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Math Help - When to use absolute value in square root of a variable?

  1. #1
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    When to use absolute value in square root of a variable?

    In the first example, the textbook says we need the absolute value symbol because x might be negative
    \sqrt{25x^2} = 5\left|x\right|.

    In the second example, the textbook doesn't include the absolute value symbol
    \sqrt{12x^3}=2x\sqrt{3x}.

    What's the difference? When do we use absolute value and when not?
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  2. #2
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    Re: When to use absolute value in square root of a variable?

    Quote Originally Posted by mathDad View Post
    In the first example, the textbook says we need the absolute value symbol because x might be negative
    \sqrt{25x^2} = 5\left|x\right|.

    In the second example, the textbook doesn't include the absolute value symbol
    \sqrt{12x^3}=2x\sqrt{3sx}.

    What's the difference? When do we use absolute value and when not?
    If memory serves the convention with the absolute value is only true for numbers. Using variables we would have \sqrt{25x^2} = \sqrt{25} \sqrt{x^2} = 5 \cdot \pm x = \pm 5x.

    Note that in the case of \sqrt{12x^3} the odd power on the x forces the condition that x \geq 0 so there is no need for the +/-.

    -Dan
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    Re: When to use absolute value in square root of a variable?

    Quote Originally Posted by topsquark View Post
    If memory serves the convention with the absolute value is only true for numbers. Using variables we would have \sqrt{25x^2} = \sqrt{25} \sqrt{x^2} = 5 \cdot \pm x = \pm 5x.

    Note that in the case of \sqrt{12x^3} the odd power on the x forces the condition that x \geq 0 so there is no need for the +/-.

    -Dan
    Sorry Dan, but \sqrt{25}\ne\pm5 because \sqrt{25}=5.

    However, \sqrt{25x^2} is defined for x=-3 therefore \sqrt{25x^2}=5|x|

    As you correct observed for the odd case no absolute values are needed.
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    Re: When to use absolute value in square root of a variable?

    Yes, in order that \sqrt{3x^3} exist (we are talking about real numbers here) x itself must non-negative so |x|= x.

    For example, in the first problem, \sqrt{25x^2}, if x= 3 or -3, we would have \sqrt{25(3)^2}= \sqrt{25(9)}= \sqrt{225}= 15 as well as \sqrt{25(-3)^2}= \sqrt{25(9)}= \sqrt{225}= 15. To cover both possibilities we must say \sqrt{25x^2}= 5|x|.

    But in the second problem, \sqrt{12x^3}, if x= -3, that would be \sqrt{12(-3)^3}= \sqrt{12(-27)}= \sqrt{-324} which is not a real number.
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    Re: When to use absolute value in square root of a variable?

    Quote Originally Posted by Plato View Post
    Sorry Dan, but \sqrt{25}\ne\pm5 because \sqrt{25}=5.

    However, \sqrt{25x^2} is defined for x=-3 therefore \sqrt{25x^2}=5|x|

    As you correct observed for the odd case no absolute values are needed.
    Thanks for the catch.

    -Dan
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    Re: When to use absolute value in square root of a variable?

    (25)^1/2 = +/- 5
    (25x^2)^1/2=+/- 5*abs value x
    if x =-3
    (25)*(-3)^2 =225
    225^1/2 = +/- 15
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    Re: When to use absolute value in square root of a variable?

    Quote Originally Posted by bjhopper View Post
    (25)^1/2 = +/- 5
    (25x^2)^1/2=+/- 5*abs value x
    if x =-3
    (25)*(-3)^2 =225
    225^1/2 = +/- 15
    That is simply incorrect!

    \sqrt{25}\ne\pm 5 We have \sqrt{25}=(25)^{\frac{1}{2}}=5
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