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Problem concerning square roots

  1. Sep 20, 2011 #1
    I had this assignment in math class and it doesn't just add up.

    x + square(5x + 10) - 8 = 0

    Nothing too difficult to solve.

    And the answer is:

    x1 = 18 if square(5 * 18 + 10) = -10
    x2 = 3 if square(5 * 3 + 10) = 5

    But teacher says that x1 is not correct. To me it is not logical.

    18 + square(5 * 18 + 10) - 8 = 18 + square(100) - 8 =

    = 18 + 10 - 8 = 20 <- it doesn't fit into the equation
    &
    = 18 - 10 -8 = 0 <- it fits into the equation

    I find it logical but teacher says that only positive numbers can come from the square(...).

    How so?

    10 * 10 = 100

    and

    -10 * -10 = 100

    so

    square from 100 is

    10

    and

    -10.

    Where's the catch?

    Am I missing something?
     
  2. jcsd
  3. Sep 20, 2011 #2
    "x+ square(5x + 10) - 8 = 0"

    is it square root or

    x+(5x+10)^2-8=0 ?
     
  4. Sep 20, 2011 #3
    By square(...) I meant that -> square root from (...)
     
  5. Sep 20, 2011 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    By definition, √(a) or sqrt(a) is never negative, so [itex]\sqrt{5x+10}[/itex] can not be -10.

    ehild
     
  6. Sep 20, 2011 #5

    Mark44

    Staff: Mentor

    It's a good thing you have "square root" in the title, or I wouldn't have been able to understand what you are doing here. I had to mentally translate "square" into "square root" to make sense of what you're doing.
    What you have above is incorrect.
    If x = 18, then
    [itex]18 + \sqrt{90 + 10} = 18 + 10 = 28 \neq 8[/itex]

    The square root of a nonnegative real number is always a nonnegative real number. The expression [itex]\sqrt{100}[/itex] has one value, 10. It is NOT also equal to -10.
    No. The square root of 100 is 10. Period.
     
  7. Sep 20, 2011 #6
    Ok, but what do I have to take square root from to get -10?
     
  8. Sep 20, 2011 #7

    Mark44

    Staff: Mentor

    [itex]-\sqrt{100} = -10[/itex]
     
  9. Sep 20, 2011 #8

    ElijahRockers

    User Avatar
    Gold Member

    I understand that when dealing with physical measurement problems it usually does not make sense to keep a negative value. Correct me (and all my professors) wrong, but through several calculus courses that I have taken, we have always considered two values when taking the square root of any nonnegative real number, even if we end up throwing one out.

    I do not have master's degree or anything, I am just a student, but I find it hard to take the above statement seriously after having found so many instances of the opposite case fascinating and logical.

    Your apparent authority on the subject has confused me, so I did a little research in my text books.. As I expected, every positive number x has two square roots: [itex]\sqrt{x}[/itex] which is positive, and [itex]-\sqrt{x}[/itex] which is negative... right? Collectively they are referred to as [itex]\pm\sqrt{x}[/itex] I think this is what Remos is alluding to...
     
  10. Sep 20, 2011 #9

    Mark44

    Staff: Mentor

    It's true that every positive real number has two square roots, one positive and one negative, and that might be where you are confused.

    An expression such as [itex]\sqrt{9}[/itex] represents the principal square root, and has a single value, 3. The principal square root of a positive number is positive.
    Right.
    Right.

    Here's from the wikipedia page for Square Root - http://en.wikipedia.org/wiki/Square_root

     
  11. Sep 20, 2011 #10

    ElijahRockers

    User Avatar
    Gold Member

    I have tossed these sentences against each other for a few minutes and I think I've figured out the semantics. So, every positive real number can have two square roots, but performing the square root operation can only result in the principal root of the number?
    I think it's all coming together, especially after reading the wiki.
    Thanks!

    Sorry to steal your thread Remos, but I hope maybe our discussion has clarified your question a little bit?
     
    Last edited: Sep 20, 2011
  12. Sep 20, 2011 #11

    Mark44

    Staff: Mentor

    Right, that's exactly it. Similarly, every real number has three cube roots, but performing the cube root operation results in the principal cube root. Same with fourth roots, fifth roots, etc.
    I hope so. This was one of the things that he/she was unclear on.
     
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