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Having $ \frac{|x-3|}{x} + |x^2-2x+1| + x > 0$ how can I arrive to the solution:

$ x < \frac 13 \left( 1-\sqrt[3]{\frac{2}{79-9\sqrt{77}}} - \sqrt[3]{\frac 12 \left(79 - 9\sqrt{77}\right)} \right) $

?

Of course, $x \neq 0$, and then I've tried all the possible ways I could think of by splitting up the real numbers in three intervals, $x < 0$, $(x>0 \land x<3)$ and $x > 3$, but I couldn't come up with anything as close to the solution above.

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    *I'll end up with different intervals (to unify)*... Once again, since every $x\gt0$ is solution, the only interval to study is $(-\infty,0)$. On which $|x-3|=3-x$ and $|x^2-2x+1|=x^2-2x+1$, so I wonder what you want to *unify*.2012-11-06

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Alrighty, here is my attempt at a solution. Bear with me, I'm terrible with LaTeX, so if the formatting is awful, someone please feel free to fix it.

  • Building off of what @Did said.

First, let $f(x) = x^2-2x+1$

  • Show that the absolute value around $f(x)$ is unnecessary (needs Calculus, but for simple proof, plot it. :)).

$\min(f(x)) \Rightarrow f'(x) = 2x-2 = 0, \Rightarrow x = 1$ ($x = 1$ is the point where the slope goes from negative to positive, or vice versa)

$f''(x) = 2 > 0 \Rightarrow$ concave upwards, and thus the only minima.
$f(1) = 0$, so $\min(f(x)) = 0$, so the $\text{abs}()$ is not necessary --- we thus remove it.

  • Simplify remaining equation

We are now left with:
$\frac{|x-3|}{x} + x^2-2x+1 + x > 0 \Leftrightarrow |x-3| + x\cdot(x^2-x+1) > 0 \Leftrightarrow $ $|x-3| + x^3 -x^2 + x > 0$

This means, that there are only two possible domains that need searching, so why not search them exhaustively?

For all $x \geq 3$, we have: $|x-3| +x^3-x^2+x > 0 \Leftrightarrow x^3-x^2+2x-3 > 0.$

Therefore, to find the locations of interest, look at where that inequality equals zero.

$x^3-x^2+2x-3 = 0$

To solve this, you need to make use of the cubic formula, which is huge. Suffice it to say (aka Wolfram Alpha) gives the exact real solution as: $\frac{1}{3} \left(1-5^{2/3} \left( \frac{2}{13+3 \sqrt{21}} \right) ^ {1/3} + \left( \frac{5}{2}(13+3 \sqrt{21}) \right) ^ {2/3} \right),$ which is not what we are looking for.

Therefore, let's try the other possibility, $x < 3$: $-(x-3)+x^3-x^2+x > 0$ $x^3-x^2+3 > 0$

Same methodology: $x^3-x^2+3 = 0$

Wolfram Alpha gives: $\frac{1}{3} \left(1-\left(\frac{2}{79-9 \sqrt{77}}\right)^{1/3}-\left(\frac{1}{2} (79-9 \sqrt{77})\right)^{1/3}\right)$

Which is exactly the expression you were looking for!

Now you just have to try values on either side, and figure out which side works. Best bet is a rounded numerical approximation of the last value --- run through a calculator.

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    Means, the "first step of the simplification". There is exactly one negative root of $x^3-x^2+3$ by counting sign variations in the coefficient sequence $(-1,-1,3)$ and x^3-x^2+3<0 is obviously true for all $x$ that are smaller than this negative root. To get an idea of the size of the expression, note that $9\sqrt{77}=\sqrt{77\cdot 81}=\sqrt{79^2-4}$.2014-05-27