No, as an operation, a sqrt can only give one answer But if you were solving an equation, like x = √16, then you can say x = ±4, because you are solving possible solutions for x He did a video about this, lemme see if I can find EDIT: HERE'S THE VIDEO, STOP REPLYING, YOU'RE ALL GETTING ANYWHERE ruclips.net/video/NBOlK_rSsm4/видео.htmlsi=c-udZ_iYdnBvrS2V
@@nxt6341x=sqrt(16) is not +/-4. it is only 4. Square root is an operation, so it gives one and only one answer. Perhaps you were thinking of cases such as x**2=16, which is an equation, and therefore may have multiple solutions (4**2 is 16 and (-4)**2 is also 16)
This had me incredibly confused, I was never taught the difference between principle square root and square root in high school. Just read about it now and I think I can understand it at least.
So sqrt((-4)^2) = ((-4)^2)^0.5 = (16)^0.5 = 4, but that is not equal to (-4)^(2/2) = (-4)^1 = -4. I always thought you can simply multiply the exponents.
but a^(m/n) is ⁿ√(a^m) and (a^m)^n is a^(mn) so ((-4)^2)^0.5 is (-4)^1 if we using the rule and (-4)^(2/2) is ²√((-4)^2) is there something im missing here?
@@film-gq1wg, If you apply the rule (a^m)^n=a^(mn) to ((-4)^2)^0.5 you will get (-4)^1=-4. However, the rule is incorrect here, as to calculate ((-4)^2)^0.5 you must first calculate (-4)^2 and then raise the result to the power 0.5. Now (-4)^2=16 and 16^0.5=4. You may ask: can't 16^0.5=-4? The answer is that -4 is a square root of 16, however when we write 16^0.5 or √16 we want it to mean a single number, and by convention we take the non-negative square root. This convention is very convenient as it means that a^x (for a>0) has a nice smooth graph. The graph of 16^x would not be smooth if at x=0.5 we said 16^x is negative!
@@MichaelRothwell1 thank you for sharing this knowledge. Now one can say: why not pick the negative for all values of x? Well we can't do that because with some values we have to pick the positive. For example 16^⅓ can't be -2.51984... it has to be 2.51984...
I think it's easier to process this when you stop thinking of the radical function as the inverse of the square function, because that's really not what it is. The inverse of x^2 is x^(1/2), while √x=|x^(1/2)|. It's a subtle difference, but it is important. (x^2)^(1/2) is x^1, but √(x^2) is |(x^2)^(1/2)|=|x^1|. In other words, x^(1/2) can have positive and negative solutions (in fact, for any given x value, it has one of each), but √x can only have positive solutions.
I 100% agree with what you are saying, but note that √x=|x^(1/2)| is only true for real numbers. You might be better off thinking of it as giving the positive root, so it generalises to imaginary numbers.
U are talking about the principal root, by definition the square root of x is the number that multiplied by himself results in x, leading to multiple solutions for the same x. y×y=x as (-y)×(-y)=x so if we don't ask exclusively for the principal root √x=(-y) and √x=(y)
@@znkr53no, the sqrt(x) implies the positive principal root. What you’re talking about is x^2=k which has the positive and the negative root. You can’t get a negative result out of the square root of a number.
I remember once the math teacher told me if (√(-4))^2 then u can bring the power inside . so √((-4)^2)=(√(-4))^2 I guess he had problems with imaginary numbers & liked teaching the wrong methods XD
Well, I don't know why, but everyone seems to forget, that positive values are not the only outcome of square roots. We can get 4 by multiplying 2 by 2 as well as -2 by -2, so √4 = ±2. And it's ALWAYS this way, square roots always have two solutions: positive and negative. That's the reason there's ± in quadratic formula. It's not that somebody just put it there, it's basically to remind us, that square roots have 2 solutions and because of that so do the equations. So √((-4)^2) and (√(-4))^2 are actually both equal to ±4 and not 4 and -4 respectively. That means you CAN just cancel out √ and ^2, as long as you remember to put ± to the solution. That also means that your teachers were right, you CAN bring the power inside, as well as outside for that matter. So in conclusion √((-x)^2)=(√(-x))^2=√((x)^2)=±x
@@sniper7269the square root operator has one value. It's defined as the positive root. There wouldn't need to be a ± in the quadratic formula if the √ itself had 2 results.
@@sniper7269 The reason ± is there is because the square root of something returns only the positive value, not the negative. That's why you'd have to explicitly add ±. It's more useful to define it this way, in math. Now, how about x^2 = 4? You can raise both sides by 1/2. (x^2)^(1/2) = 4^2(1/2) That gives us x = ±2
The reason why you can't do it that way is because you need to follow the order of operations whenever you attempt to evaluate an expression. When you take a square root of something, you need to finish everything that is under the square root first before you do anything outside of it. Therefore, canceling something that is under the root symbol with something that is outside that root symbol would never be allowed. Also, in this case, only the principal square root of the expression under the root symbol is allowed.
@@kalebjohns7715 Well, that is the way it is with a root symbol. I thought before that both roots were part of the answers involving square root symbols, but I was later told differently. The author of this video would tell you that, too. So, 4 ^ 0.5 = 2 and only 2. However; in the expression x ^ 2 = 4, x = 2, -2.
What is the general form of this? What if it was a cube root or 8th root? Is the power rule not valid? How do we know when the power rule should not be applied?
Why was the answer for the first one positive and not just +-4? I was taught that it should be +-4 in pretty much all square roots, not just a positive answer
@@thomasfleming8169 I'm assuming that you only take the positive value when simplifying expressions (don't quote me on this). When you're solving for x you'll have to take both positive and negative solutions.
@@thomasfleming8169 No. The square root function is injective. Otherwise it wouldn't be a function. Just look up "graph of sqrt(x)" and you'll see it only gives the principal root.
This is a good one, cause it brings out the subtleties and the nuances of operations, and the care that must be taken when performing a mathematical derivation. In place of a number, if it was x, one still cannot afford to simply cancel, again depending on the circumstance, yes if x is always non- negative, not otherwise.
X∈ℝ needs to be in every question (x is an element of the reals). Anytime random assumptions are made then solutions are lost. Assuming your solution is in the natural numbers, like bprp did here, is insane.
