Inequality

**Lil** · October 20th 2010, 10:52 AM

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

What I could do

**Ackbeet** · October 20th 2010, 10:55 AM

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

**Lil** · October 20th 2010, 10:59 AM

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?

**earboth** · October 20th 2010, 10:59 AM

Originally Posted by Lil

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

What I could do

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

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

Can you take it from here?

@Ackbeet: Thanks for spotting the typo.

**Ackbeet** · October 20th 2010, 11:00 AM

RHS means "Right Hand Side". Similarly, LHS means "Left Hand Side".

**Lil** · October 20th 2010, 11:00 AM

Originally Posted by earboth

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

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

Can you take it from here?

Yes, I can.

**Ackbeet** · October 20th 2010, 11:02 AM

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

Probably meant

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

Corrected in the next line, though.

**Lil** · October 20th 2010, 11:05 AM

Originally Posted by Ackbeet

Probably meant

Corrected in the next line, though.

Yes, I notice it.

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