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#36 to #24

bronybox (07/12/2013) []
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.
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.
#135 to #75

anon (07/13/2013) []
It depends on whether one considers the principal square root function or the multivalued relation. Choosing between R and C is irrelevant (for what anyway? domain? codomain?). You can map from reals to reals with a real image and still have both the positives and negatives. It just won't be a function. R vs C is irrelevant.
#60 to #7

shredheadxd (07/12/2013) []
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.
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.
#137 to #60

anon (07/13/2013) []
Not really. In this case, it's not really clear whether or not we're talking about the just the principal value or not. Unlike written out mathematical notation, words can be ambiguous on this matter. So we clarify,
You start with 4, take the (principal) square root, and you will only have 2. Likewise, if you start with the (principal) square root of 4, then you will still only have 2.
You start with 4, take the (principal) square root, and you will only have 2. Likewise, if you start with the (principal) square root of 4, then you will still only have 2.
#143 to #137

shredheadxd (07/13/2013) []
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.
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.
#204 to #202

shredheadxd (07/14/2013) []
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!
±√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!
#206 to #205

shredheadxd (07/15/2013) []
that's correct! I wasn't sure what you were talking about, but the math wasn't ambiguous so I just reiterated it.
#33 to #7

bronybox (07/12/2013) []
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.
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.
#199 to #198

bronybox (07/13/2013) []
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.
#54 to #7

mrloverlover (07/12/2013) []
everyone is replying to this with math equations etc.
and i'm just sat here thinking what the **** is wrong with that dog
and i'm just sat here thinking what the **** is wrong with that dog
#19

trazyntheinfinite (07/12/2013) []
Square root of 4 is 2.
Half of 2 is 1.
1+2=3
Half Life 3 Confirmed.
Half of 2 is 1.
1+2=3
Half Life 3 Confirmed.
#122 to #9

ningyoaijin (07/13/2013) []
The code in the program is only made to recognize a certain number of significant figures. The algorithm they use to calculate square roots must therefore produce very slightly incorrect results.
#102

ybigballz (07/13/2013) []
wait there are 6 2s kneeling then 6/2 is three well do you see where this is going?
#108 to #107

pebar (07/13/2013) []
to back up my claim
mathforum.org/library/drmath/view/74846.html
mathforum.org/library/drmath/view/74846.html
#176 to #163

pebar (07/13/2013) []
this should explain it better
Every positive real number has two square roots, one positive and one negative. For example, the two square roots of 25 are 5 and −5. The positive square root is also known as the principal square root, and is denoted with a radical sign.
en.wikipedia.org/wiki/Nth_root#Square_roots
Every positive real number has two square roots, one positive and one negative. For example, the two square roots of 25 are 5 and −5. The positive square root is also known as the principal square root, and is denoted with a radical sign.
en.wikipedia.org/wiki/Nth_root#Square_roots
#211 to #165

demandred (07/17/2013) []
plus/minus sign indicates two solutions. The square root of a number has two real solutions. What you wrote has therefore 2*2=4 solutions, it's just that (in the case of x=4) the solutions ++2 and 2 are identical, just as +2 and +2. therefore, in the case you wrote, the plus/minus sign adds no information, it merely reminds us that square roots do indeed have two real solutions  one positive and one negative.
#178 to #165

AeroChic (07/13/2013) []
I know that the plus or minus is used to demonstrate that's it's both solutions, but when we first learned about square roots in middle school or whenever, we were taught that they always had two solutions without the plus or minus. However, once into harder math courses, the plus or minus sign was used for clarification. I thought that technically it still had two solutions, but you just used the principal value.