Missing so many crucial remarks here: First of all include the field you are working in. Consering real numbers then the left expression is properly defined however the right expression is not. As a square root only is defined from [0,inf) to [0,inf) Now if you regard both expressions in the complex field then sqrt((-4+0i)^2)= sqrt(16) = 16^(1/2) = (16e^(2*k*pi*i))^(1/2) = 4e^(k*pi*i) = +-4 However Sqrt(-4) = (-4+0i)^(1/2) = (4e^(i*pi + 2k*i*pi))^(1/2) = 2e^(i*pi/2 + k*i*pi) = +- 2i Note this complex square root has always two complex roots. So (sqrt(-4))^2 = -4 since (-2i)(-2i) = 4i^2 = -4 and (2i)(2i) =4i^2 as well. Tl;DR Define the field you are working in and check function domains!
Gaiz, its a common misconception that sqr root of 16= ±4 Square root always gives a positive ans √16 is = 4 But if x²= 16 then we put ± Because acc to this x can take +4 or -4 value So its like X²= 16 X= ±√16 (When we remove square from a variable side we put ±) Then we write x=±4
I guess that second one can be done without using complex number or iota method by simply simplifying it as (√-4 )^2 , √ can also be written as 1/2 , so (-4^1/2 )^2. While 1/2 × 2 will be 1 , so the answer will be simply -4
I'd consider square roots of negative numbers as something not very well-defined. Maybe I can call it a singularity. You can't find it in unininions. Here, you use binions. In them, it's not very clear how you define square root. If you rotate it and try to keep it continuous, you'll get a negative number as a supposed square root of a positive one. You'd have to choose where you switch to the other side. But you can use a 4D version. In the probably most useful and well-known one, the switch to negative main part happens only at negative numbers (and 0). But in (√-4)², we have that exact ray. And if you go to it, you'll end up on a point of a sphere depending on the direction you came from
I'm confused. We're always been told (x^m)^n = x^(mn) = x^(nm) = (x^n)^m Let x = -4, m = 2, n = 1/2 (x^m)^n = ([-4]^2)^(1/2) = (16)^(1/2) = 4 Yet (x^n)^m = ([-4]^[1/2])^2 = (2*i)^2 = 2^2*i^2 = 4(-1) = -4 So (x^m)^n =/= (x^n)^m? Does the rule not hold with complex and imaginary numbers?
put another way, roots are not the inverses of powers. they're both the same thing, one being a fractional exponent and the other being an integer exponent. thus they don't actually undo each other. logarithms are the actual inverse of these guys.
Why was the answer for the first one positive and not just +-4? I was taught that it should be +-4 in pretty much all square roots, not just a positive answer
@@thomasfleming8169 okay so when you do a square root it will always be a positive number, but if u have a^2 = 4 then a is both ±4. So no square root of something is never a negative outcome
@@montyyardleyEvery nonzero number has two square roots, and three cube roots. Normally the square root symbol indicates that we're picking the positive one, but that's just a convention.
@@montyyardleyIt's a convention to use the particular definition of square root that makes it positive. It's often important to understand the distinction between things which are mathematically necessary, and those things that are convenient for communication.
If you are using math to score points on tests, oh yeah, positive for sure. If you are doing physics or engineering though, you better be sure or your electric circuit might not work or your antimatter might be a unpleasant reminder that reality doesnt care about conventions.
If the question were solve x² = 16, then yes, there are two solutions: x = ±√16 = ±4. But, "√16" refers to only the principal value of the square root function, which is the positive value. It's a notation thing. We're not asked to find all values that equal √16, just one, and only one (the principal value by convention). Extra confusion comes from how math is taught in many schools: nobody teaches the concept of multivalued functions and branches. Normally, teachers just say "square root of a number is both the positive and negative", and then skip writing the intermediate step of "x = ±√16" when solving x² = 16. They jump straight from x² = 16 to x = ±4.
Square roots are always positive. But if you have x^2 = 4 and you are solving that you would get +- 2 because those are the two solutions to x. Thats why there is +- square root so you can get both answers to quadratics
@@cwsurvivor5434 "Square roots are always positive." thats blatantly false. Thats how you get complex numbers - by squaring -1, the result of which is NOT a positive number. It's i.
@@empireempire3545 okay bitch, the square root of all positive numbers are positve. Yeah obviously imaginary exist but have no relevance to the problem in the video dip shit
No square root as a function can only give you one answer I think what u mean to say is x^2=16 is either +4 or -4 this is true because your finding possible values of x and both 4 and -4 square is 16 but for normal equations you can't do it this way
square root isn't always positive, or at least that is not how I learned it. I was taught that sqrt() can be positive or negative, ie sqrt(16)= +/- 4, because it is the inverse of the square function and both positive and negative square to the same value. That is to say, sqrt(x^2)=+/-x by definition. So in this specific case: sqrt((-4)^2) by defintion is +/- (-4) or +/- 4 were as (sqrt(-4))^2 by definition is (+/-2i)^2 which is strictly -4 as 1^2 and -1^2 both equal 1.
And how would you arrive at those two answers? I forget what the official terminology is, but I recall being told that sqrt is not a 'true' function as any given input can have multiple output, ie one-to-many. Taking the absolute value to have only output>=0 does collapse it to a one-to-one function, but it is literally only half the picture.@@birutologo
@@birutologo Hey sorry but the two equations that you gave are unfortunantly the same equation just slightly rearranged. one being x = sqrt(16) and the other being x^2=16 which just changed where the square is. in the right scenario since x can be both 4 and -4, then as both equations are the same the left scenario must also have the answers 4 and -4. when dealing with a square root it is important to note that any answer you recieve from them will be both positive and negative and result in two solutions. if you are still confused i would be happy to explain another way :). edit: in fact this whole video is technically incorrect because, as explained above, the left scenario has answers -4 and 4. while the right scenario has, when square rooting 2i^2 and (-2i)^2 which equal -4 and 4 respectively. I will clarify that the video is not incorrect that you cant just remove the square root sign as that would imply one solution when in reality there are two solution to both equations.
@@quinnbell2388 Incorrect; you're getting the inputs and outputs backwards. A given input CAN have multiple outputs, so long as they do not have the same x value - this is how we get parabolas in a simple quadratic equation; do a search for "vertical line test" for determining what is and is not a function, graphically, because the rule for functions is that any given *output* can have one and only one input. For example, sqrt(x) passes the vertical line test because a vertical line drawn on the graph of sqrt(x) passes through one and only one point for any value of x. In contrast, a formula which produces a circle is not a function because a vertical line will pass through two points on the circle.
