Thread: 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.

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

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 .

Express as,



To be positive we require both factors to be positive or negative.

1)Both positive. That means,


Thus,


Another way of writing this is,

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,


Thus,


Another way of writing this is,

Because if the bottom inequality is true then for certainly the top one, for it is contained in the bottom one.

Thus,

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.