\(\displaystyle \sqrt{4} =2\)
\(\displaystyle -\sqrt{4} =-2\)
The square root is, by definition, positive. So the square root of 4 is 2 not -2.Can you explain in words?
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 \(\displaystyle (-2)^2 = 4\)
Also:
\(\displaystyle \sqrt {x^2} = |x|\) not \(\displaystyle x\)
eg
\(\displaystyle \sqrt {(-3)^2} = 3\) not \(\displaystyle -3\)
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!
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.
Do you mean (anything)^2?How about sqrt{(anything^2)^2}?
This yields anything, right?
Do you mean (anything)^2?
You were taught wrongly. (It happens.)I was taught long ago that the sqrt{(anything)^2} = anything ...
You were taught wrongly. (It happens.)
\(\displaystyle \sqrt{\text{anything}^2} \;=\; \big| \; \text{anything} \; \big|\)
?
Please note that I have gone back to you very first post(the O.P.)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?
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.
Mahdad forgets to review his "notes".@mathdad - You may recall our multiple discussions about this on other sites. :|