@@Corn_Fiendsorry but you’re incorrect here. They are not the same equations. When you solve x^2=16 you get to x=+/-sqrt(16) which is different to just x=sqrt(16) as the sqrt sign only takes the positive root, if you want to have both positive and negative roots you need to add the +/- sign in front
My issue with this is, is sqrt(x) always equal to (x)^(1/2)? If so, then is x^y^z also always equal to x^yz? If so, then (x^(1/2))^2 and (x^2)^(1/2) should both always just be equal to x^(2/2), or x. Something is clearly wrong with the way my high school teacher explained rules of exponentials here
You: Is sqrt(x) always equal to x^(1/2)? Answer: Yes. For example, if x≥0, then both bits of notation mean the non-negative square root of x. E.g. √4=4^½=2 (and not -2 or ±2). In other cases, both bits of notation mean the principal square root of x. E.g. √(-1)=(-1)^½=i (and not -i or ±i). However, there are people who consider that (in the complex case), x^½ represents the two values ±√x. You: Then (x^(1/2))^2 and (x^2)^(1/2) should both always be equal to x^(2/2) or x. Comment: Here you are using the rule of exponents (aᵇ)ᶜ=aᵇᶜ. However, this rule is only guaranteed to work if the base a is positive or the exponents b and c are integers. In the case √(x²) or (x²)^½ this is not the case if x
@@MichaelRothwell1 that seems like such an arbitrary rule to just say "The Power of a Power rule only applies when the base is positive". It feels more like an afterthought stated for the purpose of making the notation they already HAD work, rather than an actual explanation of WHY that's the case. For example, one could make ((-1)^1/2)^2 COULD equal ((-1)^2)^1/2 if you consider the non-principal second root of unity. So (-1)^1/2 = i, and (-1)^2 = 1. Now the second step is i^2 is -1, and the non-principal second root of 1 is -1, resulting in them both being the same. But, because of the seemingly arbitrary decision that we must always only consider the most "regular" answer from a root and that the Power of a Power rule no longer works for negative bases, we end up with a notation that just completely stops making sense in some contexts. To me that just doesn't feel like a good reason.
@@GameJam230 You're not alone in feeling that way. As I said in my comment, there are people who consider that (in the complex case), x^½ represents the two values ±√x. However, in the world of real numbers, the convention is to consider the principal square root only, so that √x and x^½ take a definite value for each non-negative value of x. This convention is really useful, as it means that √x and x^½ are functions.
@@MichaelRothwell1awesome, thanks for that. I saw a different comment arguing that sqrt(x^2)=(x^2)^(1/2), et. all, and seriously was confused, because at first (educated) look, that actually made sense, but now I understand why it’s not the case 👍🏻
You are comparing warm with soft (or apples with oranges). Left one is done over R and right one done over C. Proper way to compare would be explicitly choosing R or C for both sides 1st and then doing calculations in correct order. R: left = 4, right = none C: left = +/- 4, right = ... also +/- 4 (+ periodicity) In general, a "root of power N" over C will have N number of principal values +i*2*pi*k periodicity.
(-4^2) = (-4 * -4)=16 and root 16 aims to finde the number that times it self gives 16 so the anser is root(-4)^2=4 You can actually calculate distance in as many dimensions as you want YouSee peace functions for example in 3d root((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2)
That's easy: *√(−4) = ±2𝒊 => (√(−4))² = (±2𝒊)² = −4* - one answer *(−4)² = 16 => √((−4)²) = √16 = ±4* - two answers In both cases a multivalued complex root is considered (just because I want so).
but wouldn't (sqrt(-4))^2 equal 4 too? Since it's basically sqrt(-4)×sqrt(-4), and when you multiply square roots you can insert both terms under the same radical?
The ans is +4 and -4(±4) both you van write √16 as ±4 or √(-4)^2 like (-4^2)^1/2 so its -4 so yes you can cancel square root and square but it will incomplete but correct ans
So it's not ok to combine the square root and exponent into a single rational exponent before evaluating anything? Or would you only be able to do that if there were no parenthesis being used? This video made me super paranoid of evaluating these situations incorrectly for years!
Yes in this cause x would be +4 or -4, and that's what people often get confused in. Working with quadratic equations and working with square root equations mean different things, and have different answer.... Like x^2 = 9 has 3 and -3 as roots, but x = √9 has only 3 as it's root
@@gametimewitharyan6665 yes because to get the -3 as the root you must first write it as -√9=-3 or in other words, to get negative part of the roots, just put a negative symbol infront of the sqrt
@@gheffz Exactly!!!! You can check its definition on Wikipedia (I know it isn't always a trustworthy source but in this case it is) If you are lazy like me then I will tell you the definition √(x^2) = |x| Hence, sqrt((3)^2) = 3 And sqrt((-3)^2) = 3
Well, I don't know why, but everyone seems to forget, that positive values are not the only outcome of square roots. We can get 4 by multiplying 2 by 2 as well as -2 by -2, so √4 = ±2. And it's ALWAYS this way, square roots always have two solutions: positive and negative. That's the reason there's ± in quadratic formula. It's not that somebody just put it there, it's basically to remind us, that square roots have 2 solutions and because of that so do the equations. So √((-4)^2) and (√(-4))^2 are actually both equal to ±4 and not 4 and -4 respectively. That means you CAN just cancel out √ and ^2, as long as you remember to put ± to the solution. That also means that your teachers were right, you CAN bring the power inside, as well as outside for that matter. So in conclusion √((-x)^2)=(√(-x))^2=√((x)^2)=±x
The real root of 16 is in fact only 4 but the complex root of 16 is both +4 and -4. So if you work in the field of complex numbers in the second example, you should also do it in the first one. Complex and real roots are defined differently.
@@empireempire3545 the square root function can not return multiple values. The roots of a functions is not the same thing as the square root function, which is meant to return one value
Can’t we argue that (sqrt-4)^2 is the same as sqrt-4 * sqrt-4, and we know that when we’re multiplying, we can combine the two radicals and multiply inside so it’ll be the same as saying sqrt of (-4)^2 so also equal to 4?
Clarity is perhaps the real problem. If the problem stated f(x)= then each problem would only have one answer by definition, preventing much of the debate. This could be confirmed graphically. I teach my students that a square root symbol, like any number, is assumed to be asking for the positive root, otherwise the problem would have been written with a +-. However they must remember to include both if they introduce a root into a problem like solving x^2=9. Personally, when simplifying the original expressions I would not introduce an equal sign into process if it was not in the original problem.
we're learning this in algebra 2 right now and the general rule is that if you are simplifying a radical and the number is to an even power, and the root symbol is also to an even power, then when you simplify it if its an odd exponent then you must add an absolute value bracket around the variable or number
For those trying to understand the inconsistency: He’s not really explaining math. He’s explaining some standard definition square root function of a symbol.
The second example is fascinating. Is it considered to have a real solution? Even though through the order of operations, the first term calculated does not have a real solution? That is weird.
