inequalities help!

**calchelp** · June 22nd 2007, 10:23 PM

how do u solve this inequality?

a) x^2 - x - 8 < 0

b) x ^3 + 3x^2 - 18x - 22 < and equal to 0

**earboth** · June 23rd 2007, 12:58 AM

Originally Posted by calchelp

how do u solve this inequality?

a) x^2 - x - 8 < 0
...

Hello,

to a):

$x^2-x+\frac{1}{4} < \frac{1}{4}+8$ ...................complete the square
$\left(x-\frac{1}{2}\right)^2 < \frac{33}{4}$ ...............calculate the square-root

$\left|x-\frac{1}{2}\right| < \frac{1}{2} \cdot \sqrt{33} \Longleftrightarrow x-\frac{1}{2} < \frac{1}{2} \cdot \sqrt{33}\ \wedge \ -\left(x-\frac{1}{2}\right) < \frac{1}{2} \cdot \sqrt{33}$

$\Rightarrow x < \frac{1}{2}+\frac{1}{2} \cdot \sqrt{33} \ \wedge \ x > \frac{1}{2}-\frac{1}{2} \cdot \sqrt{33}$

**calchelp** · June 23rd 2007, 06:11 AM

im sorry could u explain what step u took...im confused how that went to the next

**topsquark** · June 23rd 2007, 06:57 AM

Originally Posted by calchelp

a) x^2 - x - 8 < 0

Let's try it this way.

Solve
$x^2 - x - 8 = 0$

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-8)}}{2 \cdot 1}$

$x = \frac{1 \pm \sqrt{33}}{2}$

So
$x^2 - x - 8 = \left ( x - \frac{1 + \sqrt{33}}{2} \right ) \left ( x - \frac{1 - \sqrt{33}}{2} \right )$

Thus we need to solve:
$\left ( x - \frac{1 + \sqrt{33}}{2} \right ) \left ( x - \frac{1 - \sqrt{33}}{2} \right ) < 0$

Now, break the real numbers into three intervals and check the inequality on these intervals:
$\left ( -\infty, \frac{1 - \sqrt{33}}{2} \right )$ ==> $\left ( x - \frac{1 + \sqrt{33}}{2} \right ) \left ( x - \frac{1 - \sqrt{33}}{2} \right ) > 0$ (No!)

$\left ( \frac{1 - \sqrt{33}}{2}, \frac{1 + \sqrt{33}}{2}\right )$ ==> $\left ( x - \frac{1 + \sqrt{33}}{2} \right ) \left ( x - \frac{1 - \sqrt{33}}{2} \right ) < 0$ (Check!)

$\left ( \frac{1 + \sqrt{33}}{2}, \infty \right )$ ==> $\left ( x - \frac{1 + \sqrt{33}}{2} \right ) \left ( x - \frac{1 - \sqrt{33}}{2} \right ) > 0$ (No!)

So the inequality is only true on
$\left ( \frac{1 - \sqrt{33}}{2}, \frac{1 + \sqrt{33}}{2}\right )$
as Earboth said.

-Dan

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