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Manipulation of negative square root of a negative term/#

daigo

Member
Jun 27, 2012
60
Suppose I have to solve for y:

[tex]x\leq 1[/tex]

[tex](x - 1)^{2} = y[/tex]

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

[tex]-\sqrt{(x - 1)^{2}} = -\sqrt{y}[/tex]

Am I to understand that this is the same as:

[tex]-1 \cdot \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}[/tex]
[tex]=[/tex]
[tex]-1 \cdot (x - 1) = -1 \cdot \sqrt{y}[/tex]
[tex]=[/tex]
[tex]-x + 1 = -1 \cdot \sqrt{y}[/tex]

Or how does this work?
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
Suppose I have to solve for y:

[tex]x\leq 1[/tex]

[tex](x - 1)^{2} = y[/tex]

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

[tex]-\sqrt{(x - 1)^{2}} = -\sqrt{y}[/tex]

Am I to understand that this is the same as:

[tex]-1 \cdot \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}[/tex]
[tex]=[/tex]
[tex]-1 \cdot (x - 1) = -1 \cdot \sqrt{y}[/tex]
[tex]=[/tex]
[tex]-x + 1 = -1 \cdot \sqrt{y}[/tex]

Or how does this work?
Do you have another example? The equation you've given already has y as the subject.

For the equation $y = (x-1)^2$ where $x \leq 1$ you can deduce that $y \geq 0$ (in other words the range of the function is $f(x) \geq 0$)

As you said you must take the negative root of $x$ but not of $y$ because $y$ is always non-negative so we take the positive square root. In essence I'm not sure what you want to do with the equation? Do you want to solve for x (and basically find the inverse function)? If you do then remember the domain and range are swapped.

One of the easier things to do is consider and ordered pair.
$f(x) = (x-1)^2$
$f(0) = (0-1)^2 = 1$

So (0,1) is an ordered pair. If we try this in your final equation (which you've done correctly bar taking the negative root of y): $-0+1 = -1 \cdot \sqrt{1} \rightarrow 1 \neq -1$ so that equation doesn't hold.

If you took the positive root of y (ending up with $-x+1 = \sqrt{y} \rightarrow -0+1 = \sqrt{1} $ which works. Try it for another ordered pair (but remember that $y \geq 0$)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Suppose I have to solve for y:

[tex]x\leq 1[/tex]

[tex](x - 1)^{2} = y[/tex]

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

[tex]-\sqrt{(x - 1)^{2}} = -\sqrt{y}[/tex]

Am I to understand that this is the same as:

[tex]-1 \cdot \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}[/tex]
[tex]=[/tex]
[tex]-1 \cdot (x - 1) = -1 \cdot \sqrt{y}[/tex]
This is an error [tex]\sqrt{(x-1)^2}= |x- 1|[/tex] not x- 1. Here, since x-1< 0, |x- 1|= -(x- 1) so what you should have is
[tex]-1(-(x-1))= x-1= -\sqrt{y}[/tex]

[tex]=[/tex]
[tex]-x + 1 = -1 \cdot \sqrt{y}[/tex]
This can't be right because if x< 1, -x+ 1> 0 and [tex]-\sqrt{y}[/tex] is positive.

Or how does this work?
Again, [tex]\sqrt{x^2}= |x|[/tex].
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Suppose I have to solve for y:

[tex]x\leq 1[/tex]

[tex](x - 1)^{2} = y[/tex]
This is already "solved for y". Did you mean "solve for x"?

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

[tex]-\sqrt{(x - 1)^{2}} = -\sqrt{y}[/tex]
What?? You have squared one side and taken the square root of the other side! From [tex]\sqrt{4}= 2[/tex] it does NOT follow that [tex]4= \sqrt{2}[/tex]
What you need to do, to solve for x, is square both sides: [tex]x- 1= y^2[/tex] and form that [tex]x= 1+ y^2[/tex]. It follows automatically that x is less than of equal to 1 because, with [tex]x\le 1[/tex], [tex]x- 1\le 0[/tex] so y is an imaginary number, or 0, and [tex]y^2[/tex] is less than or equal to 0.

the rest of this is relevant.
Am I to understand that this is the same as:

[tex]-1 \cdot \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}[/tex]
[tex]=[/tex]
[tex]-1 \cdot (x - 1) = -1 \cdot \sqrt{y}[/tex]
[tex]=[/tex]
[tex]-x + 1 = -1 \cdot \sqrt{y}[/tex]

Or how does this work?