I love it when you focus on common mistakes. This is where your many years of teaching experience becomes really useful, and the most helpful to the rest of us.
I'm happy with x = 49/3. √3x = -7 (√3x)^2 = (-7)^2 3x = 49 x = 49/3 √3(49 / 3) = n √49 = +-7 n = +7 =/= -7 (false) n = -7 = -7 (true) The original equality specified -7 as the answer, square root of 49 is -7, +7 is your extraneous solution. BAM!
@@ZoidVERSE I would recommend you not listen to this guy and follow what your actual teacher wants and says. Otherwise, you may very well get that angry face on your next test. Also, extraneous solutions are actually the ones that don't work. Not the negative roots. It all mostly depends on how your teacher/professor wants the answers written down.
I'm aware of the positive root as the principal root. However, it seems to me that this is simply an agreed upon convention. That is to say, if you view the problem as a quadratic you get a +- solution, but also if you look at the square root as the question "what number times itself gives me this radicand?" you will get a +- solution. So, why is the positive root, a.k.a the principal root, considered more legitimate or real of a number than the negative root? I have yet to see a proof detailing why the answer must necessarily be positive. It's always just, "well the positive is the principal root". But why is that the case? The only answer seems to be that we take it this way as a matter of convention. Is that convention similar to the one where elementary teachers tell children that "you can't take a bigger number from a smaller one"? Or is there an actual, provable, mathematical reason that the negative root is disrespected in this way? Or just convention?
Agree. We need another video on the “why” behind “principle square roots”. I am having a very hard job understanding the rationale behind them. You worded it perfectly: is it merely convention, or is there a mathematical explanation of why the square root of 4 does not equal “-2” as well as “2” ?
I think the positive root convention began when educators decided that students would always have calculators (which return just the positive root on their limited displays) and so no longer needed to understand how arithmetic works, look at the awful confusion in the comments (and the appalling performance of the 'teacher' throwing out red herrings in all directions)
Where I learned mathematics (at a German university) a square root of a positive number greater than zero can be both a positive and a negative real number, both answers are correct. And if any square root has to be positive, why even start transforming the equation, totally unnessary if that where the case, because as the square root of no number according to him can be negative, so need to do any work on the equation.
For sqrt(x) to be a function of x, it can only have at most one value for each value of x. And it makes more intuitive sense to return the positive value than the negative.
My thought is it’s basically a learning measure as uses in real life will have context. At the learning/base level, without context, we see it as a given, as a positive, unless we’re given context toward it being something to be taken away. As we advance in learning, problems gain context, & that context will (hopefully) lead you in the right direction. When we’re at the base level, just numbers, we’re dealing with baser numbers, real (thus the throw out of i), positive, maybe even whole (significant digits may be down the road).
Well, it depends. If the task is to find a solution for x in real numbers, then there is no solution, as posed in the beginning of the video. This can also be quickly checked by the fact that the range of a root function is always equal or greater than zero, which implies it cannot be a negative number as in the initial equation. Now, if the task is to find a solution for x in complex numbers, then there are 2 solutions, according to Demoivre's theorem. In this case, the solutions expressed in exponential form are: x = (49/3) e^[i 2π (k-1)] where k=0,1
Exactly, once again make up a stupid video and not be clear on what you want. I hated teachers like that. This all just depends on your math level and whatever dump rule your teacher wants to add in. All I have to say is, no you are wrong be clear on what you want from the start.
@@jacobgoldman5780 Why don't you look-up what the word "solution" means. Then go look up what "extraneous solution" means. Extraneous solution does not mean negative roots as he leads people to believe. So yes if the answer comes out to be an imaginary number and works. That is a solution. This based on what the definitions say. Now if you do not want to confuse your kids in lower math or you do not want the people in your class to learn about these solutions. Then yes place rules that state give me only these. But give it from the start. Not, solve the problem; then your wrong because I only want this one.
..or infinite many solutions x=(49/3)i^2 = (49/3) e^[i 2π (k-1)] = ( (49/3)cos(n2π) + isin(n2π) )^n wich gives for n between 0 and infinity, infinite many solutions
It doesn't depend at all. 49/3 works fine if you plug it in to the equation. Square roots of numbers has 2 answers, the positive and negative number. For example squaring -2 is 4, so -2 is quare root of 4; as well as +2.
It is simply an *accepted convention* that the radical equation √(N^2)=|N| (meaning mod(N)) for N an integer. There is no fundamental reason why this has to be, we could easily have chosen √(N^2)= -|N| and there would be no issues if we adopted this definition. Some talk about the function f(x)=√x having a range y=f(x) ≥ 0 being the inverse to the surjective function x^2 (with domain x.∊ℝ) thus forcing us to restrict to a particular branch to maintain the inverse function property. However you argue this, it is just a handy agreed convention so that we can teach primary level maths that √4=2, √9=3 etc. We then move into uncharted territory talking about radicals used on the complex field (as hinted at by the host) as to what to do with √(-4). Is it √(-4)= +2i, or -2i, which one should I choose?
@@tomctutor I think there is a bubble of acceptance of this but that it is not universal. Are you American? I'm looking for references to books or papers that state the authority for this. A governing body perhaps. Even the op's other videos state there are two roots for a sqrt. I'm also looking for an explanation as to why negative roots are excluded. The only explanation I can think of is that it makes teaching easier?
@@ScottM7209 No im British. The solution to x^2-4=0 is x={-2,2} but √4 =2 (Wolfram alpha will return this specifically). Wo behold if you write √4 =-2, they will attack you in droves! 🙄
Good explanation of this common misunderstanding but do be careful when you extract the square root of a negative number like in the example given. The square root of minus forty nine is just seven times “i”and not plus or less seven times “i”. Plus or less seven times “i” is the solution of an equation like x^2 = - 49, thanks!
@@lucifer9273 …49.. but the square root of 3 times 49/3, which is simply 49, does not and cannot equal a negative number, -7 in this case, in the set of real numbers! The square root of 49 equals 7…, and 7 ≠ -7! ..think of the number line, perhaps..! In other words, x = 49/3 is not a solution for x which would make the equation true in the set of real numbers. I’ve read other opinions and arguments on here that are worth considering, and worth me reconsidering my position on the matter. So I’ll suspend my above opinion and position for the moment…
First I thought it would be 49/3 but I had difficulty coming with -7 when x was substituted by that value. It looks like no solution but I was wrong before.
After a walk in the park thinking about this problem I figured out there could not be a solution in R. My HP 50g Calculator even refused to evaluate this equation. No error, no answer in i, just nothing. Like your videos that challenge intuitive thinking.
