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Square root and cube root question

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    why is it possible to take the cube root of a negative number and not a square root of a negative number?
    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 3, 2010 #2


    Staff: Mentor

    It is possible to take the square root of a negative number, but you don't get a real number.

    If you can write the number whose square root you want as the product of two equal factors, the square root of that number will be one of those factors. For example, 4 = 2*2, so the square root of 4 is 2. Every positive number has two square roots: a positive square root and a negative square root. The symbol [itex]\sqrt{n}[/itex] is taken to mean the principal or positive square root.

    The cube root of a number (positive or negative) is one of three equal factors of that number, so [tex]\sqrt[3]{27} = 3[/tex], since 27 = 3*3*3. Similarly, [tex]\sqrt[3]{-8} = -2[/tex] since -8 = (-2)(-2)(-2). Real numbers have only 1 real cube root.

  4. Feb 3, 2010 #3
    Make an assume that it is possible , and work in reverse.. you will realise something tat contradicts from what you've learnt.
  5. Feb 3, 2010 #4
    oh I see so for a cube root you a real number answer no matter if its negative or positive ,but for a square root you get an imaginary number if its negative. Mark and icy thanks now I understand.
  6. Feb 3, 2010 #5


    Staff: Mentor

    And the same idea can be extended to odd and even roots. An even root (square root, fourth root, sixth root, etc.) of a nonnegative real number gives you a nonnegative real number. An odd root (cube root, fifth root, etc.) of a real number gives you a real number with the same sign.
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