[tex] \sqrt{x^2} = |x| [/tex] ??? Watch

Ano9901whichone
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So we have  \sqrt{x^2}= |x| right?
'It makes sense try x=-1,' ok so  \sqrt{(-1)^2} = \sqrt 1 = 1 .
But wait a second let's try that again but square root first.
 \left ( \sqrt{-1} \right )^2 = i^2 = -1 . Is there some rule or convention that I do not know about.
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Jammy Duel
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(Original post by Ano9901whichone)
So we have  \sqrt{x^2}= |x| right?
'It makes sense try x=-1,' ok so  \sqrt{(-1)^2} = \sqrt 1 = 1 .
But wait a second let's try that again but square root first.
 \left ( \sqrt{-1} \right )^2 = i^2 = -1 . Is there some rule or convention that I do not know about.
You're looking at two separate functions, the bottom line is (\sqrt{x})^2=x and not \sqrt{x^2}=\lvert x \rvert as in the first instance.
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lllllllllll
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(Original post by Ano9901whichone)
So we have  \sqrt{x^2}= |x| right?
'It makes sense try x=-1,' ok so  \sqrt{(-1)^2} = \sqrt 1 = 1 .
But wait a second let's try that again but square root first.
 \left ( \sqrt{-1} \right )^2 = i^2 = -1 . Is there some rule or convention that I do not know about.
any square root has two solutions

 



\sqrt{x^2} = \pm x

\n

and

\n

\sqrt{-1^2} = \pm1
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Ano9901whichone
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(Original post by lllllllllll)
any square root has two solutions

 



\sqrt{x^2} = \pm x

\n

and

\n

\sqrt{-1^2} = \pm1
That's fundamentally wrong.
The square root sign returns you the principal square root I.e the positive square root. This is just so wrong what you said.
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Ano9901whichone
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(Original post by Jammy Duel)
You're looking at two separate functions, the bottom line is (\sqrt{x})^2=x and not \sqrt{x^2}=\lvert x \rvert as in the first instance.
But are they not the same?
 \sqrt{x^2}=(x^2)^{1/2} right?
 (\sqrt{x})^2 =(x^{1/2})^2
What's the difference?
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Jammy Duel
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(Original post by Ano9901whichone)
That's fundamentally wrong.
The square root sign returns you the principal square root I.e the positive square root. This is just so wrong what you said.
Except what he said was right, it is convention to take the positive root unless otherwise stated, but that does not mean the negative root does not exist as it is also a solution to the equation, in this case x^2=1
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Jammy Duel
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(Original post by Ano9901whichone)
But are they not the same?
 \sqrt{x^2}=(x^2)^{1/2} right?
 (\sqrt{x})^2 =(x^{1/2})^2
What's the difference?
 \sqrt{x^2}=(x^2)^{1/2}=x=(x^{1/2})^2 = (\sqrt{x})^2
taking the absolute value changes the function
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Ano9901whichone
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(Original post by Jammy Duel)
Except what he said was right, it is convention to take the positive root unless otherwise stated, but that does not mean the negative root does not exist as it is also a solution to the equation, in this case x^2=1
No, he was in fact wrong. Yes the solution to the equation  x^2 =4 are  x=\pm \sqrt{4} = \pm 2 .
Sure that is right, but the is not the same as saying  \sqrt{4} = \pm 2 which would be incorrect.
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Jammy Duel
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(Original post by Ano9901whichone)
No, he was in fact wrong. Yes the solution to the equation  x^2 =4 are  x=\pm \sqrt{4} = \pm 2 .
Sure that is right, but the is not the same as saying  \sqrt{4} = \pm 2 which would be incorrect.
The square root function is the inverse of the squaring function, when you square root something, x, you are asking the question "what, when multiplied with itself, gives x", as said, convention dictates that the positive root, where it exists, is quoted unless specified otherwise, that does not mean that the negative root does not exist.

Let us take your 4 example, what, when multiplied by itself, gives 4?
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Ano9901whichone
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(Original post by Jammy Duel)
The square root function is the inverse of the squaring function, when you square root something, x, you are asking the question "what, when multiplied with itself, gives x", as said, convention dictates that the positive root, where it exists, is quoted unless specified otherwise, that does not mean that the negative root does not exist.

Let us take your 4 example, what, when multiplied by itself, gives 4?
I'm right. You and the other poster are wrong.
Sketch the graph of  y=\sqrt x and you will see why? Please draw the graph and post it without cheating/checking it before you post if you're so confident.
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lllllllllll
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(Original post by Ano9901whichone)
That's fundamentally wrong.
The square root sign returns you the principal square root I.e the positive square root. This is just so wrong what you said.
the square root function is just the inverse of x^2. for negative numbers you just cant see the roots because they are imaginary
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Jammy Duel
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(Original post by lllllllllll)
the square root function is just the inverse of x^2. for negative numbers you just cant see the roots because they are imaginary
Stop prodding the troll
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Ano9901whichone
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(Original post by lllllllllll)
the square root function is just the inverse of x^2. for negative numbers you just cant see the roots because they are imaginary
That graph of  x=y^2 is not, I repeat not the same as the graph of  y=\sqrt x I promise you. Search it and you will see.
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Ano9901whichone
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(Original post by Jammy Duel)
Stop prodding the troll
I'm not a troll. I made this account to ask what uni I should make my insurance choice. But while I'm here I had a question. I am not troll I assure you.
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Jammy Duel
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(Original post by Ano9901whichone)
I'm not a troll. I made this account to ask what uni I should make my insurance choice. But while I'm here I had a question. I am not troll I assure you.
Then listen to what you're being told.
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morgan8002
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(Original post by Jammy Duel)
The square root function is the inverse of the squaring function, when you square root something, x, you are asking the question "what, when multiplied with itself, gives x", as said, convention dictates that the positive root, where it exists, is quoted unless specified otherwise, that does not mean that the negative root does not exist.

Let us take your 4 example, what, when multiplied by itself, gives 4?
(Original post by lllllllllll)
the square root function is just the inverse of x^2. for negative numbers you just cant see the roots because they are imaginary
See http://www.thestudentroom.co.uk/show....php?t=3173187

A function cannot be multiple valued.

edit: Also, \sqrt{-1^2} = \sqrt{-1} = i.
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Ano9901whichone
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(Original post by Jammy Duel)
Then listen to what you're being told.
You're wrong mate. I will attach a picture of the  y=\sqrt x graph.
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Ano9901whichone
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This is  y=\sqrt x . Based on the 2 posters above logic, how can this be?? Good luck explaining it.
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Ano9901whichone
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(Original post by morgan8002)
See http://www.thestudentroom.co.uk/show....php?t=3173187

A function cannot be multiple valued.

edit: Also, \sqrt{-1^2} = \sqrt{-1} = i.
Thank you. How many times do you see people confused about this? All the time right?
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Ano9901whichone
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(Original post by Jammy Duel)
Then listen to what you're being told.
Can you use your superior knowledge and explain my last post to me?
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