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.
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
You have, for $\displaystyle a>0$.
$\displaystyle x^2 > a $
Express as,
$\displaystyle x^2 - a >0$
$\displaystyle x^2 - (\sqrt{a})^2 > 0$
$\displaystyle (x-\sqrt{a})(x+\sqrt{a})>0$
To be positive we require both factors to be positive or negative.
1)Both positive. That means,
$\displaystyle x-\sqrt{a}>0$
$\displaystyle x+\sqrt{a}>0$
Thus,
$\displaystyle x>\sqrt{a}$
$\displaystyle x>-\sqrt{a}$
Another way of writing this is,
$\displaystyle x>\sqrt{a}$
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,
$\displaystyle x-\sqrt{a}<0$
$\displaystyle x+\sqrt{a}<0$
Thus,
$\displaystyle x<\sqrt{a}$
$\displaystyle x<-\sqrt{a}$
Another way of writing this is,
$\displaystyle x<-\sqrt{a}$
Because if the bottom inequality is true then for certainly the top one, for it is contained in the bottom one.
Thus,
$\displaystyle x>\sqrt{a} \mbox{ or }x<-\sqrt{a}$