To find a extremal point of a function with parameters

Question

I have a function $$f(x) = (x-5m)(x+m)^2$$ I have tried to find the extremal points of this function (and then find if it's local maxima or minima). That means I need to find the x of derivative. The derivative is: $$-9m^2 - 6m + 3x^2$$ From there I tried to find what's the value of x: \begin{align} \ -9m^2 - 6mx + 3x^2 = 0\\ \ 3x^2 - 6mx = 9m^2 \\ \ x^2 - 2mx = 3m^2 \\ \ x(x-2m) = 3m^2 \\ \ x= \frac{3m^2}{x-2m} \end{align}

I checked with WolframAlpha that x needs to be 3m or -m, but I don't success to solve it. Someone that can help to get to this two answers? WolframAlpha problem.

Answer 1

From the quadratic formula, the solutions of $ax^2+bx+c=0$ are given by $$x_\pm = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ and in your case $a=1, b=-2m, c=-3m^2$ so you get $$x_\pm = \frac{2m \pm \sqrt{4m^2+12m^2}}{2} = \frac{2m \pm \sqrt{16m^2}}{2} = \frac{2m \pm 4m}{2} = -m \text{ or } 3m. $$

Answer 2

You have made a mistake.

$\Large x=\frac{3m^{2}}{x-2m}$ does not really give a value of $x$.

Both sides have $x$.

This is $x^{2}-2mx-3m^{2}=0$.

Now use Sridharacharya's Formula for quadratic equation $$x=\frac{-b\pm {\sqrt{b^{2}-4ac}}}{2a}$$ for an equation like $ax^{2}+bx+c=0$

and you will get $3m$ or $-m$ as answers.