inb4 trip thumbs

But still its =2.

Depends if you are on R or on C

2 OR -2, not AND, OR.

If you start with 4, then take the square root, then you are right. The answers are 2, and -2.

But if you start with the square root of 4, then you are wrong. The only answer is 2.

Sincerely,
A 4th year math major, not that anybody cares.

On the contrary,
When you do the action of square rooting, you must consider the positive and negative cases. Since you do not know whether the square root was principal or not, you must consider both cases. The positive case is the one shown in the content - the principal square root. The negative case will give you -2.

If you start with simply "square root of 4", you have skipped the action of square rooting, and jump right to evaluating the positive case (the principal square root).

It is quite extremely crystal clear.

Then would you (should you?) not write in the notation "±√4" as opposed to simply writing the notation for the principal square root?

Lemme break it down.

±√4 = +√4 AND -√4 (it's "AND" or "OR" depending on the situation)
+√4 = √4 = 2
-√4 = -2

So if you just look at the above statements, you can see that √4 = 2. Not -2. Math don't lie!

Isn't that what I've been saying?
√4 = 2 and only 2
-√4 = -2 and only -2

That's been my point throughout this entire discussion.

that's correct! I wasn't sure what you were talking about, but the math wasn't ambiguous so I just reiterated it.

I stopped looking at porn for this?

I usually get off to my own comments too!

everyone is replying to this with math equations etc.

and i'm just sat here thinking what the **** is wrong with that dog

square root is a function that is always equal or above 0.
mathematically put : sqrt(x) >= 0

Get ready to learn ************ (s).

As displayed by the graph of a square root function (shown on the left), there are no negative y values, and thus f(x) can never be negative [ sqrt(2) can never equal - 2 ]. This is because the square root operation IS A function, and thus their can only be 1 y value for every x value (but not necessarily the other way around).

You are stating one of the most common misconceptions people have once they are taught to solve quadratic equations, observe.

1. x^2 = 4
2. x = sqrt (4)
3. x = ± 2

You make the assumption that the square root of 4 is equal to +/- 2, however, by actual mathematical definition, line 2 holds a mistake. This is because x DOES NOT equal sqrt(4). The correct solution is:

1. x^2 = 4
2. x = ± sqrt(4)
3. x = ± 2

Just as you cannot take the square root of a negative number, the square root of a positive number can never be negative. The radical symbol by definition is only the principal square root, i.e., always positive.

Wait wait, you're totally wrong. If the square root of a positive number can never be negative, why is -2 squared 4? That one you have up is only the positive, not negative roots.

You're not understanding the definition of a square root. I even put a graph there and everything.

I am. If you take the root of 4, its both -2 and 2. But if you already have the square root of four, its just two. I do understand the definition of a square root, with is a number that when multiplied by itself gives you the original. -2 times -2 is 4.

I'm saying the square root operation signified by the radix will only yield the positive roots of a number by definition.

If you start with a square root yeah. But if I take a square root it will be both.

But when you take the square root, you are taking the positive and negative root, and you SHOULD technically be using the notation "±sqrt (a)", otherwise what you are writing is simply incorrect by means of notation. You may be doing the right thing in your head, but you're not writing down what you should be.

I never said anything about not having the plus or minus sign. But yeah I see what you mean.

>using a badly coded graphing software to demonstrate an incorrect point

Type the square root of x into any graphing software and I assure you it will either give you that exact graph (might also include the imaginary part as well, a reflection along the y-axis)

Here's an example from Wolfram|Alpha

why are you thumbed down, this is correct

Good god that's terrifying

Actually by definition it's only positive 2. Even though -2 squared is 4, the square root of 4 is 2.

-2 * -2 = 4
2 * 2 = 4

(sqrt) 4 = +/- 2

both results are valid, the correct answer is: +/− 2

Maybe it's a cultural thing, when I took math I was required to express the results to a square root always as +/- (unless imaginary number) and if the root was part of a bigger operation record both results (when the value is negative and when positive)

Don't say these things.

math is "universal"

x² = 4
√x² = √4
│x│= 2
±x = 2
so,
x = ±2

You are 100% correct, there are two solutions to x^2 = 4, ±2
HOWEVER.

You are wrong in the fact that √x² = √4
It should be ±√x² = ±√4

Because √x is a function (let's denote it f(x) = √x)
f(x) can never equal both 2 and -2 when x = 4 by the definition of a function.

ever heard of an injective function?
f(x) = f(y) => x=y

there are, however, functions that are not injective. quite many.

An injective function is the basic function we always think of, I don't understand what you mean. The square root of x does happen to be an injective function by it's definition.

Or rather, you aren't wrong that √x² = √4 , it is simply an incomplete statement.

thats only if you bring in the square root yourself

You might want to reconsider