Please allow him to explain to those of us who are not as quick as you are. Patience is a trait needed in mankind . Not everyone learn at the same pace.
Sure it can take 5 min because you understand the concept. He takes the time to explain clearly to those who don't get it. Pat yourself on the back that you get it. Lucky you. Don't begrudge the tutorial to the rest
Starting to get hooked on your videos and almost 50 😂 Anyways, quick question about answer. When you square both sides, couldn’t the right side be written as -1(49), then working that forward I get an answer of x = -16 1/3 or -49/3. Assuming I messed something up. Is it wrong to break it out that way?
@@brucewellman The imaginary unit and complex numbers are not calculus, they are very much part of algebra. They are an extension of the "real" numbers, enabling solutions to polynomials where otherwise no solutions would exist. That is not to say that they don't crop up all the time in calculus, but equally there is a lot of calculus that doesn't feature complex numbers at all. Different topics, but with an overlap in places. You could perfectly well learn about complex numbers without ever touching differentiation and integration, although they are topics that are broadly at similar level and likely to be looked at in proximity to each other on a curriculum, typically (in the UK) at 16-18 which we call "A-Level".
@@eudyptes5046 Then this should be made clear. The math teacher should say “principal square roots” are a convention. There is no mathematical reason for them.
Follow what your Teacher or Professor want. Otherwise you will be getting some angry faces on your tests. Yes the PSR is only a convention. Also, "Extraneous Solutions" refer to the solutions which do not work. Do not follow just one math Teacher, but follow what your math Teacher asks.
This video has made the unwarranted assumption that the square root sign √ refers only to the "principal" square root.A positive number real has two square roots. One of the square roots of 3x (where x = 49/3) is 7 and _the other is -7._ So it's simply _not true_ that the given equation has no solutions.
Solution Make both sides multiply itself once (x²) (√3x)² = -7² (note negative number becomes positive when it multiplying itself even times) >>> 3x = 49 Now divide by 3 on both sides 3x/3 = 49/3 Hence, x = 49/3
@@Nerthus2010 There is an essential difference between - 7^ (^= 2) and (- 7)^ . In the first example the result = - 49 and in the second it is + 49. I agree that John could have explained the question more clearly.
@@Kleermaker1000 What should (-7)^ even mean? (-7)*(-7) equals 49 and 7*7 equals 49. And because of that the square root of 49 is both 7 and -7. The square root of any number is defined to be any number which multiplied with itself gives that number. At least that is the mathematical meaning of the word square root. if you are looking for a number, who multiplied with itself gives a negative amount as result you have to use imaginary numbers, which are seldom used in schools.
So what would be SQRT(i^2). +1 or -1? If it is -1, then would not x=(49/3)*i^2 be a solution? If you rewrote the question as (3x)^(1/2) = -7, would that change the answer?
That would be i. i^2 is -1, SQRT(-1) is i. Also, x^(1/2) is just a different way of writing a square root, so it would change absolutely nothing except the notation. Also, SQRT(i^2) = i^2^(1/2) = i^(2/2) = i.
I immediately tried to throw in i, x=(49*i^4)/3 But been out of math for a while so could be confusing something or missed the clarification of looking for only a real number; thus, the null or N/A
That would work. Basically a proof that a square root has two valid answers. But honestly 49/3 would work too unless you constrain the problem properly.
You nailed me on this one. I was thinking quadratic eq with two solutions. Thanks for reminding me about the imaginary i sq root of -1. Good review. Thanks for the explanation.
√(- 1)² = i and √x < 0 are two completely different things. The latter is undefined. You can take the square root of a negative number (which is a complex number), but the square root of a number can't be negative.
You can't even take the square root of a negative number - you can only solve the "corresponding" polynomal equation. Squareroot(-1) does not exist, but x^2=-1 has i as a solution.
@@carl2872 But it is commonly recognized that ±√(−1) = ± i. The problem is that x² = −1 has two solutions, i and −i. And i ⋅ (−i) = 1, just as x ⋅ (−x) = −x²
@@Nikioko yes, i is A solution, not the solution. -i is another one - that can be concluded from "i is a solution" and the axioms of a field (C is a field)
Greetings. On the surface it would appear that the answer is + or - 49/3. However, when these values are factored into the equation neither one will result into -7.
-7 times -7 would always be positive 49. So it cannot be + OR - 49. Just + 49. A quadratic squared root always has two outcomes. Both the negative number and the positive number will work. The trick here, is that it's not a quadratic squared root and so you Must come up with the positive only. But the answer is - 7. Which in itself (without doing the actual math) can answer the question for you, namely 'false'. ONLY quadratic squared roots have a negative outcome as well.
@@johnmaguire2185 He explained that, when you see a principal square root, that the answer is always just the positive one. And that when it's quadratic you get both positive and negative. So now your telling me, that I used the wrong vocabulary. But at least I did understand the principle. But this is really taking the joy right out of it. Thanks for ruining it for me.
@@gardenjoy5223 why is it removing the joy? No point in using incorrect vocabulary especially in maths. At the “root” of this video is that 7 squared and - 7 squared both = 49 7i squared and -7i squared both = -49 The square root of 49 is 7. The positive root only, this is by convention not by mathematics. The square root of -49 is not 7i the positive complex root by mathematics as root(-49) = root(49) x root(-1) = 7 x root(-1) , but root(-1) is not equal to i. Root (-1) has its own complex roots.
The video is too long-winded. The usual convention is that sqrt(a) for positive a is evaluated to be a positive number. But you can choose to accept or reject this convention any time you wish. So when you ask someone to solve the equation sqrt(3x)=-7, it appears to you reject the convention. When someone gives you the solution 49/3, it is grossly unfair for you to come back and say, "you are wrong, because I chose to accept the convention, so there are no solutions." It is a stupid trick you are doing, and not teaching anybody anything.
@@attica7980 Ehh ... I know he's long winded -- it's because he is used to teaching teenagers who need a lot of words to stick in their heads. I'm a teacher too so I know how it is. But he is making an absolutely vital point: Taking the square root of a number, it is always the principle square root that is expected one is working with equations that exceed (up or down) x¹ -- or in other exponential equations. Unless you are working with exponents (or logarithms, a different interpretation of exponents) one should always discard anything other than the principle square root.
@@danieldennis9831 If you are a teacher, then you absolutely know that correctly "principal square root" and not "principle square root." Just saying. The difference between principle and principal is very easy to learn.
