I need explanation.
I have to solve irrational equation:
I have tried to solve it like this:
But in the book stands that solution is actually interval:
which is true.
How do I find that interval? Can someone explain me solution.
I need explanation.
I have to solve irrational equation:
I have tried to solve it like this:
But in the book stands that solution is actually interval:
which is true.
How do I find that interval? Can someone explain me solution.
Start with .
As you noticed, this is:
.
But when we take the square root, we get the answer, so:
, where the signs are independant.
Using the "+ +" version we get x=1.
Using the "+ -" version we get -2=2. Nonsense.
Using the "- +" version we get 2=2. Nothing new.
Using the "- -" version we get x=-1.
I admit that I'm baffled by the x=0 solution. I tried squaring twice to get rid of the radicals (the most general method I know of) and I get (unless I made a mistake) which doesn't help at all.
-Dan
Exactly right, and what I've just been scratching out on paper.Originally Posted by CaptainBlack
I now recommend not looking at the original equation, but the function:
This is the function related to the given equation, so we are looking for the zeros of f(x).
The simplest solution is to graph it. The zeros are obvious...x = [-1,1] as originally advertised. To explain this, note that (as CaptainBlack noted) the convention is to take positive. So, taking a look at f(x):
where we now take the +/- signs to require that the term is positive.
So, on the interval x=(-infinity,-1):
(Since (x+1) is negative on this interval we keep the "-" sign to make it positive...similarly (x-1) is negative and we again keep the "-" sign.)
and thus f(x)=2x+2.
On the interval x=(1,infinity):
and thus f(x)=-2x+2.
Finally, on the interval x=[-1,1]:
and thus f(x)=0.
This is an interesting problem. It gives a strong reminder that we can't always simplify an equation and get the same solution set as we had in the unsimplified equation. It's always a good idea to plug your solution back into the original equation to see if you've got everything. (I'm going to note this problem for when I teach College Algebra again!)
-Dan