# Algebraic anomaly

• terryds

## Homework Statement

I want to find x that satisfies ##x-\sqrt{(x+2)^2} = 0##
Using algebric method, I find x = -1
But, x = -1 does not satisfy that equation...
What's wrong with the calculation?

(Basic algebra)

## The Attempt at a Solution

##x-\sqrt{(x+2)^2} = 0 \\
x = sqrt{(x+2)^2} ##
By squaring both sides , I\ get
##x^2 = (x+2)^2 \\
x^2 = x^2 + 4x + 4 \\
0 = 4x + 4 \\
x = -1 \\
##

What's wrong with this ?

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The equation you want to solve does not have a solution to begin with.

The equation you want to solve does not have a solution to begin with.

So, does it mean that it's necessary to plug the solution that I get from algebric calculation into the equation to confirm it's true??

## Homework Statement

I want to find x that satisfies ##x-\sqrt{(x+2)^2} = 0##
Using algebric method, I find x = -1
But, x = -1 does not satisfy that equation...
What's wrong with the calculation?

(Basic algebra)

## The Attempt at a Solution

##x-\sqrt{(x+2)^2} = 0 \\
x = sqrt{(x+2)^2} ##
By squaring both sides , I\ get
##x^2 = (x+2)^2 \\
x^2 = x^2 + 4x + 4 \\
0 = 4x + 4 \\
x = -1 \\
##

What's wrong with this ?
When you square both sides of an equation, there's the possibility that you'll get an equation whose solutions are not the same as the original equation. In this case, squaring introduced an extraneous solution.

So, does it mean that it's necessary to plug the solution that I get from algebric calculation into the equation to confirm it's true??
Yes

• terryds
When you square both sides of an equation, there's the possibility that you'll get an equation whose solutions are not the same as the original equation. In this case, squaring introduced an extraneous solution.

Hmmm..
So squaring both sides is a wrong attempt to find the solution ?
I wonder if there are any article about to-dos and not-to-dos in solving particular algebraic equations...

Hmmm..
So squaring both sides is a wrong attempt to find the solution ?
No, this is the right approach. It's just that when you apply any non-reversible operation (such as squaring, raising to fourth power, etc.), you need to check to see if you have introduced extraneous solutions.
terryds said:
I wonder if there are any article about to-dos and not-to-dos in solving particular algebraic equations...
You might try doing a search for "extraneous solutions". Most algebra textbooks mention this in the section where equations involving radicals are presented.

• terryds
For this equation, you can actually still work in a way which does not introduce any extra solution. Note that you can cast the original equation into ##x-|x+2|=0##. Solve this equation for ##x<-2## and ##x>-2## to see if there is a solution in each region.

No, this is the right approach. It's just that when you apply any non-reversible operation (such as squaring, raising to fourth power, etc.), you need to check to see if you have introduced extraneous solutions.

You might try doing a search for "extraneous solutions". Most algebra textbooks mention this in the section where equations involving radicals are presented.

Thanks a lot!

For this equation, you can actually still work in a way which does not introduce any extra solution. Note that you can cast the original equation into ##x-|x+2|=0##. Solve this equation for ##x<-2## and ##x>-2## to see if there is a solution in each region.

x-x+2 = 0 (No solution)
x+x+2=0
2x+2=0
x=-1 (Still an extraneous solution)

x-x+2 = 0
x-x-2=0
x+x+2=0
2x+2=0
x=-1 (Still an extraneous solution)
You forget that you are looking for a solution in ##x<-2##. The solution you find there lies outside this range.

x-x-2=0

You forget that you are looking for a solution in ##x<-2##. The solution you find there lies outside this range.

So it means no solutions, right ?

Yes.

## Homework Statement

I want to find x that satisfies ##x-\sqrt{(x+2)^2\,} = 0##
Using algebraic method, I find x = -1
But, x = -1 does not satisfy that equation...
What's wrong with the calculation?
Suppose you were to solve ##\ x+\sqrt{(x+2)^2\,} = 0\ ## and you used the same method.

That is equivalent to ##\ \sqrt{(x+2)^2\,} = -x\ .##

Squaring both sides gives ##\ (x+2)^2 = x^2\ ##, the same as in your case, giving ##\ x=-1\ .## Here the solution is valid.