# Proving Inequalities

• Jan 14th 2007, 09:19 AM
symmetry
Proving Inequalities
I needed to type the questions again because there was a typing mistake this morning when I originally posted the math questions.

QUESTION 1:

If a > 0, show that the solution set of the inequality x^2 < a consists of all numbers x for which -(sqrt{a}) < x < (sqrt{a}).

QUESTION 2:

If a > 0, show that the solution set of the inequality x^2 > a consists of all numbers x for which x > (sqrt{a}) OR x < -(sqrt{a}).

I hope this makes sense now.

Thanks.
• Jan 14th 2007, 10:48 AM
CaptainBlack
Quote:

Originally Posted by symmetry
QUESTION 2:

If a > 0, show that the solution set of the inequality x^2 > a consists of all numbers x for which x > (sqrt{a}) OR x < -(sqrt{a}).

Both ImPerfectHacker and CaptainBlack did this one in the other thread
as far as I can recall, and the other is a direct consequence of this result.

RonL
• Jan 14th 2007, 01:06 PM
symmetry
ok
So, basically there is no need to go over the second part of the question?

Yes? No?