Thread: Irrational 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.

Originally Posted by DenMac21

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.

To show that the book is wrong set and then evaluate:



Does it equal ?

RonL

Originally Posted by CaptainBlack

To show that the book is wrong set and then evaluate:



Does it equal ?

RonL

Yes, you are right. But, -1 and 0 are also solutions, so author maybe had in mind only integers (even though it is not mentioned anywhere).
How to come to those solutions?

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

Originally Posted by topsquark

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

Except by (a common) convention denotes the the
positive root.

Also , so you should have ,
(add and signs to taste if you don't want to use the
positive root convention).

RonL

Originally Posted by CaptainBlack

Except by (a common) convention denotes the the
positive root.

Also , so you should have ,
(add and signs to taste if you don't want to use the
positive root convention).

RonL

Exactly right, and what I've just been scratching out on paper.

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

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,

Notice not a finite number of solutions!