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

#### daigo

##### Member
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?

#### SuperSonic4

##### Well-known member
MHB Math Helper
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?
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
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:

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

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

Or how does this work?
Again, $$\sqrt{x^2}= |x|$$.

#### HallsofIvy

##### Well-known member
MHB Math Helper
Suppose I have to solve for y:

$$x\leq 1$$

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

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

the rest of this is relevant.
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?