anonymous
  • anonymous
8-6). Assume all variables are positive, and find the following root. sqrt((x+b)^2)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
sqrt((x+b)²)=x+b
anonymous
  • anonymous
wherever there is a Root + Square they cancel each other so sqrt ((x+b) ^ 2 = x+b
anonymous
  • anonymous
Notice that the assumption about positive variables is important! For example: \[ \sqrt{(-2)^2} = 2 \neq -2 \] It is not always true that root and square root "cancel".

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anonymous
  • anonymous
yeah i know , if there is an addition function which is square we can't cancel it with the sqrt right ?
anonymous
  • anonymous
Sorry, I meant "square and square root" not "root and square root".
anonymous
  • anonymous
Can you clarify Atooty? I don't know what you're asking.
anonymous
  • anonymous
hmm my english is bad ><" and i don't know how to explain my point .. am sorry ><"
anonymous
  • anonymous
No problem with your english. I'll explain a little more. The "cancelling" of the square root and the square is actually the property of being inverse functions. The trick is though that they are not inverses everywhere, only for positive numbers. In general \[ (\sqrt{x})^2 = x \] is true but only valid for \[ x>0. \] In general \[ \sqrt{x^2} = |x| \] which is true for all values of x. Since \[ |x| = x \] for x>0, we have \[ \sqrt{x^2} = x \] for positive values of x.
anonymous
  • anonymous
=D thats what i meant + i heard that if there is smthing like \[\sqrt{a ^{2}+ b^{2}}\] it will not cancel it right ?
anonymous
  • anonymous
Right, it won't cancel it, because of the plus. That's how algebra works. My original point was more about how functions and inverses work.

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