MHB Can You Find the Real Solutions to This Complex Equation?

[anemone]

Here is this week's POTW:

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Solve for the real solution(s) for the equation below:

$$\sqrt[3]{2+3x^2-15x^3}-x=1+71\sqrt{16x^3+3x-1}$$

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

 

[anemone]

Congratulations to lfdahl for his correct solution, which you can find below::)

Spoiler

Given the equation:

$\sqrt[3]{2+3x^2-15x^3}-x=1+71\sqrt{16x^3+3x-1} \;\;\;\;\;\;\;\;\;\;\;\;(1).$

Since, we´re looking for real solutions only, the polynomial on the RHS must obey the inequality:

$16x^3+3x-1 \ge 0\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\; (2).$

Equality holds when $x=\frac{1}{4}$ (the two other roots are complex).

The LHS must obey the inequality:

$\sqrt[3]{2+3x^2-15x^3}-x \geq 1 \\\\ \Rightarrow 2+3x^2-15x^3 \geq (1+x)^3=1+3x+3x^2+x^3 \\\\ \Rightarrow 16x^3+3x-1\leq 0\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\;\;\;\; (3).$

Both $(2)$ and $(3)$ must hold, therefore: $16x^3+3x-1 = 0$ is required,

thus $x=\frac{1}{4}$ is the only real solution to $(1)$.