@@attica7980 I teach recording, audio and music production, not English. Though I do have a very large vocabulary, tend to be verbose and am often pedantic about word use, I concentrate more on the nomenclature of my subject rather than common English words, despite extensive study of such to absurd levels. I occasionally will make mistakes such as mixing principle and principal or capital and capitol as my typing tends to be on automatic when not paying attention to that aspect of word use. Oooops. . I was just commenting from the perspective of a person who has been an educator for over 35 years. (I almost confused myself when I was reading my answer before hitting reply. For a second I thought "Should that have been prospective?" and then 2 seconds later decided I had used the right word. Eh. Just trying to be funny despite the serious nature of the subject at hand.)
The curve y=sqrt(3x) is a sideways parabola and not a function. The plot of the principal value of sqrt(3x) is a function above the x-axis that is half a parabola.
@@samwong3123 So the square root is not a function. A function needs a one to one mapping of inputs to values. So we split it into two and call one the "principal square root" which are all the values of the square root that are above the x-axis. And we use this for a lot of math. But if you plot the actual square root you will get the same plot as if you plotted x^2 and turned it 90 degrees clockwise. Square root as a function becomes important in calculus so that's what we use most often and most calculators and computers assume. In fact, we assume the principal square root most of the time, which is the point here. I honestly understand what he's getting at, but I'm an engineer with 15+ years of math under my belt, and I was a little confused as to his reasoning, so I think it could be explained a bit more clearly. I tend to not like "because we say so" answers in math because sometimes that leads us down bad paths. For instance, antimatter's existence was theorized by Dirac by using the negative answer of the square root.
@@MarkEmerAndersonII As you include your background, I will include mine. I have been teaching collegiate math since the first Bush Administration. I have earned a doctorate in math in the field of algebraic number theory. I don't know what you are saying when you write that the square root is not a function. It is a well defined mapping from [0,∞) to [0,∞) providing a single output for every valid input. Note: functions are not required to be a one-to-one mapping (or surjection). The function f(x)=√x is what we view as the principal square root or the positive square root. We could easily have constructed the square root function to be based on negative outputs, namely g(x) as a well defined mapping from [0,∞) to (-∞,0] which would correspond to the bottom half of the parabola with vertex at the origin opening to the right. There's nothing stopping us from doing that. The main reason why we chose the positive square root to be the default definition is more about applications. Most applications using square roots (like computing distance) needs the output to be positive. It does not make practical sense to have to compensate for that negative in the majority of situations. This is akin to why the range of arccos(x) is [0,π] and not [87π,88π]. There's nothing to stop us from using [87π,88π]. Mathematics could have been restructured with [87π,88π] as the default range, but then it would have been more head-scratching, complicated, and just plain stupid. We use [0,π] for practicality, simplicity, and convenience. The thing is that once the selection is made as to how to define an inverse like the square root, consistency must be adhered to. This is where the "because we say so" doctrine comes into play. We need to have a standard to enable a continuous understanding across mathematics in order. This doesn't mean that the other choices are invalid. I don't know anything about Dirac's theory of antimatter. But I assume that in his theory he states that the negative inverse is being used. He's communicating that the non-standard yet still valid square root is the appropriate and useful choice. Those that are learning his theory understand this context. So what's going on? Why does it seem two answers to x^2=49 but √49 is only one number? The simple answer is that there are two possible inverses to x^2, the positive (or principal) square root g(x)=√x and the negative square root h(x)=-√x. So when determining the solution to x^2=49, we need to consider both. Now when you write √49, you have just chosen which inverse you are using. I tell my students that when the √ is written the choice has already been made, such as √16=4 or -√36=-6. But, if you introduce the √ into the problem, the choice of the square root has not been made; both possibilities are still valid and need to be considered, such as x^2=25 yields x = ±√25 = ±5. I'm sorry for the very long winded explanation here.
It's not only for quadratic equations: +/- results, but also for all of the even degree polynomial equations such as quartic, sextic, octic and so on etc... These are all even root polynomial equations...
You are committing a error of logic when you use the term "the square root of". It is an error of logic to use the definite article when the term does not refer to a unique entity. See Kalish & Montague in their book Logic for a discussion of your error.
The committed error is writing an equation that intentionally can't be solved by left and right side equation manipulating. 🙄. If √(3x) = -7 is written: √(3x)√(3x) = 3x = -7√(3x) holds ... -(1/7)(3x) = √(3x) holds ... But that equalled -7 at the start. So -(1/7)(3x) = -7 and continuing to x = 49/3 manipulating left side and right sides without doing division by zero elimination cheating watching what I was doing every time. It is the first time hearing we can't manipulate to solve equations or tidy up an equation mathematics in my life teachings. 😫
At the University of Wisconsin, where I attended for my BSEE, no matter what, A SQUARE ROOT CANNOT EVER BE A NEGATIVE NUMBER in the real number system. Of course, it can be a multiple of SQRT(-1) or "j" (engineering use), engineering does exist in the space of complex numbers but going back to basics, in real terms, a square root cannot be a negative number simply because any number squared, including a squared square-root, must be positive. HOWV=EVER now going into the complex number system; SQRT(3x) = j*(7), 3x (-1)*(49)/3 x= -49/3 Note that there is no solution of +49/3 because the complex "j" squared is a factor in the expression.
A video on waffling and time-wasting. now it's clear why some don't like math. if the principal solution is always positive, but the answer is negative only a fool would choose the principal solution I used to be a PhD Cambridge math postgraduate but I feel dizzy after this exposition on such a trivial matter. Americans certainly know how to obscure clarity
This is where I don't agree: √16 = 4. (only the principle root) But ±√16 = ±4. So I'm saying, you always need to put the ± in front of the root as well. And that happens in a quadratic equation: If x² = 16 Then x = ±√16 = ±4. That's why the quadratic formula says x = (-b±√(b²-4ac))/2a. The ± in front of the root tells us we need both the positive and the negative roots, not just the principle root.
This is what I keep telling people. Answer, how your Teacher and Professor want it answered. Otherwise you may get some of them angry faces on your test's. Also, "Extraneous Solutions" refer to the solutions which "DO NOT" work. Not just non "Principal Square Roots (PSR)." The practice of only using "PSR" is a convention, because it makes teaching easier.
@@danv2888It doesn't just make teaching easier. It's much more fundamental than that. The √ sign is defined to mean principal square root because that makes it easy to use the √ symbol clearly and precisely. Clarity and precision are important in mathematical notation. The principal square root of x is √x. The other square root of x is -√x. And if you want both square roots of x (like, for example, the formula for solving a quadratic equation) you write ±√x.
@@gavindeane3670 That is not true. I do not know when you went to school. However, when I was going the same symbol is used for both, principal and non-principal. Also, in the directions is where it stated whether both or just one was needed. Also, it was more likely to be accepted that both where wanted, and where only principal was wanted that would be stated. Overall, I am just telling people to listen to what there Teacher or Professor are asking and saying. Not to what this guy is saying is the correct way. Because I have had things not stated and changed on me and then they claim your wrong. Then you get that angry/sad Face he is talking about.
