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Math Help - Manipulation of negative square root of a negative term/#

  1. #1
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    Manipulation of negative square root of a negative term/#

    Suppose I have to solve for y:

    x\leq 1

    (x - 1)^{2} = y

    So I know that (x - 1) will always be 0 or a negative, therefore I must take the negative square root of (x - 1):

    -\sqrt{(x - 1)^{2}} = -\sqrt{y}

    Am I to understand that this is the same as:

    -1 \cdot  \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}
    =
    -1 \cdot (x - 1) = -1 \cdot \sqrt{y}
    =
    -x + 1 = -1 \cdot \sqrt{y}

    Or how does this work?
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  2. #2
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    Re: Manipulation of negative square root of a negative term/#

    Quote Originally Posted by daigo View Post
    Suppose I have to solve for y:
    x\leq 1
    (x - 1)^{2} = y
    So I know that (x - 1) will always be 0 or a negative, therefore I must take the negative square root of (x - 1):
    -\sqrt{(x - 1)^{2}} = -\sqrt{y}
    Am I to understand that this is the same as:
    -1 \cdot  \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}
    =
    -1 \cdot (x - 1) = -1 \cdot \sqrt{y}
    =
    -x + 1 = -1 \cdot \sqrt{y}
    It is a bit awkward, and you missed a sign.
    You you could have done this.
    (x - 1)^{2} = y  \Rightarrow~|x-1|=\sqrt{y}
    But x\le 1  \Rightarrow~1-x=\sqrt{y}
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    Re: Manipulation of negative square root of a negative term/#

    Hello, daigo!

    \text{Solve for }y\!:\;\;\begin{Bmatrix}x\:\leq\: 1 & [1] \\ (x - 1)^2 \:=\: y & [2]\end{Bmatrix}

    From [1]: . x\,\le\,1 \quad\Rightarrow\quad x - 1\,\le\,0 \quad\Rightarrow\quad (x-1)^2\,\ge\,0

    Substitute into [2]: . y \:=\:(x-1)^2 \:\ge\:0

    . . Therefore: . y \:\ge\:0
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    Re: Manipulation of negative square root of a negative term/#

    I don't really want to focus on solving the problem, I just need some insight on the negative square root operation.

    I'm trying to understand the way it was done here:

    y = (x - 1)^{2}

    -\sqrt{y} = -\sqrt{(x - 1)^{2}}

    -\sqrt{y} = x - 1

    -\sqrt{y} + 1 = x

    Is this correct or incorrect?
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    Re: Manipulation of negative square root of a negative term/#

    Incorrect.
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    Re: Manipulation of negative square root of a negative term/#

    Quote Originally Posted by daigo View Post
    Suppose I have to solve for y:

    x\leq 1

    (x - 1)^{2} = y

    So I know that (x - 1) will always be 0 or a negative, therefore I must take the negative square root of (x - 1):

    -\sqrt{(x - 1)^{2}} = -\sqrt{y}

    Am I to understand that this is the same as:

    -1 \cdot  \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}
    =
    -1 \cdot (x - 1) = -1 \cdot \sqrt{y}
    =
    -x + 1 = -1 \cdot \sqrt{y}

    Or how does this work?

    In this problem, it does not matter what value of x is because the quantity  (x-1)^2 will always be positive. So  y\geq0.
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    Re: Manipulation of negative square root of a negative teronfusiom/#

    Quote Originally Posted by thevinh View Post
    In this problem, it does not matter what value of x is because the quantity  (x-1)^2 will always be positive. So  y\geq0.
    That may be a problem. But it is not the problem which is the OP refuses to see that:
    \sqrt{(x-1)^2}=|x-1|.

    That is the source of all this confusion.
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    Re: Manipulation of negative square root of a negative term/#

    Is there anything wrong with this.
    (x-1)= + or -root y So rooty=(x-1) or (1-x) For example if x=-2 y=9 and root y =+3 or -3
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    Re: Manipulation of negative square root of a negative term/#

    Can I join in ? What's wrong with this ?

    x\leq1, so either x-1<0 or x-1=0.

    If x-1=0 then y=0, so concentrate on the first case.

    If (x-1)^{2}=y, then x-1=\pm\sqrt{y}, ( y is positive), but we know that x-1<0 so x-1=-\sqrt{y}.
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