SOLUTION: 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 figured it out

Algebra ->  Inequalities -> SOLUTION: 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 figured it out       Log On


   

Question 660769: If +a+%3E+0+ show that the solution set of the inequality +x%5E2+%3E+a+ consists of all numbers x for which +x+%3C+-+sqrt%28a%29+ or +x+%3E+sqrt%28a%29+.
I figured it out all the way to the point +%28x+%2B+sqrt%28a%29%29%28x+-+sqrt%28a%29%29+%3E+0+ from here ,when I try to solve for x by setting each factor as greater than 0, I get the opposite result of what I need, i.e. +x+%3E+-+sqrt%28a%29+ instead of +x+%3C+-+sqrt%28a%29+ and I don't know why. Thanks so much again!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+%28x+%2B+sqrt%28a%29%29%28x+-+sqrt%28a%29%29+%3E+0+
Just looking at the signs of the 2 factors,
Neither factor can be +0+
These 2 cases work:
(+) * (+) >0
( - ) * ( - ) > 0
------------
These don't work:
(+) * (-)
( - ) * (+)
--------
The 1st case:
(+) * (+) >0
+x+%2B+sqrt%28a%29+%3E+0+
+x+%3E+-sqrt%28a%29+
and
+x+-+sqrt%28a%29+%3E+0+
+x+%3E+sqrt%28a%29+
----------------
Both of these have to be true. They are only
true if the 1st is true, so +x+%3E+-sqrt%28a%29+
must hold
---------------
The 2nd case
(-)*(-) >0
+x+%2B+sqrt%28a%29+%3C+0+
+x+%3C+-sqrt%28a%29+
and
+x+-+sqrt%28a%29+%3C+0+
+x+%3C+sqrt%28a%29+
-------------
Both of these have to be true. They are only true
if +x+%3C+-sqrt%28a%29+
----------------
so, +x+%3C+-sqrt%28a%29+ or +x+%3E+sqrt%28a%29+
since either of the 2 cases works