Finding the domain for inequality

1. Feb 7, 2012

solar nebula

1. The problem statement, all variables and given/known data
The problem is:
for all 0≤a≤1

so i need to find the domain

2. Relevant equations
N/A

3. The attempt at a solution

I tried it like this:

yet my solution is wrong,i am not so sure why.
wolfram gives me this;

2. Feb 7, 2012

micromass

Staff Emeritus
Here are some problems:

1) In the second step you wrote $\frac{1}{a}$. This is only well-defined if a is nonzero. So you should look at the case a=0 seperately.

2) In the last step you essentially did $x=\sqrt{x^2}$. This is true only for positive x. Indeed: if x=-1, then $x^2=1$ and $\sqrt{x^2}=-1$. So you need to look at the case where x is positive and x is negative.

Moderator note: since this has nothing to do with calculus, I'm moving it to precalculus.

3. Feb 8, 2012

solar nebula

Just to make it simple to solve for now, i am not going to look at case a=0 (only for now)

so i tired to do this:

Again my solution is wrong, i am now covering the entire number line (-inf,+inf), i feel like i am messing up basic algebra. Please help

4. Feb 8, 2012

micromass

Staff Emeritus
No, you again did $\sqrt{x^2}=x$. This is not true!!!!!

You need to consider two cases:

1) x is positive.

2) x is negative

And please don't use things like $\pm$, it's confusing. Just taking the square root (which BY DEFINITION is positive) will suffice.

5. Feb 8, 2012

solar nebula

I think i got it now:, Also when i take a square root of a number, isn't that i should take the positive and negative root?. I understand for
Code (Text):
\sqrt{x^2}=x
, because its basically
Code (Text):
\abs{x}

6. Feb 8, 2012

micromass

Staff Emeritus
No. And in fact, taking the negative square root is wrong. Indeed, we have that 4>1, but if we take the negative square root of both sides then we have $-\sqrt{4}>-\sqrt{1}$. This is the same as -2>-1 which is not true.

So if a and b are positive than a<b implies $\sqrt{a}<\sqrt{b}$, but it doesn't imply that $-\sqrt{a}<-\sqrt{b}$.