What is the solution set for -x^2+9?

Question

What is the solution set for -x^2+9>?

anonymous · Answer

$$-x^2+9>$$

anonymous · Answer

Yea but do you know what the solution set is?

anonymous · Answer

> than what? ...

anonymous · Answer

Oops. Sorry Kreshnik. >0

lgbasallote · Answer

go @Kreshnik ! hehe

anonymous · Answer

that's better ;) .... we have: $$\Large -x^2+9>0$$ $$\Large 9-x^2>0\quad \longrightarrow \[3^2-x^2>0$$\] now we use this formula... $$\Large a^2-b^2=(a-b)(a+b)$$ and we have.. $$\Large 3^2-x^2>0 \longrightarrow (3-x) (3+x )>0$$ from here we have two parts 3-x>0 or 3+x>0 so we get that... -x>-3 or x>-3 multiply left inequality by -1 and swap the sight > into < (everytime you multiply\divide by a negative number in inequalities you change sign !) and finally we have: x<3 or x>-3 so $$\LARGE x\in (-\infty ,3) \cup (-3, +\infty)$$ from here we get that... $$\LARGE x\in (-3,-2,-1,0,1,2,3)$$

anonymous · Answer

Wow you're great Kreshnik. I really appreciate all of your help.

anonymous · Answer

@lgbasallote  :P   am I right or not ? :P

anonymous · Answer

@mathsux4real  Glad to help ;)    (I got to go to school, I'm out ! see you later ;)   )

anonymous · Answer

Cool, see ya around!

lgbasallote · Answer

yes :D very good job @Kreshnik *thumbs up* ;D

anonymous · Answer

bye ! ;)   @lgbasallote  thanks ;)

anonymous · Answer

actually it should be...  

$$\LARGE x\in (-\infty,3)\cap(-3,+\infty)$$
  
the rest is ok.