Hi, I was just curious to know what happens if we write sqrt(-4)^2 = [(-4)^2]^1/2 ? Won't the twos just cancel out in the exponent and we'll be left with just -4? I'm just learning exponents, so correct me if I am wrong!! :)
square root itself is an operation that requires the input to be non-negative (so positive or 0) AND outputs only non-negatives. So no, you'll *have* to take absolute value
Hey, your intuition is correct, however note that exponential notation requires a non-negative base as well. So while it "works out" in this case it is not technically correct. Maybe to clarify for a nonnegative real number x it holds indeed that (Sqrt(x))^2 = (x^(1/2))^2 = x^1 = (x^2)^(1/2) = sqrt(x^2))
Just canceling the sqrt(x)^2 got me all the way through a BSci. in Applied Physics, thank Christ I didnt have to unlearn that… it would be an absolute BITCH in ODEs/PDEs 😬
THAT! That right there! Thats why you dont teach nonsense like bprp is doing here! Later on it gets tough and your basis needs to be correct from the start. You can slap in x is an element of (whole, natural, real, complex, whatever) and play these stupid games, but you cant just sing a song like the snake was singing to mowgli and say trussssst me... rootssssss are posssitive.....
That what I was thinking as well, I figured it would just end up being -4^1. The only thing is that you're missing the positive 4 (because the square root of 16 is ±4). I think the same rule would apply when taking square roots though because you would still have to add the ±
I do not agree; a root of order y had y roots/solutions. Both of roots are of order 2, so both have two valid solutions. In both cases the solutions are +4 and -4.
I'm sorry, I know that somehow the consensus is that √-1 = i, but this is something I am simply unwilling to accept as a definition for i. If instead we used i^2=-1 and left √x as undefined for x
So, I encounter this in a community post, and normally I prefer to work it out in the comments section, but sqrt(-4) is breaking my brain.... Although, it doesn't take much these days, though.... I'm gonna watch the video now.
I don't agree with this process. We should use algebraic/radical mathematical solving first. The approach its done in this video though its computational, from inside or outside out first. In pure maths though raising in either way the power goes to 2/2 FIRST, then resulting only -4.
@@forbidden-cyrillic-handleyour formula doesn't even apply here. (-4) can be considered in both cases in brackets and it is negative in both cases. Then exponent a half times two or two times a half. You fell into technical and subsequential computation. Like root mean square to get rid of negatives which is done purposely and technically
the "note" section is an actual definion of a square power under the square root, and the square power on top of the square root will always cancel out, again by definition... idk why is he complicating 4th grade math with imaginary number lmao
Q: 2^2^3 = ?
Answer here ruclips.net/video/AXtbpKoWHZU/видео.html
A: 256
Right-associativity here.
or 64
256
256
"But check this out-"
Top 10 cliffhangers
Is the pokeball your mic or does it just help you focus?
possibly both
both
Yes its his mic but he also said he likes to keep his other hand occupied
dude the way you switch pens while writing soo cool
Do you know how to eat with chopsticks? I guess this youtuber applicated it on those pens. (It is pretty easy, you can try it)
@@massimilianohu asian entered the chat
Isn't sqrt of 16=±4?
I was gonna say that ! Although his point stands, it really triggered me because my calc teacher drilled this into me day one
No, as an operation, a sqrt can only give one answer
But if you were solving an equation, like x = √16, then you can say x = ±4, because you are solving possible solutions for x
He did a video about this, lemme see if I can find
EDIT: HERE'S THE VIDEO, STOP REPLYING, YOU'RE ALL GETTING ANYWHERE
ruclips.net/video/NBOlK_rSsm4/видео.htmlsi=c-udZ_iYdnBvrS2V
ruclips.net/video/N30hLAa-to8/видео.html
@@nxt6341x=sqrt(16) is not +/-4. it is only 4. Square root is an operation, so it gives one and only one answer. Perhaps you were thinking of cases such as x**2=16, which is an equation, and therefore may have multiple solutions (4**2 is 16 and (-4)**2 is also 16)
@@Anonymous4045 yeah, that's it. Thanks!
Check what out, blackpenredpen?
My ass
Spunge
Check this out:
|z| = magnitude of z. So |i| = 1.
But sqrt(i^2) = i ≠ -1.
So the only time sqrt(x^2) = |x| is when x is real.
@@nanamacapagal8342 shouldn’t it be the other way around?
@@nanamacapagal8342sqrt(z^2) has to be in the first quadrant
Fun fact: sqrt((x)^2) is the same as taking its absolute value (symbol: |x|). Which is making x positive no matter what (except for complex numbers)
id expect that
sqrt(x*conj(x))
This works for Real numbers _and_ Complex numbers. It even works for a few other number systems like Quaternions.
But isn't sqrt(x) = ±y
Also for his example
sqrt(16) = ±4
@@redcrafterlppa303
other way around
@@redcrafterlppa303 No, sqrt(x) = y. Do not confuse with sqrt(x^2) = ±y, wich is true
This had me incredibly confused, I was never taught the difference between principle square root and square root in high school. Just read about it now and I think I can understand it at least.
So sqrt((-4)^2) = ((-4)^2)^0.5 = (16)^0.5 = 4, but that is not equal to (-4)^(2/2) = (-4)^1 = -4. I always thought you can simply multiply the exponents.
You can always multiply the exponents if the base is positive or the exponents are integers.
but a^(m/n) is ⁿ√(a^m)
and (a^m)^n is a^(mn)
so ((-4)^2)^0.5 is (-4)^1 if we using the rule
and (-4)^(2/2) is ²√((-4)^2) is there something im missing here?
@@film-gq1wg, If you apply the rule (a^m)^n=a^(mn) to
((-4)^2)^0.5 you will get (-4)^1=-4. However, the rule is incorrect here, as to calculate ((-4)^2)^0.5 you must first calculate (-4)^2 and then raise the result to the power 0.5. Now (-4)^2=16 and 16^0.5=4. You may ask: can't 16^0.5=-4? The answer is that -4 is a square root of 16, however when we write 16^0.5 or √16 we want it to mean a single number, and by convention we take the non-negative square root. This convention is very convenient as it means that a^x (for a>0) has a nice smooth graph. The graph of 16^x would not be smooth if at x=0.5 we said 16^x is negative!
@@MichaelRothwell1 thank you for sharing this knowledge. Now one can say: why not pick the negative for all values of x? Well we can't do that because with some values we have to pick the positive.
For example 16^⅓ can't be -2.51984... it has to be 2.51984...
