# Thread: What could x be

1. ## What could x be

What could x be in here?
$\displaystyle x(x-6) < -6$

2. Originally Posted by klik11
What could x be in here?
$\displaystyle x(x-6) < -6$

x^2 - 6x + 6 < 0

Draw a graph of y = x^2 - 6x + 6 and get it's x-intercepts. You want the values of x for which y < 0 (that is, the parabola is below the x-axis).

3. Originally Posted by klik11
What could x be in here?
$\displaystyle x(x-6) < -6$

$\displaystyle x^2 - 6x < -6$

$\displaystyle x^2 - 6x + 9 < -6 + 9$

$\displaystyle (x - 3)^2 < 3$

$\displaystyle |x-3| < \sqrt{3}$

$\displaystyle -\sqrt{3} < x-3 < \sqrt{3}$

$\displaystyle 3-\sqrt{3} < x < 3 + \sqrt{3}$

4. Thank you both

5. 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)!