@@danv2888 It is true. You can look up the definition of the √ symbol in the international standard for mathematical notation if you want to. The principal square root of x is √x. The other square root of x is -√x. And if you need both square roots (like, for example, in the formula for solving a quadratic equation) you write ±√x. There is no sign indicator in front of the radical symbol in this question so therefore the question explicitly requires the principal square root and only the principal square root. You may not have been taught this. It does seem to be a point that isn't always taught very well - as is evident whenever this topic comes up. But that doesn't change the fact that it's true. Undoubtedly, the symbol has been defined this way so that it can be used clearly and precisely. Clarity and precision are important in mathematical notation. Yes, if someone's goal is to pass a test then they should answer questions in the way they were told to answer by the teacher whose job it was to prepare them for that test. But if someone is being taught incorrectly then explaining to them how it really works is a good thing. They might need the incorrect understanding on test day, but it's the correct understanding that matters on every other day.
I have solved the underlying problem in general - it's very tricky and this video can't even scratch the surface (question underlying: why and when do extra solution show up?) Solution in general: check all the solutions you get in the original equation (and cross your fingers or better check that every step you made was allowed for all x you consider)
So basically the square root of 3X does not equal -7. Would it actually be an equation if the two sides are not equal. for example I would say that 2+2=7 is not an equation.
It's really intuitive actually in its penultimate form. You would never assume the sq rt of 49 to be -7. If you did that while within a larger equation you will always get the wrong answer.
I ALWAYS test my answers in the original equations as I certainly am fallable in my math manipulations. So I would have caught the impossibility of a real number solution to this equation.
✓(x^2) = |x| = the absolute value of x The equation ✓(3x) = -7 has no solution On the other hand, there is one solution to the equation (-7)^2 = 3x x = 49/3
I do not get it. When you resolve this equation you upgrade to square the left side of the equation and the right side. So -7 becomes +49 and the solution is x=49/3 But is true the checking is not working so what is wrong? What is the correct answer?
Only a quadratic squared root has two outcomes, being both the negative and the positive number. He explained, that there exists a RULE, that if a squared root is NOT quadratic, you may not alternate it into a quadratic one to get both answers. When you have a simple squared root, the answer MUST be a positive number. Since the number here is a negative, namely - 7, that alone tells you, that the answer cannot be given. For with the simple squared root the answer ALWAYS has to be positive. You don't even have to complete the math here to know that. (But I bet math teachers still prefer you to write everything out anyway...)
Silvia: the checking IS working though. If you plug in “49/3” as the value of “X”, you get 2 possible answers, one of which is “-7”. It is true that “-7” is not the only answer, but it is not wrong.
i guess my not so much mathematical schooled brain helped me because i immediately thought that the squareroot of something cannot be a negative number as a negative number multiplied by itself becomes positive. and of course a positive number stays positive. So i got to the right solution by only considering the mathematical basics i learned in primary school.
There's no convention barring imaginary and complex numbers. But that doesn't help - there is no solution, real or complex, to this equation. (49/3)i⁴ is not an imaginary number. It's a real number. It's just a fancy way of writing 49/3. If x = (49/3)i⁴ the 3x = 49i⁴ = 49. And √49 does not equal -7. It equals 7.
We see the problem so is there an answer? After watching this I was still waiting for an answer or is it impossible to square root a number to make a negative answer?
If someone gave me that on a test (I’m a teacher so that would probably be another teacher, or the people who put together standardised tests) I’d school them on how to use notation. The square-root symbol written like that is by definition the positive root, and the whole equation is just nonsense; it’s not merely a thing where “no solutions” is an appropriate response, but “that’s not a question” works too. 👍🏻😂
sq rt (3x) = 7*(i^2); where i^2 = -1 (sq rt(3x))^2 = 49*(i^2)*(i^2) 3x = 49*(-1)*(-1) 3x = 49 So, x = 49/3 or 16(1/3) Can't this be the solution for x? If not, why? Please explain. Thank you.
x = [-7(sq rt of 3x)] / 3 I like it. It's crazy wrong, but fun. So... (3x)^2 = 9x^2 [ sq rt (3x) ] X [ sq rt (3x) ] = -7^2 OR [ sq rt (3x) ] X [ sq rt (3x) ] = -7[ sq rt (3x) ] .... hmmmm.... 9x^2 = 49 x^2 = 49/9 some crazy number... or maybe... x might be negative... oh sheeeeeit.... LOL LMAO!!! These are THE BEST equations!!!!!
So -7 is not a valid root of 49? Who re-wrote maths in the 50-odd years since I took my A-levels? We'd have lost marks, or even been given zero, if we only gave the positive root as the answer!
Is the full answer to the problem X= (49/3) which is not true? What is the full answer ie. how would the answer be fully verbalized? I can follow the math, but it seems at the end when you plug in the answer it is not correct.
I understand why -49/3 wouldn’t work, because would lead to an imaginary answer of + or - 7i. But I don’t understand why +49/3 would not work. That would yield the answers +7 and -7. Granted, when you think of square root of 49, you usually think of the positive square root of 7, but I don’t see how -7 would be incorrect because (-7) squared yields 49.
You can, of course, have a square root of -49 which is 7i where (i) is the square root of -1, an imaginary number. This, however, is not the point of this video.
No (49/3)i doesn't work. That would work if √i equalled -1. But that's backwards. It's not √i = -1, it's √(-1) = i. So bringing i into this doesn't help. There's still no solution to this equation.
1. The variable x domain has not been established. 2. The squaring operation on equations introducing unexpecting values to the variables and should be avoided. 3. No square root can be taken of a negative number within the system of real numbers, because squares of all real numbers are non-negative.
Ya got me pretty good there guy, or did I just get myself, hard to say. Anyway I was blasting verbosity at my monitor as I was watching this video. "Of course there is a solution you dolt and it's not the nul set, How styuupid can you be?" So to prove how smart I am and how dumb you are I wrote the equation down and solved it. Then solved it. Then solved it. Then solved it. Then solved it. Then solved it. Then solved it. Tail appropriately tucked, thank you as much for the leason in humility than the math leason. Great video.
If I can be honest about this, the explanation confused me. This is definitely "NEW" Math for me. When I went to High School, way back in the dark ages, it was always the case that a positive number could have both positive and negative square roots; a negative number could NOT have a square root! Seemed unfair for the negative number, but there you go! Nobody in those days said that Math had to be fair! I really do not understand this unique explanation. I do like the term "unsolvable" which we were encouraged to use back in the "dark ages" as an answer to a negative square root problem. Though I suspect true mathematicians dislike the word unsolvable!