The rule only holds for positive real numbers in the base
What we have to check sir👍👍❤️❤️
I think it's easier to process this when you stop thinking of the radical function as the inverse of the square function, because that's really not what it is. The inverse of x^2 is x^(1/2), while √x=|x^(1/2)|. It's a subtle difference, but it is important. (x^2)^(1/2) is x^1, but √(x^2) is |(x^2)^(1/2)|=|x^1|. In other words, x^(1/2) can have positive and negative solutions (in fact, for any given x value, it has one of each), but √x can only have positive solutions.
That's all I needed to hear thanks alot
I 100% agree with what you are saying, but note that √x=|x^(1/2)| is only true for real numbers. You might be better off thinking of it as giving the positive root, so it generalises to imaginary numbers.
U are talking about the principal root, by definition the square root of x is the number that multiplied by himself results in x, leading to multiple solutions for the same x.
y×y=x as (-y)×(-y)=x so if we don't ask exclusively for the principal root √x=(-y) and √x=(y)
@@znkr53no, the sqrt(x) implies the positive principal root. What you’re talking about is x^2=k which has the positive and the negative root. You can’t get a negative result out of the square root of a number.
@@blablabla7796 yes u can if u dont use the √ symbol
I remember once the math teacher told me if (√(-4))^2 then u can bring the power inside . so √((-4)^2)=(√(-4))^2
I guess he had problems with imaginary numbers & liked teaching the wrong methods XD
All teachers teach it that way. Maybe it's curriculum or they don't know better.
Well, I don't know why, but everyone seems to forget, that positive values are not the only outcome of square roots.
We can get 4 by multiplying 2 by 2 as well as -2 by -2, so √4 = ±2. And it's ALWAYS this way, square roots always have two solutions: positive and negative.
That's the reason there's ± in quadratic formula. It's not that somebody just put it there, it's basically to remind us, that square roots have 2 solutions and because of that so do the equations.
So √((-4)^2) and (√(-4))^2 are actually both equal to ±4 and not 4 and -4 respectively. That means you CAN just cancel out √ and ^2, as long as you remember to put ± to the solution. That also means that your teachers were right, you CAN bring the power inside, as well as outside for that matter.
So in conclusion √((-x)^2)=(√(-x))^2=√((x)^2)=±x
@@sniper7269the square root operator has one value. It's defined as the positive root. There wouldn't need to be a ± in the quadratic formula if the √ itself had 2 results.
@@sniper7269 The reason ± is there is because the square root of something returns only the positive value, not the negative. That's why you'd have to explicitly add ±. It's more useful to define it this way, in math.
Now, how about x^2 = 4? You can raise both sides by 1/2. (x^2)^(1/2) = 4^2(1/2)
That gives us x = ±2
The second one can be done this way. Since sqrt equals power 1/2 so it'll be come (4^1/2)^2. So power multiply 1/2×2 which is one which is -4^1=-4
why is it not +(or)- 4 for the left side solution?
It makes sense now. Thank you
The reason why you can't do it that way is because you need to follow the order of operations whenever you attempt to evaluate an expression. When you take a square root of something, you need to finish everything that is under the square root first before you do anything outside of it. Therefore, canceling something that is under the root symbol with something that is outside that root symbol would never be allowed. Also, in this case, only the principal square root of the expression under the root symbol is allowed.
why is only the principal root allowed?
@@kalebjohns7715 Well, that is the way it is with a root symbol. I thought before that both roots were part of the answers involving square root symbols, but I was later told differently. The author of this video would tell you that, too. So, 4 ^ 0.5 = 2 and only 2. However; in the expression x ^ 2 = 4, x = 2, -2.
What is the general form of this? What if it was a cube root or 8th root? Is the power rule not valid? How do we know when the power rule should not be applied?
Square roots are kind of skimmed over in pemdas, thanks for the heads up on the correct order with exponents and roots.
Some people say gemdas instead of pemdas, so they can say groupings instead of parenthesis. You could say 4^2 was grouped inside of the square root.
Why was the answer for the first one positive and not just +-4? I was taught that it should be +-4 in pretty much all square roots, not just a positive answer
@@thomasfleming8169 I'm assuming that you only take the positive value when simplifying expressions (don't quote me on this). When you're solving for x you'll have to take both positive and negative solutions.
@@foureyedgrayp ok tha k you so much!
@@thomasfleming8169 No. The square root function is injective. Otherwise it wouldn't be a function. Just look up "graph of sqrt(x)" and you'll see it only gives the principal root.
This is a good one, cause it brings out the subtleties and the nuances of operations, and the care that must be taken when performing a mathematical derivation. In place of a number, if it was x, one still cannot afford to simply cancel, again depending on the circumstance, yes if x is always non- negative, not otherwise.
X∈ℝ needs to be in every question (x is an element of the reals). Anytime random assumptions are made then solutions are lost. Assuming your solution is in the natural numbers, like bprp did here, is insane.
Missing so many crucial remarks here:
First of all include the field you are working in. Consering real numbers then the left expression is properly defined however the right expression is not. As a square root only is defined from [0,inf) to [0,inf)
Now if you regard both expressions in the complex field then sqrt((-4+0i)^2)= sqrt(16) = 16^(1/2) = (16e^(2*k*pi*i))^(1/2) = 4e^(k*pi*i) = +-4
However
Sqrt(-4) = (-4+0i)^(1/2) = (4e^(i*pi + 2k*i*pi))^(1/2) = 2e^(i*pi/2 + k*i*pi) = +- 2i
Note this complex square root has always two complex roots.
So (sqrt(-4))^2 = -4 since
(-2i)(-2i) = 4i^2 = -4 and
(2i)(2i) =4i^2 as well.
Tl;DR
Define the field you are working in and check function domains!
Gaiz, its a common misconception that sqr root of 16= ±4
Square root always gives a positive ans
√16 is = 4
But if x²= 16 then we put ±
Because acc to this x can take +4 or -4 value
So its like
X²= 16
X= ±√16
(When we remove square from a variable side we put ±)
Then we write x=±4
Square root does not always lead to a positive answer. Square root of 0 is 0.
biggest mistakes in math now far away :-) thank you 🙂
I guess that second one can be done without using complex number or iota method by simply simplifying it as (√-4 )^2 , √ can also be written as 1/2 , so (-4^1/2 )^2. While 1/2 × 2 will be 1 , so the answer will be simply -4
I'd consider square roots of negative numbers as something not very well-defined. Maybe I can call it a singularity. You can't find it in unininions. Here, you use binions. In them, it's not very clear how you define square root. If you rotate it and try to keep it continuous, you'll get a negative number as a supposed square root of a positive one. You'd have to choose where you switch to the other side. But you can use a 4D version. In the probably most useful and well-known one, the switch to negative main part happens only at negative numbers (and 0). But in (√-4)², we have that exact ray. And if you go to it, you'll end up on a point of a sphere depending on the direction you came from
I'm confused.
