# Irrational equation help

• Feb 8th 2006, 03:58 AM
DenMac21
Irrational equation help
I need explanation.

I have to solve irrational equation:
$\sqrt {x^2 - 2x + 1} + \sqrt {x^2 + 2x + 1} = 2$

I have tried to solve it like this:
$\sqrt {(x - 1)^2 } + \sqrt {(x + 1)^2 } = 2$
$x-1+x+1 = 2$
$x = 1$

But in the book stands that solution is actually interval:
$- 1 \le x \le 1$
which is true.

How do I find that interval? Can someone explain me solution.
• Feb 8th 2006, 08:10 AM
CaptainBlack
Quote:

Originally Posted by DenMac21
I need explanation.

I have to solve irrational equation:
$\sqrt {x^2 - 2x + 1} + \sqrt {x^2 + 2x + 1} = 2$

I have tried to solve it like this:
$\sqrt {(x - 1)^2 } + \sqrt {(x + 1)^2 } = 2$
$x-1+x+1 = 2$
$x = 1$

But in the book stands that solution is actually interval:
$- 1 \le x \le 1$
which is true.

How do I find that interval? Can someone explain me solution.

To show that the book is wrong set $x=1/2$ and then evaluate:

$\sqrt {x^2 - 2x + 1} + \sqrt {x^2 + 2x + 1}$

Does it equal $2$?

RonL
• Feb 8th 2006, 08:50 AM
DenMac21
Quote:

Originally Posted by CaptainBlack
To show that the book is wrong set $x=1/2$ and then evaluate:

$\sqrt {x^2 - 2x + 1} + \sqrt {x^2 + 2x + 1}$

Does it equal $2$?

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?
• Feb 8th 2006, 09:10 AM
topsquark
Start with $\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}=2$.

As you noticed, this is:
$\sqrt{(x-1)^2}+\sqrt{(x+1)^2}=2$.

But when we take the square root, we get $\pm$ the answer, so:
$\pm (x-1) + \pm (x+1) = 2$, where the $\pm$ 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 $(x+1)^2=(x+1)^2$ (unless I made a mistake) which doesn't help at all.

-Dan

• Feb 8th 2006, 09:24 AM
CaptainBlack
Quote:

Originally Posted by topsquark
Start with $\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}=2$.

As you noticed, this is:
$\sqrt{(x-1)^2}+\sqrt{(x+1)^2}=2$.

But when we take the square root, we get $\pm$ the answer, so:
$\pm (x-1) + \pm (x+1) = 2$, where the $\pm$ 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 $(x+1)^2=(x+1)^2$ (unless I made a mistake) which doesn't help at all.

-Dan

Except $\sqrt{x}$ by (a common) convention denotes the the
positive root.

Also $\sqrt{u^2}=|u|$, so you should have $|x-1| + |x+1| = 2$,
(add $\pm$ and $\mp$ signs to taste if you don't want to use the
positive root convention).

RonL
• Feb 8th 2006, 09:48 AM
topsquark
Quote:

Originally Posted by CaptainBlack
Except $\sqrt{x}$ by (a common) convention denotes the the
positive root.

Also $\sqrt{u^2}=|u|$, so you should have $|x-1| + |x+1| = 2$,
(add $\pm$ and $\mp$ 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:
$f(x)=2-\sqrt{x^2+2x+1}-\sqrt{x^2-2x+1}$
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 $\sqrt{x}$ positive. So, taking a look at f(x):
$f(x)=2- \pm (x+1) - \pm (x-1)$
where we now take the +/- signs to require that the term is positive.

So, on the interval x=(-infinity,-1):
$f(x)=2- -(x+1) - -(x-1)$
(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):
$f(x)=2- +(x+1) - +(x-1)$
and thus f(x)=-2x+2.

Finally, on the interval x=[-1,1]:
$f(x)=2- +(x+1) - -(x-1)$
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
• Feb 8th 2006, 06:31 PM
ThePerfectHacker
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,
$-1\leq x\leq 1$ :eek:
Notice not a finite number of solutions!