# Inequality

• Oct 20th 2010, 10:52 AM
Lil
Inequality
$\displaystyle \left (\frac{\pi-3 }{2} \right )^{x^2-x}+\frac{6\pi}{4}-\frac{\pi^{2}+9}{4}> 0$

What I could do:confused:
• Oct 20th 2010, 10:55 AM
Ackbeet
I would throw everything over to the RHS that doesn't have an x in it. What does that give you?
• Oct 20th 2010, 10:59 AM
Lil
Quote:

Originally Posted by Ackbeet
I would throw everything over to the RHS that doesn't have an x in it. What does that give you?

'RHS' - what does it mean?
• Oct 20th 2010, 10:59 AM
earboth
Quote:

Originally Posted by Lil
$\displaystyle \left (\frac{\pi-3 }{2} \right )^{x^2-x}+\frac{6\pi}{4}-\frac{\pi^{2}+9}{4}> 0$

What I could do:confused:

$\displaystyle \left (\dfrac{\pi-3 }{2} \right )^{x^2-x}>-\dfrac{6\pi}{4}+\dfrac{\pi^{2}+9}{4}$

$\displaystyle \left (\dfrac{\pi-3 }{2} \right )^{x^2-x}>\left(\dfrac{\pi-3}2\right)^2$

Can you take it from here?

@Ackbeet: Thanks for spotting the typo.
• Oct 20th 2010, 11:00 AM
Ackbeet
RHS means "Right Hand Side". Similarly, LHS means "Left Hand Side".
• Oct 20th 2010, 11:00 AM
Lil
Quote:

Originally Posted by earboth
$\displaystyle \left (\dfrac{\pi-3 }{2} \right )^{x^2-x}>\dfrac{6\pi}{4}-\dfrac{\pi^{2}+9}{4}$

$\displaystyle \left (\dfrac{\pi-3 }{2} \right )^{x^2-x}>\left(\dfrac{\pi-3}2\right)^2$

Can you take it from here?

Yes, I can.
• Oct 20th 2010, 11:02 AM
Ackbeet
Quote:

$\displaystyle \displaystyle\left (\dfrac{\pi-3 }{2} \right )^{x^2-x}>\dfrac{6\pi}{4}-\dfrac{\pi^{2}+9}{4}$
Probably meant

Quote:

$\displaystyle \displaystyle\left (\dfrac{\pi-3 }{2} \right )^{x^2-x}>\dfrac{\pi^{2}+9}{4}-\dfrac{6\pi}{4}$
Corrected in the next line, though.
• Oct 20th 2010, 11:05 AM
Lil
Quote:

Originally Posted by Ackbeet
Probably meant

Corrected in the next line, though.

Yes, I notice it.