whether you started with $f(x) = x^3 -2x + C$ (again, C simply represents a constant, which disappears after finding $f'(x).$ $\endgroup$ – Namaste Mar 15 '18 at 18:43
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    $\begingroup$ On posting the original question, it is clear that you made multiple errors. Firstly, you differentiated $f$ wrongly. Secondly, you seem to not know how to interpret the roots of $f'$ to tell you the stationary points of $f$. You should try to get a teacher or textbook that can explain that properly to you. $\endgroup$ – user21820 Mar 16 '18 at 15:06
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There are no solutions in the $\mathbb{R}$.

The solutions are in $\mathbb{C}$

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

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

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

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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.

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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 $i = \sqrt{-1}$?

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