We're always been told (x^m)^n = x^(mn) = x^(nm) = (x^n)^m
Let x = -4, m = 2, n = 1/2
(x^m)^n = ([-4]^2)^(1/2) = (16)^(1/2) = 4
Yet (x^n)^m = ([-4]^[1/2])^2 = (2*i)^2 = 2^2*i^2 = 4(-1) = -4
So (x^m)^n =/= (x^n)^m? Does the rule not hold with complex and imaginary numbers?
What is that watch you are wearing tho
put another way, roots are not the inverses of powers. they're both the same thing, one being a fractional exponent and the other being an integer exponent. thus they don't actually undo each other.
logarithms are the actual inverse of these guys.
Why was the answer for the first one positive and not just +-4? I was taught that it should be +-4 in pretty much all square roots, not just a positive answer
@@thomasfleming8169 okay so when you do a square root it will always be a positive number, but if u have a^2 = 4 then a is both ±4. So no square root of something is never a negative outcome
@@montyyardleyEvery nonzero number has two square roots, and three cube roots. Normally the square root symbol indicates that we're picking the positive one, but that's just a convention.
@@montyyardleyIt's a convention to use the particular definition of square root that makes it positive. It's often important to understand the distinction between things which are mathematically necessary, and those things that are convenient for communication.
If you are using math to score points on tests, oh yeah, positive for sure.
If you are doing physics or engineering though, you better be sure or your electric circuit might not work or your antimatter might be a unpleasant reminder that reality doesnt care about conventions.
A question, doesn't the first question have two solutions? Both 4 and -4, as both 4*4=16 and -4*(-4) is 16?
If the question were solve x² = 16, then yes, there are two solutions: x = ±√16 = ±4. But, "√16" refers to only the principal value of the square root function, which is the positive value. It's a notation thing. We're not asked to find all values that equal √16, just one, and only one (the principal value by convention).
Extra confusion comes from how math is taught in many schools: nobody teaches the concept of multivalued functions and branches. Normally, teachers just say "square root of a number is both the positive and negative", and then skip writing the intermediate step of "x = ±√16" when solving x² = 16. They jump straight from x² = 16 to x = ±4.
Square Root of 16 are +√16 and -√16. If you ask √16 you only ask for positive one.
but in the first example isn't sqrt 16 equal to +-4?
Square roots are always positive. But if you have x^2 = 4 and you are solving that you would get +- 2 because those are the two solutions to x. Thats why there is +- square root so you can get both answers to quadratics
@@cwsurvivor5434 "Square roots are always positive." thats blatantly false. Thats how you get complex numbers - by squaring -1, the result of which is NOT a positive number. It's i.
@@empireempire3545 okay bitch, the square root of all positive numbers are positve. Yeah obviously imaginary exist but have no relevance to the problem in the video dip shit
@@empireempire3545 he obviously meant for positive numbers
@@cwsurvivor5434 Hov can the set of complex numbers have no relevance ? Did you forget that this includes the set of real numbers ?
I have a question. Isn't √16=±4?
No square root as a function can only give you one answer I think what u mean to say is x^2=16 is either +4 or -4 this is true because your finding possible values of x and both 4 and -4 square is 16 but for normal equations you can't do it this way
Yes, you are right, in the set of complex numbers any square root has two solutions.
square root isn't always positive, or at least that is not how I learned it. I was taught that sqrt() can be positive or negative, ie sqrt(16)= +/- 4, because it is the inverse of the square function and both positive and negative square to the same value. That is to say, sqrt(x^2)=+/-x by definition. So in this specific case: sqrt((-4)^2) by defintion is +/- (-4) or +/- 4 were as (sqrt(-4))^2 by definition is (+/-2i)^2 which is strictly -4 as 1^2 and -1^2 both equal 1.
When doing sqrt(16), you only take the positive awnser, 4. But, if we have x^2=16, in that case we would have 2 awnsers.
And how would you arrive at those two answers? I forget what the official terminology is, but I recall being told that sqrt is not a 'true' function as any given input can have multiple output, ie one-to-many. Taking the absolute value to have only output>=0 does collapse it to a one-to-one function, but it is literally only half the picture.@@birutologo
@@birutologo Hey sorry but the two equations that you gave are unfortunantly the same equation just slightly rearranged. one being x = sqrt(16) and the other being x^2=16 which just changed where the square is. in the right scenario since x can be both 4 and -4, then as both equations are the same the left scenario must also have the answers 4 and -4. when dealing with a square root it is important to note that any answer you recieve from them will be both positive and negative and result in two solutions. if you are still confused i would be happy to explain another way :).
edit: in fact this whole video is technically incorrect because, as explained above, the left scenario has answers -4 and 4. while the right scenario has, when square rooting 2i^2 and (-2i)^2 which equal -4 and 4 respectively. I will clarify that the video is not incorrect that you cant just remove the square root sign as that would imply one solution when in reality there are two solution to both equations.
@@quinnbell2388 Incorrect; you're getting the inputs and outputs backwards. A given input CAN have multiple outputs, so long as they do not have the same x value - this is how we get parabolas in a simple quadratic equation; do a search for "vertical line test" for determining what is and is not a function, graphically, because the rule for functions is that any given *output* can have one and only one input. For example, sqrt(x) passes the vertical line test because a vertical line drawn on the graph of sqrt(x) passes through one and only one point for any value of x. In contrast, a formula which produces a circle is not a function because a vertical line will pass through two points on the circle.
@@Corn_Fiendsorry but you’re incorrect here. They are not the same equations. When you solve x^2=16 you get to x=+/-sqrt(16) which is different to just x=sqrt(16) as the sqrt sign only takes the positive root, if you want to have both positive and negative roots you need to add the +/- sign in front
Man sideways math always throws me
My issue with this is, is sqrt(x) always equal to (x)^(1/2)? If so, then is x^y^z also always equal to x^yz? If so, then (x^(1/2))^2 and (x^2)^(1/2) should both always just be equal to x^(2/2), or x. Something is clearly wrong with the way my high school teacher explained rules of exponentials here
You: Is sqrt(x) always equal to x^(1/2)?
Answer: Yes.
For example, if x≥0, then both bits of notation mean the non-negative square root of x. E.g. √4=4^½=2 (and not -2 or ±2).
In other cases, both bits of notation mean the principal square root of x. E.g. √(-1)=(-1)^½=i (and not -i or ±i). However, there are people who consider that (in the complex case), x^½ represents the two values ±√x.
You: Then (x^(1/2))^2 and (x^2)^(1/2) should both always be equal to x^(2/2) or x.
