How Do You Solve the Equation \(x^3(x+1) = 2(x+a)(x+2a)\) for Real \(a\)?

[anemone]

Here is this week's POTW:

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Solve the equation $x^3(x+1)=2(x+a)(x+2a)$ where $a$ is a real parameter.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to Opalg for his correct solution(Cool), which you can find below:

Spoiler

If $x^3(x+1) = 2(x+a)(x+2a)$ then $x^4 + x^3 - 2x^2 - 6ax - 4a^2 = 0$. That factorises as $$x^4 + x^3 - 2x^2 - 6ax - 4a^2 = (x^2-x-2a)(x^2+2x + 2a) = 0.$$ If $x^2-x-a = 0$ then $x = \frac12\bigl(1\pm\sqrt{1+8a}\bigr).$ If $x^2+2x + 2a = 0$ then $x = -1\pm\sqrt{1-2a}$.

Trade secret: where did the factorisation $x^4 + x^3 - 2x^2 - 6ax - 4a^2 = (x^2-x-2a)(x^2+2x + 2a)$ come from?
[sp]Replace the parameter $a$ by $y$, and get Desmos to graph the curve $x^4 + x^3 - 2x^2 - 6xy - 4y^2 = 0$:
[DESMOS]advanced: {"version":7,"graph":{"squareAxes":false,"viewport":{"xmin":-5,"ymin":-5,"xmax":5,"ymax":5}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"x^4+x^3-2x^2-6xy-4y^2=0"}]}}[/DESMOS]
This clearly splits into two separate parabolas, which both go through the origin. The upwards-opening parabola has its vertex at $\bigl(\frac12,-\frac18\bigr)$. The downwards-opening parabola has its vertex at $\bigl(-1,\frac12\bigr)$. From that it is easy to find that the first parabola has equation $y = \frac12\bigl(x-\frac12\bigr)^2-\frac18$, or $x^2 - x - 2y = 0$. And the second parabola has equation $y = -\frac12(x+1)^2 + \frac12$, or $x^2 + 2x + 2y = 0.$ Therefore $$x^4 + x^3 - 2x^2 - 6xy - 4y^2 = (x^2 - x - 2y)(x^2 + 2x + 2y).$$ Now replace $y$ by $a$ to get the required factorisation.[/sp]