Positive/negative square roots question

[gcooper]

(x - 2)^2 = 16

It does not matter whether I take the positive or negative square root on that whole equation to solve it, I always get to the same result:

positive:

x - 2 = 4
x = 6

negative:

-x + 2 = -4
-x = -6
x = 6

But if I combine (take positive for the left side, and negative square root for the right side), I get a different result(which is still valid):

(positive left, negative right)
x - 2 = -4
x = -2

Isn't it a little bit odd? Shouldn't I be taking the exact root of both sides, since its an equation? And what happens when you end up having 4 square roots in an equation and you start mangling them, won't you have a lot more solutions then(due to the negative on one side/positive on the other of the = sign)?

 

[pka]

(x - 2)^2 = 16

Let's find solutions to the equation. We are not using steps.

Is \(\displaystyle x=6\) a solution? Why is it? How do we know?
Note that we used no more than \(\displaystyle (6-2)^2=16\).

\(\displaystyle x=-6\) is not a solution! Why is it not ? How do we know?

Is \(\displaystyle x=-2\) a solution? Why is it? How do we know?

 

[gcooper]

I'm sorry, but that approach doesn't ring any bells to me.

How do I know that a number is a solution? By plugging it in, into the equation to check. My actual question is how does it make any sense to apply positive root on one side, while applying negative root on the other(and actually get a good solution), and how doesn't that affect big equations in which there are many roots.

 

[pka]

that approach doesn't ring any bells to me.

How do I know that a number is a solution? By plugging it in, into the equation to check. My actual question is how does it make any sense to apply positive root on one side, while applying negative root on the other(and actually get a good solution), and how doesn't that affect big equations in which there are many roots.

Well of course it rings no bells with you. That is is exactly why you cannot work the problem. You do not understand what is means to be a solution. It has absolutely nothing to do with the method use, but does the result work in the question?

Here is a simple example: Any one in middle-school student should be able to solve \(\displaystyle (x-5)^2=36\) without any real thought. Does \(\displaystyle (\pm 6)^2=36~?\) Does not any middle-schooler know that?

 

[gcooper]

"(x - 2)^2 = 16

x^2 - 4x + 4 = 16
x^2 - 4x - 12 = 0
(x - 6)(x + 2) = 0"

Yes Dennis, using the polynomial way DOES make sense, but I was going the fast root way, the one that was actually bothering me with the result, which I got by using positive root on one side, negative root on the other. Was it just a coincindence it worked, or...?!





"Does not any middle-schooler know that?", thanks a lot! It just so happens I'm trying to learn algebra by myself, and you sir, just managed to convince me not to ever post here again, just because I'll look stupid, not knowing stuff you knew while you were in your mother's womb.

Of course that something^2 = 36, would immediately get you to think 6. What I was asking though, was whether to always consider taking both the square root AND the negative square root, why? Because x^2 = (-x)^2.

Does [FONT=MathJax_Main]([FONT=MathJax_Main]±[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]36[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main]?[/FONT][/FONT] Does not any middle-schooler know that?

If you read my issue correctly, I'm bothered more by the second answer, which is -2, but I guess you were not paying attention to such non trivial stuff.

Anyhow, thank you for taking the time to try and disappoint me regarding math, YOU ARE one of the reasons people quit math, people like you make them afraid to ask questions, have a nice day!

 

[HallsofIvy]

(x - 2)^2 = 16

It does not matter whether I take the positive or negative square root on that whole equation to solve it, I always get to the same result:

positive:

x - 2 = 4
x = 6

negative:

-x + 2 = -4
-x = -6
x = 6

This is your fundamental problem. By changing signs on both right and left you have not changed the actual equation. "-x+ 2= -4" is the same as "-(x- 2)= -4" and multiplying on both sides by -1 gives you x- 2= 4 again.
What you are trying to use, but not remembering properly, is that \(\displaystyle x^2= 4\), then either x= 2 or x= -2. The right side is changed, but it is still "x", not "-x" that is equal to. Here you should have x- 2= 4 and x- 2= -4.

But if I combine (take positive for the left side, and negative square root for the right side), I get a different result(which is still valid):

(positive left, negative right)
x - 2 = -4
x = -2

Isn't it a little bit odd? Shouldn't I be taking the exact root of both sides, since its an equation? And what happens when you end up having 4 square roots in an equation and you start mangling them, won't you have a lot more solutions then(due to the negative on one side/positive on the other of the = sign)?

 

[Dale10101]

?

(x - 2)^2 = 16

It does not matter whether I take the positive or negative square root on that whole equation to solve it, I always get to the same result:

positive:

x - 2 = 4
x = 6

negative:

-x + 2 = -4
-x = -6
x = 6

But if I combine (take positive for the left side, and negative square root for the right side), I get a different result(which is still valid):

(positive left, negative right)
x - 2 = -4
x = -2

Isn't it a little bit odd? Shouldn't I be taking the exact root of both sides, since its an equation? And what happens when you end up having 4 square roots in an equation and you start mangling them, won't you have a lot more solutions then(due to the negative on one side/positive on the other of the = sign)?

I think I see what you are thinking.

eq1: (y)^2 = 16 => y = +4 or -4

eq2: (-y)^2 = 16 => y = +4 or -4

Conjecture:

If y = x-2 shouldn’t it follow that

eq3: (x-2)^2 =16, and

eq4: (-(x-2))^2 = (-x+2)^2 = 16 will have the same solution set as well?

That would be the case if (-(x-2))^2 = ((-1)(x-2))^2 = (x-2)^2 but that is not the case because

((-1)(x-2))^2 /= (-1)^2(x-2)^2 = (1)(x-2)^2 = (x-2)^2… BTW am using “/=” to symbolize “NOT EQUAL”..

The initial conjecture does seem reasonable from a distance … perhaps that is what you were thinking … or not.