Comment: Here you are using the rule of exponents (aᵇ)ᶜ=aᵇᶜ. However, this rule is only guaranteed to work if the base a is positive or the exponents b and c are integers. In the case √(x²) or (x²)^½ this is not the case if x
@@MichaelRothwell1 that seems like such an arbitrary rule to just say "The Power of a Power rule only applies when the base is positive". It feels more like an afterthought stated for the purpose of making the notation they already HAD work, rather than an actual explanation of WHY that's the case. For example, one could make ((-1)^1/2)^2 COULD equal ((-1)^2)^1/2 if you consider the non-principal second root of unity. So (-1)^1/2 = i, and (-1)^2 = 1. Now the second step is i^2 is -1, and the non-principal second root of 1 is -1, resulting in them both being the same. But, because of the seemingly arbitrary decision that we must always only consider the most "regular" answer from a root and that the Power of a Power rule no longer works for negative bases, we end up with a notation that just completely stops making sense in some contexts.
To me that just doesn't feel like a good reason.
@@GameJam230 You're not alone in feeling that way. As I said in my comment, there are people who consider that (in the complex case), x^½ represents the two values ±√x.
However, in the world of real numbers, the convention is to consider the principal square root only, so that √x and x^½ take a definite value for each non-negative value of x. This convention is really useful, as it means that √x and x^½ are functions.
@@MichaelRothwell1awesome, thanks for that. I saw a different comment arguing that sqrt(x^2)=(x^2)^(1/2), et. all, and seriously was confused, because at first (educated) look, that actually made sense, but now I understand why it’s not the case 👍🏻
@@neohavic6012 you're welcome!
what if you were to rewrite the sqrt as being to the 1/2 power and then distributing to the 2 making it (-4)^1? how would that work?
Everywhere is look it says the power rule is (x^a)^b = x^ab..... So is this only true for whole numbers of a and b?
You are comparing warm with soft (or apples with oranges). Left one is done over R and right one done over C.
Proper way to compare would be explicitly choosing R or C for both sides 1st and then doing calculations in correct order.
R: left = 4, right = none
C: left = +/- 4, right = ... also +/- 4 (+ periodicity)
In general, a "root of power N" over C will have N number of principal values +i*2*pi*k periodicity.
(-4^2) = (-4 * -4)=16 and root 16 aims to finde the number that times it self gives 16 so the anser is root(-4)^2=4
You can actually calculate distance in as many dimensions as you want YouSee peace functions for example in 3d
root((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2)
That's easy:
*√(−4) = ±2𝒊 => (√(−4))² = (±2𝒊)² = −4* - one answer
*(−4)² = 16 => √((−4)²) = √16 = ±4* - two answers
In both cases a multivalued complex root is considered (just because I want so).
it shoulbe plus or minus four right?
but wouldn't (sqrt(-4))^2 equal 4 too? Since it's basically sqrt(-4)×sqrt(-4), and when you multiply square roots you can insert both terms under the same radical?
However, is x^(1/2)^2 not equal to x? And is x^(1/2) the same as sqrt(x)?
The ans is +4 and -4(±4) both you van write √16 as ±4 or √(-4)^2 like (-4^2)^1/2 so its -4 so yes you can cancel square root and square but it will incomplete but correct ans
So it's not ok to combine the square root and exponent into a single rational exponent before evaluating anything? Or would you only be able to do that if there were no parenthesis being used? This video made me super paranoid of evaluating these situations incorrectly for years!
If it were x^2 = (-4)^2 and you solve for x, then it would be x = +4 and x = -4, right?
Yes in this cause x would be +4 or -4, and that's what people often get confused in. Working with quadratic equations and working with square root equations mean different things, and have different answer....
Like
x^2 = 9 has 3 and -3 as roots, but
x = √9 has only 3 as it's root
@@gametimewitharyan6665 yes because to get the -3 as the root you must first write it as
-√9=-3
or in other words, to get negative part of the roots, just put a negative symbol infront of the sqrt
You can also look at it graphically and see that x^2 will always yield two solutions which is why it’s not a one-to-one function.
@@gametimewitharyan6665 So it's by definition?
@@gheffz Exactly!!!! You can check its definition on Wikipedia (I know it isn't always a trustworthy source but in this case it is)
If you are lazy like me then I will tell you the definition
√(x^2) = |x|
Hence,
sqrt((3)^2) = 3 And
sqrt((-3)^2) = 3
Well, I don't know why, but everyone seems to forget, that positive values are not the only outcome of square roots.
We can get 4 by multiplying 2 by 2 as well as -2 by -2, so √4 = ±2. And it's ALWAYS this way, square roots always have two solutions: positive and negative.
That's the reason there's ± in quadratic formula. It's not that somebody just put it there, it's basically to remind us, that square roots have 2 solutions and because of that so do the equations.
So √((-4)^2) and (√(-4))^2 are actually both equal to ±4 and not 4 and -4 respectively. That means you CAN just cancel out √ and ^2, as long as you remember to put ± to the solution. That also means that your teachers were right, you CAN bring the power inside, as well as outside for that matter.
So in conclusion √((-x)^2)=(√(-x))^2=√((x)^2)=±x
If Sqrt(-4)^2 is 16, so after expression, it will 4 or -4?
both
@@empireempire3545 my 28 days
But turn them into exponent notation you get 2/2 right?
Of course.
The real root of 16 is in fact only 4 but the complex root of 16 is both +4 and -4. So if you work in the field of complex numbers in the second example, you should also do it in the first one. Complex and real roots are defined differently.
The √ implies you take the absolute value which i get because its a function
But also i think its pretty dumb sometimes
no such implication is made. Square root can and often does return multiple values.
@@empireempire3545 the square root function can not return multiple values. The roots of a functions is not the same thing as the square root function, which is meant to return one value
So
(-4)^(1/2)×2 doesn't equalls to (-4)^2×1/2
Interesting so can we really use the roots as a power of 1/a in math ? Or it is wrong ?
What abouut cube root or other index ?
Can’t we argue that (sqrt-4)^2 is the same as sqrt-4 * sqrt-4, and we know that when we’re multiplying, we can combine the two radicals and multiply inside so it’ll be the same as saying sqrt of (-4)^2 so also equal to 4?
Clarity is perhaps the real problem. If the problem stated f(x)= then each problem would only have one answer by definition, preventing much of the debate. This could be confirmed graphically.
I teach my students that a square root symbol, like any number, is assumed to be asking for the positive root, otherwise the problem would have been written with a +-. However they must remember to include both if they introduce a root into a problem like solving x^2=9.