You are good when you do not accept his explanation, because it is just wrong. The square root of 49 has two solutions + 7 and - 7. And has for a very long time now, in like for centuries.
"Unsolvable" was expressed as "doesn't belong to the field of x" where x is integer, real, rational, etc. When I was in high school. I think there's no "unsolvable" in mathematics, just undetermined like when there's a division by 0. Of course, the calculator can't show you "undetermined" in a 8 digits display, so they just went with "error" for anything the limited programming (and possibly understanding of the programmer) couldn't solve.
So, it takes over 14 minutes to say that the square root sign means "the positive square root of." The example shown says "the positive square root of (3x) = -7" which clearly has no solution.
The square root sign always means the principal square root. . .is not a universally-agreed rule, it is a convention used by some for what purpose. For example, if I have the quadratic equation x^2=64, and the directions are "solve by taking the square root of both sides" (which in that particular problem is possible method to solve), if I do that, are you saying to solve that, you would write "x=8, x=-8, can't have a negative square root, so x=8?" My math teachers (and my brother with his undergrad degree in mathematics) would fail that answer, saying you did not understand the concept of a quadratic equation ALWAYS has two solutions. So the principal square root rule does not work everywhere in mathematics. It also does not work is if you have something like "provide the zeroes of x^2=√16" So for that because x is squared, anything squared will be greater than or equal to zero (negative times negative always makes positive) So the first step we simplify the radical, the square root of 16 is 4, -4 but only 4 will work in the equation NOT due to any principal square root convention, but SIMPLY because the SQUARE of a number can never be negative. Take the SQUARE ROOT of both sides x=2 or x=-2 Now. . .one last thing. A square rootcof a number x is any number that when multiplied by itself equals x. For example, (-2)*(-2)=4 and 2*2=4 Conventions are one thing. Proper math is another.
LHS of the equation is nothing more than exp((1/2)log(3x)) = exp((1/2)Log(3x) + kπi), k∈ℤ; RHS is exp(log(-7)) = exp(Log(-7) + 2lπi) = exp(ln(7) + Arg(-7)i + 2lπi) = exp(ln(7) + (2l+1)πi), l∈ℤ and you may proceed from here...
No, that's not what he's saying. You can have a square root of a negative number, but that's irrelevant here. What he's saying is that the principle square root of any number can't be a negative number.
Note that there are no complex solutions either. The same logic applies. Any complex number, z, whose square root was -1 would satisfy the equation z = (-1)^2 = 1. That's the only possibility and it's not a solution.
May I suggest it's a convoluted description? More simply: Square roots are all positive if not zero. Negatives of square roots are all negative if not zero. √16 =4 is true. -√16=-4 is true ±√16=±4 is true √16=±4 is false (Hence, the quadratic formula, for example, necessitates the ± symbol, not merely the +, to produce 2 solutions. If anyone thinks square roots themselves can be negative, remove the ± from all your quadratic formula statements please. And draw y=√x and y=~√x as the same thing as y=±√x). The initial equation here in this video has no solution simply because it states that a positive number (left) equals negative seven (right).
I love it when you focus on common mistakes. This is where your many years of teaching experience becomes really useful, and the most helpful to the rest of us.
I'm happy with x = 49/3.
√3x = -7
(√3x)^2 = (-7)^2
3x = 49
x = 49/3
√3(49 / 3) = n
√49 = +-7
n = +7 =/= -7 (false)
n = -7 = -7 (true)
The original equality specified -7 as the answer, square root of 49 is -7, +7 is your extraneous solution. BAM!
I bet you divide by 0 a lot.
@@thorinpalladino2826 -- Not especially frequently.
there is no solution to this question
@@ZoidVERSE
I would recommend you not listen to this guy and follow what your actual teacher wants and says. Otherwise, you may very well get that angry face on your next test. Also, extraneous solutions are actually the ones that don't work. Not the negative roots. It all mostly depends on how your teacher/professor wants the answers written down.
@@danv2888 it's actually generally accepted that there is no solution to this type of question. and thank you for your reply 😄.
I'm aware of the positive root as the principal root. However, it seems to me that this is simply an agreed upon convention. That is to say, if you view the problem as a quadratic you get a +- solution, but also if you look at the square root as the question "what number times itself gives me this radicand?" you will get a +- solution. So, why is the positive root, a.k.a the principal root, considered more legitimate or real of a number than the negative root? I have yet to see a proof detailing why the answer must necessarily be positive. It's always just, "well the positive is the principal root". But why is that the case? The only answer seems to be that we take it this way as a matter of convention. Is that convention similar to the one where elementary teachers tell children that "you can't take a bigger number from a smaller one"? Or is there an actual, provable, mathematical reason that the negative root is disrespected in this way? Or just convention?
Agree. We need another video on the “why” behind “principle square roots”. I am having a very hard job understanding the rationale behind them. You worded it perfectly: is it merely convention, or is there a mathematical explanation of why the square root of 4 does not equal “-2” as well as “2” ?
I think the positive root convention began when educators decided that students would always have calculators (which return just the positive root on their limited displays) and so no longer needed to understand how arithmetic works, look at the awful confusion in the comments (and the appalling performance of the 'teacher' throwing out red herrings in all directions)
Where I learned mathematics (at a German university) a square root of a positive number greater than zero can be both a positive and a negative real number, both answers are correct.
And if any square root has to be positive, why even start transforming the equation, totally unnessary if that where the case, because as the square root of no number according to him can be negative, so need to do any work on the equation.
For sqrt(x) to be a function of x, it can only have at most one value for each value of x. And it makes more intuitive sense to return the positive value than the negative.
My thought is it’s basically a learning measure as uses in real life will have context. At the learning/base level, without context, we see it as a given, as a positive, unless we’re given context toward it being something to be taken away. As we advance in learning, problems gain context, & that context will (hopefully) lead you in the right direction. When we’re at the base level, just numbers, we’re dealing with baser numbers, real (thus the throw out of i), positive, maybe even whole (significant digits may be down the road).
Well, it depends.
If the task is to find a solution for x in real numbers, then there is no solution, as posed in the beginning of the video. This can also be quickly checked by the fact that the range of a root function is always equal or greater than zero, which implies it cannot be a negative number as in the initial equation.
Now, if the task is to find a solution for x in complex numbers, then there are 2 solutions, according to Demoivre's theorem.
In this case, the solutions expressed in exponential form are:
x = (49/3) e^[i 2π (k-1)]
where k=0,1
Exactly, once again make up a stupid video and not be clear on what you want. I hated teachers like that. This all just depends on your math level and whatever dump rule your teacher wants to add in. All I have to say is, no you are wrong be clear on what you want from the start.
