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Thread: 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:

    $\displaystyle x\leq 1$

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

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

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

    Am I to understand that this is the same as:

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

    Or how does this work?
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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:
    $\displaystyle x\leq 1$
    $\displaystyle (x - 1)^{2} = y$
    So I know that (x - 1) will always be 0 or a negative, therefore I must take the negative square root of (x - 1):
    $\displaystyle -\sqrt{(x - 1)^{2}} = -\sqrt{y}$
    Am I to understand that this is the same as:
    $\displaystyle -1 \cdot \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}$
    $\displaystyle =$
    $\displaystyle -1 \cdot (x - 1) = -1 \cdot \sqrt{y}$
    $\displaystyle =$
    $\displaystyle -x + 1 = -1 \cdot \sqrt{y}$
    It is a bit awkward, and you missed a sign.
    You you could have done this.
    $\displaystyle (x - 1)^{2} = y \Rightarrow~|x-1|=\sqrt{y}$
    But $\displaystyle x\le 1 \Rightarrow~1-x=\sqrt{y}$
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    Re: Manipulation of negative square root of a negative term/#

    Hello, daigo!

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

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

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

    . . Therefore: .$\displaystyle y \:\ge\:0$
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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:

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

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

    $\displaystyle -\sqrt{y} = x - 1$

    $\displaystyle -\sqrt{y} + 1 = x$

    Is this correct or incorrect?
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  5. #5
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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:

    $\displaystyle x\leq 1$

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

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

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

    Am I to understand that this is the same as:

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

    Or how does this work?

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

    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 ?

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

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

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