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I'm interested in solving the inequality $$-16 x-6\sqrt{6} \sqrt{x}+\frac{30 \sqrt{6}}{\sqrt{x}}+72<0.$$ When using the Mathematica command

Reduce[-16 x - 6 Sqrt[6] Sqrt[x] + (30 Sqrt[6])/Sqrt[x] + 72 < 0]

I got back

x > Root[-675 + 918 #1 - 315 #1^2 + 32 #1^3 &, 3]

meaning that $x$ is larger than the third (and largest) root of the polynomial $$-675+918t-315 t^2+32 t^3. $$ I can't see where this polynomial came from. I would appreciate if anyone could show me how the original inequality reduces into $x$ being greater than this polynomial's root. Thanks.

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Let $\;\sqrt{x} = \sqrt{6} \,t \gt 0\;$ then, after canceling out a factor of $\,6\,$, the inequality becomes:

$$16 t^2 + \sqrt{6} \cdot \sqrt{6} \,t - \frac{5 \sqrt{6}}{\sqrt{6} \, t} - 12 \gt 0 \quad\iff\quad 16 t^3 + 6 t^2 - 12 t - 5 \gt 0$$

The latter cubic has exactly one positive root $\,t_0 \gt 0\,$ (by the rule of signs for example), so the solution set of the original inequality is $\,x \gt 6 t_0^2\,$.

I do not see offhand where the polynomial returned by Mathematica would come from.


[ EDIT #2 ]   Mathematica's polynomial is in fact the minimal polynomial of $\,6 t_0^2\,$. This can be verified by first eliminating the odd powers from $\,16 t^3 + 6 t^2 - 12 t - 5\,$:

$$ (16 t^3 + 6 t^2 - 12 t - 5)\cdot(16 t^3 - 6 t^2 - 12 t + 5) = 256 t^6 - 420 t^4 + 204 t^2 - 25 $$

Then muliplying by $\,27\,$:

$$ 27(256 t^6 - 420 t^4 + 204 t^2 - 25) = 32 (6 t^2)^3 - 315 (6 t^2)^2 + 918 (6 t^2) - 675 $$

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    $\begingroup$ Are the values of $6t^2$ where $16t^3+6t^2-12t-5=0$ the roots of the Mathematica-returned polynomial? $\endgroup$ – Carl Schildkraut Feb 27 '18 at 23:39
  • $\begingroup$ @CarlSchildkraut Good point. The largest positive one (which matters), yes. Updated the answer with that. $\endgroup$ – dxiv Feb 28 '18 at 0:05

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