Either way there is no solution as I=sqrt(-1) not -1=sqrt(i).
@@jacobgoldman5780
Why don't you look-up what the word "solution" means. Then go look up what "extraneous solution" means. Extraneous solution does not mean negative roots as he leads people to believe. So yes if the answer comes out to be an imaginary number and works. That is a solution. This based on what the definitions say. Now if you do not want to confuse your kids in lower math or you do not want the people in your class to learn about these solutions. Then yes place rules that state give me only these. But give it from the start. Not, solve the problem; then your wrong because I only want this one.
..or infinite many solutions
x=(49/3)i^2 = (49/3) e^[i 2π (k-1)] = ( (49/3)cos(n2π) + isin(n2π) )^n
wich gives for n between 0 and infinity, infinite many solutions
It doesn't depend at all. 49/3 works fine if you plug it in to the equation. Square roots of numbers has 2 answers, the positive and negative number. For example squaring -2 is 4, so -2 is quare root of 4; as well as +2.
It is simply an *accepted convention* that the radical equation √(N^2)=|N| (meaning mod(N)) for N an integer.
There is no fundamental reason why this has to be, we could easily have chosen √(N^2)= -|N| and there would be no issues if we adopted this definition.
Some talk about the function f(x)=√x having a range y=f(x) ≥ 0 being the inverse to the surjective function x^2 (with domain x.∊ℝ)
thus forcing us to restrict to a particular branch to maintain the inverse function property.
However you argue this, it is just a handy agreed convention so that we can teach primary level maths that √4=2, √9=3 etc.
We then move into uncharted territory talking about radicals used on the complex field (as hinted at by the host) as to what to do with √(-4).
Is it √(-4)= +2i, or -2i, which one should I choose?
Accepted by whom?
@@ScottM7209 The radical square root, we are told, is to be taken to mean positive root they keep telling me! Accepted by professors etc.
@@tomctutor I think there is a bubble of acceptance of this but that it is not universal. Are you American? I'm looking for references to books or papers that state the authority for this. A governing body perhaps. Even the op's other videos state there are two roots for a sqrt. I'm also looking for an explanation as to why negative roots are excluded. The only explanation I can think of is that it makes teaching easier?
@@ScottM7209 No im British. The solution to x^2-4=0 is x={-2,2}
but √4 =2 (Wolfram alpha will return this specifically).
Wo behold if you write √4 =-2, they will attack you in droves! 🙄
The answer is 2i. It is not -2i.
Indeed that reinforces... The need to check your work by plugging it in....always!
very important lesson ! Thank you
Problem is that it is incorrect.
Good explanation of this common misunderstanding but do be careful when you extract the square root of a negative number like in the example given. The square root of minus forty nine is just seven times “i”and not plus or less seven times “i”. Plus or less seven times “i” is the solution of an equation like x^2 = - 49, thanks!
What is the product of (-7)(-7)=?
I means Solution 😢
@@lucifer9273
…49.. but the square root of 3 times 49/3, which is simply 49, does not and cannot equal a negative number, -7 in this case, in the set of real numbers! The square root of 49 equals 7…, and 7 ≠ -7! ..think of the number line, perhaps..!
In other words, x = 49/3 is not a solution for x which would make the equation true in the set of real numbers.
I’ve read other opinions and arguments on here that are worth considering, and worth me reconsidering my position on the matter. So I’ll suspend my above opinion and position for the moment…
This video did help very much. thank you I got the answer I just did not know to declare it null.
First I thought it would be 49/3 but I had difficulty coming with -7 when x was substituted by that value. It looks like no solution but I was wrong before.
so? no solution?
@@timhochstetlerno solution
After a walk in the park thinking about this problem I figured out there could not be a solution in R. My HP 50g Calculator even refused to evaluate this equation. No error, no answer in i, just nothing. Like your videos that challenge intuitive thinking.
Please allow him to explain to those of us who are not as quick as you are. Patience is a trait needed in mankind . Not everyone learn at the same pace.
This guy takes forever and ever and forever and repeats himself 3 times...this video is 27 times (3 to the power of 3) LONGER THAN IT NEEDS TO BE
imagine being a student in his class...
Sure it can take 5 min because you understand the concept. He takes the time to explain clearly to those who don't get it. Pat yourself on the back that you get it. Lucky you. Don't begrudge the tutorial to the rest
So the whole problem pivots around the concept of "principal sq. roots" as you put it.
& quads
Starting to get hooked on your videos and almost 50 😂
Anyways, quick question about answer.
When you square both sides, couldn’t the right side be written as -1(49), then working that forward I get an answer of x = -16 1/3 or -49/3.
Assuming I messed something up. Is it wrong to break it out that way?
You can't . Squaring (-7) means multiplying -7 * -7 which doesn't equal what you stated "-1 (-49)"
@@CCCompiler Thank you. I think you got the best explanation that makes sense.
You are close but until you know the imaginary number i but that is calculus
@@brucewellman The imaginary unit and complex numbers are not calculus, they are very much part of algebra. They are an extension of the "real" numbers, enabling solutions to polynomials where otherwise no solutions would exist.
That is not to say that they don't crop up all the time in calculus, but equally there is a lot of calculus that doesn't feature complex numbers at all. Different topics, but with an overlap in places.
You could perfectly well learn about complex numbers without ever touching differentiation and integration, although they are topics that are broadly at similar level and likely to be looked at in proximity to each other on a curriculum, typically (in the UK) at 16-18 which we call "A-Level".
So what's the correct solution to the equation?
When you do not give reasons for your rules, you lose me.
There are no reasons, these are definitions.
@@eudyptes5046 Then this should be made clear. The math teacher should say “principal square roots” are a convention. There is no mathematical reason for them.
Follow what your Teacher or Professor want. Otherwise you will be getting some angry faces on your tests. Yes the PSR is only a convention. Also, "Extraneous Solutions" refer to the solutions which do not work. Do not follow just one math Teacher, but follow what your math Teacher asks.
This video has made the unwarranted assumption that the square root sign √ refers only to the "principal" square root.A positive number real has two square roots. One of the square roots of 3x (where x = 49/3) is 7 and _the other is -7._ So it's simply _not true_ that the given equation has no solutions.
Unless stated, the PSR is the only correct answer.
Thanks Professor!
So, the question is no good from the outset?
Solution
Make both sides multiply itself once (x²)
(√3x)² = -7² (note negative number becomes positive when it multiplying itself even times) >>> 3x = 49
Now divide by 3 on both sides
3x/3 = 49/3
Hence, x = 49/3
No.
