What could x be in here?

$\displaystyle x(x-6) < -6$

Excuse me for my bad explanation, my English is bad.

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- Sep 30th 2010, 02:13 PM #1

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- Sep 30th 2010, 02:17 PM #2

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- Sep 30th 2010, 03:17 PM #4

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- Oct 1st 2010, 02:03 AM #5

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Yet another way: x(x- 6)< -6 is the same as $\displaystyle x^2- 6x+ 6< 0$ which is the same as $\displaystyle x^2- 6x+ 6= x^2- 6x+ 9- 3= (x- 3)^2- 3= (x-3)^2- \sqrt{3}^2$$\displaystyle = (x- 3- \sqrt{3})(x- 3+ \sqrt{3})< 0$ (I used the fact that [tex]a^2- b^2= (a- b)(a+ b)).

The product of two numbers is negative if and only if one is positive and the other negative. Since the two factors are "x- 3" plus and minus the same thing, obviously $\displaystyle x- 3+ \sqrt{3}> x- 2- \sqrt{3}$. The first must be positive and the second negative: we must have $\displaystyle x- 3- \sqrt{3}< 0$ and $\displaystyle x- 3+ \sqrt{3}> 0$.

From the first $\displaystyle x< 3+ \sqrt{3}$ and from the second $\displaystyle x> 3- \sqrt{3}$. Together, those give, as before, $\displaystyle 3- \sqrt{3}< x< 3+ \sqrt{3}$.

klik11, you English is excellent. Certainly compared with my*(Put whatever language you want here)*!