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

• June 30th 2012, 06:59 AM
daigo
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?
• June 30th 2012, 07:19 AM
Plato
Re: Manipulation of negative square root of a negative term/#
Quote:

Originally Posted by daigo
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 $x\le 1 \Rightarrow~1-x=\sqrt{y}$
• June 30th 2012, 07:21 AM
Soroban
Re: Manipulation of negative square root of a negative term/#
Hello, daigo!

Quote:

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

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: . $y \:\ge\:0$
• June 30th 2012, 07:32 AM
daigo
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?
• June 30th 2012, 11:33 AM
Wilmer
Re: Manipulation of negative square root of a negative term/#
Incorrect.
• June 30th 2012, 05:06 PM
thevinh
Re: Manipulation of negative square root of a negative term/#
Quote:

Originally Posted by daigo
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 $y\geq0$.
• June 30th 2012, 06:10 PM
Plato
Re: Manipulation of negative square root of a negative teronfusiom/#
Quote:

Originally Posted by thevinh
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.
• July 1st 2012, 12:57 AM
biffboy
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
• July 1st 2012, 02:35 AM
BobP
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 $x-1=-\sqrt{y}.$