# Thread:

1. ## Show Inequality X^2 > a

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

Okay...what exactly is the question asking in simple math terms?

I don't follow the logic behind this sort of reasoning.

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

Okay...what exactly is the question asking in simple math terms?

I don't follow the logic behind this sort of reasoning.
First there is a typo in your post, the phrase:

"consists of all numbers x for which -(sqrt{a}) < a (sqrt{a})"

is probably mistyped as it is always true, there should be an x in there somewhere.

If you sketch the graph of x^2-a (with a>0) you will see that this is >0 everywhere
except between the roots of x^2-a, and when this is >0 we have x^2>a.

So x^2>0 when x<-sqrt(a), or when x>sqrt(a).

RonL

3. Originally Posted by symmetry
If a > 0, show that the solution set of the inequality x^2 > a consists of all numbers x for which -(sqrt{a}) < a (sqrt{a}).
You have, for $a>0$.
$x^2 > a$
Express as,
$x^2 - a >0$
$x^2 - (\sqrt{a})^2 > 0$
$(x-\sqrt{a})(x+\sqrt{a})>0$
To be positive we require both factors to be positive or negative.

1)Both positive. That means,
$x-\sqrt{a}>0$
$x+\sqrt{a}>0$
Thus,
$x>\sqrt{a}$
$x>-\sqrt{a}$
Another way of writing this is,
$x>\sqrt{a}$
Because if the top inequality is true then for certainly the bottom one, for it is contained in the top one.

2)Both negativei. That means,
$x-\sqrt{a}<0$
$x+\sqrt{a}<0$
Thus,
$x<\sqrt{a}$
$x<-\sqrt{a}$
Another way of writing this is,
$x<-\sqrt{a}$
Because if the bottom inequality is true then for certainly the top one, for it is contained in the bottom one.

Thus,
$x>\sqrt{a} \mbox{ or }x<-\sqrt{a}$

4. ## ok

I want to thank both for your replies.

Captainblack,

You are right in making your statement.

There is a typing error.

I will now send the correct questions.

Thanks.
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