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.

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- February 8th 2006, 04:58 AMDenMac21Irrational equation help
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. - February 8th 2006, 09:10 AMCaptainBlackQuote:

Originally Posted by**DenMac21**

Does it equal ?

RonL - February 8th 2006, 09:50 AMDenMac21Quote:

Originally Posted by**CaptainBlack**

How to come to those solutions? - February 8th 2006, 10:10 AMtopsquark
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

- February 8th 2006, 10:24 AMCaptainBlackQuote:

Originally Posted by**topsquark**

positive root.

Also , so you should have ,

(add and signs to taste if you don't want to use the

positive root convention).

RonL - February 8th 2006, 10:48 AMtopsquarkQuote:

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 - February 8th 2006, 07:31 PMThePerfectHacker
Here is a graph:

(I realized that people have a lot of difficulty when it come to solving absolute value equations, probably has to do with examaning each interval).

Thus the solution set is,

:eek:

Notice not a finite number of solutions!