@@jakemccoy The answer is correct, because a square root of any positive real number can be a positive and a negative real number.
@@Nerthus2010 There is an essential difference between - 7^ (^= 2) and (- 7)^ . In the first example the result = - 49 and in the second it is + 49. I agree that John could have explained the question more clearly.
@@Kleermaker1000 What should (-7)^ even mean?
(-7)*(-7) equals 49 and 7*7 equals 49.
And because of that the square root of 49 is both 7 and -7.
The square root of any number is defined to be any number which multiplied with itself gives that number. At least that is the mathematical meaning of the word square root.
if you are looking for a number, who multiplied with itself gives a negative amount as result you have to use imaginary numbers, which are seldom used in schools.
@@nerthusmeier7038 (-7)^ means - 7 x - 7.
So what would be SQRT(i^2). +1 or -1? If it is -1, then would not x=(49/3)*i^2 be a solution? If you rewrote the question as (3x)^(1/2) = -7, would that change the answer?
That would be i. i^2 is -1, SQRT(-1) is i.
Also, x^(1/2) is just a different way of writing a square root, so it would change absolutely nothing except the notation. Also, SQRT(i^2) = i^2^(1/2) = i^(2/2) = i.
I immediately tried to throw in i, x=(49*i^4)/3
But been out of math for a while so could be confusing something or missed the clarification of looking for only a real number; thus, the null or N/A
That would work. Basically a proof that a square root has two valid answers. But honestly 49/3 would work too unless you constrain the problem properly.
i^4=1…
You nailed me on this one. I was thinking quadratic eq with two solutions. Thanks for reminding me about the imaginary i sq root of -1. Good review. Thanks for the explanation.
√(- 1)² = i and √x < 0 are two completely different things. The latter is undefined.
You can take the square root of a negative number (which is a complex number), but the square root of a number can't be negative.
You can't even take the square root of a negative number - you can only solve the "corresponding" polynomal equation. Squareroot(-1) does not exist, but x^2=-1 has i as a solution.
@@carl2872 But it is commonly recognized that ±√(−1) = ± i. The problem is that x² = −1 has two solutions, i and −i. And i ⋅ (−i) = 1, just as x ⋅ (−x) = −x²
@@Nikioko yes, i is A solution, not the solution. -i is another one - that can be concluded from "i is a solution" and the axioms of a field (C is a field)
I think it depends on the application.... In some aplicaitons you can have an i part of the solution and others you cant.
What about complex solutions?
If x is complex, then sqrt(x) is complex. sqrt(x) is never a negative real number.
Greetings. On the surface it would appear that the answer is + or - 49/3. However, when these values are factored into the equation neither one will result into -7.
-7 times -7 would always be positive 49. So it cannot be + OR - 49. Just + 49.
A quadratic squared root always has two outcomes. Both the negative number and the positive number will work. The trick here, is that it's not a quadratic squared root and so you Must come up with the positive only. But the answer is - 7. Which in itself (without doing the actual math) can answer the question for you, namely 'false'. ONLY quadratic squared roots have a negative outcome as well.
@@gardenjoy5223there is no such thing as a quadratic square root. A quadratic is an equation where the highest power of any variable is two.
There are roots of quadratic equations. Which will be both positive both be zero or both be complex.
@@johnmaguire2185 He explained that, when you see a principal square root, that the answer is always just the positive one. And that when it's quadratic you get both positive and negative. So now your telling me, that I used the wrong vocabulary. But at least I did understand the principle.
But this is really taking the joy right out of it. Thanks for ruining it for me.
@@gardenjoy5223 why is it removing the joy? No point in using incorrect vocabulary especially in maths.
At the “root” of this video is that
7 squared and - 7 squared both = 49
7i squared and -7i squared both = -49
The square root of 49 is 7. The positive root only, this is by convention not by mathematics.
The square root of -49 is not 7i the positive complex root by mathematics as
root(-49) = root(49) x root(-1) = 7 x root(-1) , but root(-1) is not equal to i. Root (-1) has its own complex roots.
My ancient Ti-81 ( almost 30 years old). I hope replacing coin battery will bring it back to life.
Only quadratic equations can have both positive and negative roots. You learn something new everyday
If scratching my head and giving up is the same as concluding a null answer (null=indeterminable), then I get an A+. 🤔
This video was very valuable for my understanding. Thank you.
The video is too long-winded. The usual convention is that sqrt(a) for positive a is evaluated to be a positive number. But you can choose to accept or reject this convention any time you wish. So when you ask someone to solve the equation sqrt(3x)=-7, it appears to you reject the convention. When someone gives you the solution 49/3, it is grossly unfair for you to come back and say, "you are wrong, because I chose to accept the convention, so there are no solutions." It is a stupid trick you are doing, and not teaching anybody anything.
@@attica7980 Ehh ... I know he's long winded -- it's because he is used to teaching teenagers who need a lot of words to stick in their heads. I'm a teacher too so I know how it is. But he is making an absolutely vital point: Taking the square root of a number, it is always the principle square root that is expected one is working with equations that exceed (up or down) x¹ -- or in other exponential equations. Unless you are working with exponents (or logarithms, a different interpretation of exponents) one should always discard anything other than the principle square root.
@@danieldennis9831 If you are a teacher, then you absolutely know that correctly "principal square root" and not "principle square root." Just saying. The difference between principle and principal is very easy to learn.
@@attica7980 I teach recording, audio and music production, not English. Though I do have a very large vocabulary, tend to be verbose and am often pedantic about word use, I concentrate more on the nomenclature of my subject rather than common English words, despite extensive study of such to absurd levels. I occasionally will make mistakes such as mixing principle and principal or capital and capitol as my typing tends to be on automatic when not paying attention to that aspect of word use. Oooops. . I was just commenting from the perspective of a person who has been an educator for over 35 years. (I almost confused myself when I was reading my answer before hitting reply. For a second I thought "Should that have been prospective?" and then 2 seconds later decided I had used the right word. Eh. Just trying to be funny despite the serious nature of the subject at hand.)
Got to 49/3 and was happy my brain still works but thanks for the illumination
Think of the curve y=sqrt(3x) is always positive and y=-7 is always negative These two curve never meet. Hence no solution.
This is true in real numbers. In complex ones there are solutions
The curve y=sqrt(3x) is a sideways parabola and not a function. The plot of the principal value of sqrt(3x) is a function above the x-axis that is half a parabola.
@@MarkEmerAndersonII I have no idea what you are trying to say. Some context and purposes.will help.
@@samwong3123 So the square root is not a function. A function needs a one to one mapping of inputs to values. So we split it into two and call one the "principal square root" which are all the values of the square root that are above the x-axis. And we use this for a lot of math.