Personally, when simplifying the original expressions I would not introduce an equal sign into process if it was not in the original problem.
we're learning this in algebra 2 right now and the general rule is that if you are simplifying a radical and the number is to an even power, and the root symbol is also to an even power, then when you simplify it if its an odd exponent then you must add an absolute value bracket around the variable or number
I love the fundamentals videos
For those trying to understand the inconsistency:
He’s not really explaining math. He’s explaining some standard definition square root function of a symbol.
wait, I'm sorry. Where does square root fit in the order of operations (PEMDAS)?
The second example is fascinating. Is it considered to have a real solution? Even though through the order of operations, the first term calculated does not have a real solution? That is weird.
Hi, I was just curious to know what happens if we write sqrt(-4)^2 = [(-4)^2]^1/2 ? Won't the twos just cancel out in the exponent and we'll be left with just -4? I'm just learning exponents, so correct me if I am wrong!! :)
square root itself is an operation that requires the input to be non-negative (so positive or 0) AND outputs only non-negatives.
So no, you'll *have* to take absolute value
Hey, your intuition is correct, however note that exponential notation requires a non-negative base as well. So while it "works out" in this case it is not technically correct.
Maybe to clarify for a nonnegative real number x it holds indeed that
(Sqrt(x))^2 = (x^(1/2))^2 = x^1 = (x^2)^(1/2) = sqrt(x^2))
doesnt the square root output +- ? bc, for exemple, with 4, 2*2 = 4, and -2*-2=4
So the square root of 4 would be +-2
"But check this out-"
Top 10 anime betrayals
Just canceling the sqrt(x)^2 got me all the way through a BSci. in Applied Physics, thank Christ I didnt have to unlearn that… it would be an absolute BITCH in ODEs/PDEs 😬
THAT! That right there! Thats why you dont teach nonsense like bprp is doing here! Later on it gets tough and your basis needs to be correct from the start.
You can slap in x is an element of (whole, natural, real, complex, whatever) and play these stupid games, but you cant just sing a song like the snake was singing to mowgli and say trussssst me... rootssssss are posssitive.....
2:33 the note here my teacher thought me this i didnt understand why u need the absolute value well i just left it as ut is now ik why
And what if you write those two expresions in fractions of exponents, like: -4^(2/2)
In which case would this apply?
That what I was thinking as well, I figured it would just end up being -4^1. The only thing is that you're missing the positive 4 (because the square root of 16 is ±4). I think the same rule would apply when taking square roots though because you would still have to add the ±
(-4)^(2/2) = -4
In this case always simplify the fraction
@@forbidden-cyrillic-handlewhy not?
Why is is 2i? And what is "i" in the first place?
i = √(-1)
It is the primary Imaginary Number
New channel name bprpbp math basics (2nd bp means blue pen)
brb(p)
But for example if it was a function you'd have to see the domain first if it's negative then it would become 4 but for the positive -4
But we can write both of them as (-4)^(2/2) = -4, right?
Only works if the base is positive, which is not true in this case
His second problem is correct
Not sure about the third line though
lmaoooo i went through calc 1 2, linear algebra and real analysis without knowing this.
2:21 go to the complex world immediatelly, sqrt(x^2) cannot be simplified
Nice explaination (and accecnt)
Nice comment (and letters)
This is more complicated than what the math teacher taught me
wouldn't root 16 be = 4 or -4?
You do that for something like x^2 usually this is the principal square root
I love the little kirbyyyyyy
Great problem!
good trick btw what's with the pokeball???
[√(-4)^2'=√+16'=±4[YOU-CHOOSE-THE-FIRST] & (√-4')^2=(±2•i)^2=-4 & -4 AS A DOUBLE-ROOT: TWO-DI-FFERENT-WAYS.]
The way you write your 4s makes them look like the letter Psi. You should fix that.
I do not agree; a root of order y had y roots/solutions. Both of roots are of order 2, so both have two valid solutions. In both cases the solutions are +4 and -4.
I'm sorry, I know that somehow the consensus is that √-1 = i, but this is something I am simply unwilling to accept as a definition for i. If instead we used i^2=-1 and left √x as undefined for x
lol i like how hes holding a pokeball
UT doesn't root 16 equal to both positive and negative 4
So, I encounter this in a community post, and normally I prefer to work it out in the comments section, but sqrt(-4) is breaking my brain.... Although, it doesn't take much these days, though.... I'm gonna watch the video now.
why not:
(sqrt(-4))^2
==
(sqrt(-4))(sqrt(-4))
=
sqrt(16)
=
4
?
Negative numbers under sq root don't get multiplied that way.
It violates mathematical rules
I wish when I was taught index laws, the exceptions with negative numbers were layed out
I don't agree with this process. We should use algebraic/radical mathematical solving first. The approach its done in this video though its computational, from inside or outside out first. In pure maths though raising in either way the power goes to 2/2 FIRST, then resulting only -4.
He is correct both solutions
You guys don't get my point
@@forbidden-cyrillic-handle then you can assume to every exponent that has just a 1, make it to 2/2 again, then come to wrong conclusion
@@forbidden-cyrillic-handle its not about your formula, its about how you approach. You either approach technically or mathematically
@@forbidden-cyrillic-handleyour formula doesn't even apply here. (-4) can be considered in both cases in brackets and it is negative in both cases. Then exponent a half times two or two times a half.
You fell into technical and subsequential computation.
Like root mean square to get rid of negatives which is done purposely and technically
Very nice watch by the way.
U r wrong
( Sqrt (-4) )^2 = ((-4)^1/2)^2 = -4
Power of powers is distributable
You are absolutely correct - Professor of Maths here!
√ x^2 = | x | means by definition
Or, to para-phrase B. Riemman : All numbers are just a special case of imaginary "i"...
“the output of the square root is always positive”
“well we have sqrt(-1) here which is i by definition”
??
Doesn't (root(-4))² = (( -4)½)² ??
Then solving this becomes easier as 1/2 and 2 multiply to give( -4) ^ (1/2 × 2) = (-4)¹ = -4
When i get into this problem, i basically use (√a)2 = |a|
Square root look like a 2 with a - after it. You have to write it like that, then cancel the - as well
0:44 but cant sqrts be ±?
no, because principal square root will ALWAYS be positive
√x² = |x|
That's the definition of absolute value. So √(-4)²=|-4|=4
the "note" section is an actual definion of a square power under the square root, and the square power on top of the square root will always cancel out, again by definition... idk why is he complicating 4th grade math with imaginary number lmao
wrong square root of 16 is positive or negative 4
OHH nvm I just watched the video why xD my bad my bad
No one ever questioned my maths teachers