Square Root Yields Two Answers

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mathdad

mathdad

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Can you please explain, in easy terms, why the square root of, let's say, 4 yields -2 and 2?

In the same line of reasoning, why does the square root of (x^2) yield ONLY x and not -x and x?
 
mathdad said:
Can you explain in words?
Click to expand...
The square root is, by definition, positive. So the square root of 4 is 2 not -2.
When solving for example:
\(\displaystyle x^2 =4\)
we get \(\displaystyle x =\pm \sqrt{4}=\pm2\)
because
\(\displaystyle (2)^2 = 4\) and also \(\displaystyle (-2)^2 = 4\)
 
Harry_the_cat said:
The square root is, by definition, positive. So the square root of 4 is 2 not -2.
When solving for example:
\(\displaystyle x^2 =4\)
we get \(\displaystyle x =\pm \sqrt{4}=\pm2\)
because
\(\displaystyle (2)^2 = 4\) and also \(\displaystyle (-2)^2 = 4\)
Click to expand...

Perfectly explained.
 
Harry_the_cat said:
Also:
\(\displaystyle \sqrt {x^2} = |x|\) not \(\displaystyle x\)
eg
\(\displaystyle \sqrt {(-3)^2} = 3\) not \(\displaystyle -3\)
Click to expand...

I was taught long ago that the sqrt{(anything)^2} = anything.

To me, -3 is my substitute for the word anything.

This is my reasoning:

sqrt{(-3)^2} = sqrt{9} = -3 and 3.

Why can't we raise (-3) to the second power before taking the square root?
 
mathdad said:
I was taught long ago that the sqrt{(anything)^2} = anything. That's not correct, see my example in previous post.

To me, -3 is my substitute for the word anything.

This is my reasoning:

sqrt{(-3)^2} = sqrt{9} = -3 and 3. NO! \(\displaystyle \sqrt{9} = 3\) NOT \(\displaystyle -3\) .The square root is, by definition, positive.

Why can't we raise (-3) to the second power before taking the square root? You can!
Click to expand...
 

Ok. Happy to know that (-3) can be squared within the radicand as step 1.

How about sqrt{(x^2)^2}?

Solution:

sqrt{x^4} = x^2

Yes?

How about sqrt{(anything^2)^2}?

This yields anything, right?
 
Yes! Because x^2 is non-negative.

By the way, before I said the square root is, by definition, positive. I meant non-negative not positive, because it can be 0.
 
Harry_the_cat said:
Do you mean (anything)^2?
Click to expand...

No. I mean sqrt{(anything^2)^2} = anything or does it yield (anything)^2?

Let me see.

(anything^2)^2 = (anything)^4.
Taking the square of that yields (anything)^2.
You are right. I was wrong.
 
mathdad said:
Can you please explain, in easy terms, why the square root of, let's say, 4 yields -2 and 2?
In the same line of reasoning, why does the square root of (x^2) yield ONLY x and not -x and x?
Click to expand...
Please note that I have gone back to you very first post(the O.P.)
There are two square roots of any number either real or complex.
If \(\displaystyle x\) is a positive real number then there are two square roots of \(\displaystyle x\) one of which is positive and one negative.
NOW if \(\displaystyle x<0,~\&~x\mathbb{R}\) then there are still to square roots that are complex numbers.
What happened was a gross failure in mathematics. Mathematicians have always been completely lazy about notation.
The radical sign \(\displaystyle \sqrt{\;}\) is the poster-child for that statement. Saying that \(\displaystyle \sqrt{9}=\pm 3\) is an example of an absolute abuse of notation.
Least you think this is something new, in 1957 I was in ninth grade and Mrs Lee was our mathematics teacher. Her husband was Herman Lee at UTK fairly well known in mathematics education, she taught us that \(\displaystyle \sqrt{9}=\pm 3\) was incorrect.

Thus using a radical sign to mean a square root is simply wrong.
 
pka said:
Please note that I have gone back to you very first post(the O.P.)
There are two square roots of any number either real or complex.
If \(\displaystyle x\) is a positive real number then there are two square roots of \(\displaystyle x\) one of which is positive and one negative.
NOW if \(\displaystyle x<0,~\&~x\mathbb{R}\) then there are still to square roots that are complex numbers.
What happened was a gross failure in mathematics. Mathematicians have always been completely lazy about notation.
The radical sign \(\displaystyle \sqrt{\;}\) is the poster-child for that statement. Saying that \(\displaystyle \sqrt{9}=\pm 3\) is an example of an absolute abuse of notation.
Least you think this is something new, in 1957 I was in ninth grade and Mrs Lee was our mathematics teacher. Her husband was Herman Lee at UTK fairly well known in mathematics education, she taught us that \(\displaystyle \sqrt{9}=\pm 3\) was incorrect.

Thus using a radical sign to mean a square root is simply wrong.
Click to expand...

Good information here for my notes.
 
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