But if you plot the actual square root you will get the same plot as if you plotted x^2 and turned it 90 degrees clockwise.
Square root as a function becomes important in calculus so that's what we use most often and most calculators and computers assume. In fact, we assume the principal square root most of the time, which is the point here.
I honestly understand what he's getting at, but I'm an engineer with 15+ years of math under my belt, and I was a little confused as to his reasoning, so I think it could be explained a bit more clearly.
I tend to not like "because we say so" answers in math because sometimes that leads us down bad paths. For instance, antimatter's existence was theorized by Dirac by using the negative answer of the square root.
@@MarkEmerAndersonII As you include your background, I will include mine. I have been teaching collegiate math since the first Bush Administration. I have earned a doctorate in math in the field of algebraic number theory.
I don't know what you are saying when you write that the square root is not a function. It is a well defined mapping from [0,∞) to [0,∞) providing a single output for every valid input. Note: functions are not required to be a one-to-one mapping (or surjection). The function f(x)=√x is what we view as the principal square root or the positive square root.
We could easily have constructed the square root function to be based on negative outputs, namely g(x) as a well defined mapping from [0,∞) to (-∞,0] which would correspond to the bottom half of the parabola with vertex at the origin opening to the right. There's nothing stopping us from doing that.
The main reason why we chose the positive square root to be the default definition is more about applications. Most applications using square roots (like computing distance) needs the output to be positive. It does not make practical sense to have to compensate for that negative in the majority of situations. This is akin to why the range of arccos(x) is [0,π] and not [87π,88π]. There's nothing to stop us from using [87π,88π]. Mathematics could have been restructured with [87π,88π] as the default range, but then it would have been more head-scratching, complicated, and just plain stupid. We use [0,π] for practicality, simplicity, and convenience.
The thing is that once the selection is made as to how to define an inverse like the square root, consistency must be adhered to. This is where the "because we say so" doctrine comes into play. We need to have a standard to enable a continuous understanding across mathematics in order. This doesn't mean that the other choices are invalid.
I don't know anything about Dirac's theory of antimatter. But I assume that in his theory he states that the negative inverse is being used. He's communicating that the non-standard yet still valid square root is the appropriate and useful choice. Those that are learning his theory understand this context.
So what's going on? Why does it seem two answers to x^2=49 but √49 is only one number? The simple answer is that there are two possible inverses to x^2, the positive (or principal) square root g(x)=√x and the negative square root h(x)=-√x. So when determining the solution to x^2=49, we need to consider both. Now when you write √49, you have just chosen which inverse you are using.
I tell my students that when the √ is written the choice has already been made, such as √16=4 or -√36=-6. But, if you introduce the √ into the problem, the choice of the square root has not been made; both possibilities are still valid and need to be considered, such as x^2=25 yields x = ±√25 = ±5.
I'm sorry for the very long winded explanation here.
It's not only for quadratic
equations: +/- results, but also for all of the even degree polynomial equations such as quartic, sextic, octic and so on etc... These are all even root polynomial equations...
You are committing a error of logic when you use the term "the square root of". It is an error of logic to use the definite article when the term does not refer to a unique entity. See Kalish & Montague in their book Logic for a discussion of your error.
The committed error is writing an equation that intentionally can't be solved by left and right side equation manipulating. 🙄. If √(3x) = -7 is written: √(3x)√(3x) = 3x = -7√(3x) holds ... -(1/7)(3x) = √(3x) holds ... But that equalled -7 at the start. So -(1/7)(3x) = -7 and continuing to x = 49/3 manipulating left side and right sides without doing division by zero elimination cheating watching what I was doing every time. It is the first time hearing we can't manipulate to solve equations or tidy up an equation mathematics in my life teachings. 😫
(49/3) i^4😂😂😂😂😂🙃
At the University of Wisconsin, where I attended for my BSEE, no matter what, A SQUARE ROOT CANNOT EVER BE A NEGATIVE NUMBER in the real number system. Of course, it can be a multiple of SQRT(-1) or "j" (engineering use), engineering does exist in the space of complex numbers but going back to basics, in real terms, a square root cannot be a negative number simply because any number squared, including a squared square-root, must be positive. HOWV=EVER now going into the complex number system;
SQRT(3x) = j*(7),
3x (-1)*(49)/3
x= -49/3
Note that there is no solution of +49/3 because the complex "j" squared is a factor in the expression.
A video on waffling and time-wasting. now it's clear why some don't like math. if the principal solution is always positive, but the answer is negative only a fool would choose the principal solution I used to be a PhD Cambridge math postgraduate but I feel dizzy after this exposition on such a trivial matter. Americans certainly know how to obscure clarity
What application you use to make such presentation?
Webcam and a chalkboard
This is where I don't agree:
√16 = 4. (only the principle root)
But ±√16 = ±4.
So I'm saying, you always need to put the ± in front of the root as well. And that happens in a quadratic equation:
If x² = 16
Then x = ±√16 = ±4.
That's why the quadratic formula says x = (-b±√(b²-4ac))/2a.
The ± in front of the root tells us we need both the positive and the negative roots, not just the principle root.
Great presentation
2:07 nifty response.-Ernie Moore Jr.
I was taught to answer with all possible roots unless the problem or test wants answers from a particular subset of numbers.
This is what I keep telling people. Answer, how your Teacher and Professor want it answered. Otherwise you may get some of them angry faces on your test's. Also, "Extraneous Solutions" refer to the solutions which "DO NOT" work. Not just non "Principal Square Roots (PSR)." The practice of only using "PSR" is a convention, because it makes teaching easier.
@@danv2888It doesn't just make teaching easier. It's much more fundamental than that. The √ sign is defined to mean principal square root because that makes it easy to use the √ symbol clearly and precisely. Clarity and precision are important in mathematical notation.
The principal square root of x is √x. The other square root of x is -√x. And if you want both square roots of x (like, for example, the formula for solving a quadratic equation) you write ±√x.
This question explicitly tells you that it only wants the principal square root. That is the definition of the √ symbol.
@@gavindeane3670 That is not true. I do not know when you went to school. However, when I was going the same symbol is used for both, principal and non-principal. Also, in the directions is where it stated whether both or just one was needed. Also, it was more likely to be accepted that both where wanted, and where only principal was wanted that would be stated. Overall, I am just telling people to listen to what there Teacher or Professor are asking and saying. Not to what this guy is saying is the correct way. Because I have had things not stated and changed on me and then they claim your wrong. Then you get that angry/sad Face he is talking about.
@@danv2888 It is true. You can look up the definition of the √ symbol in the international standard for mathematical notation if you want to.
The principal square root of x is √x. The other square root of x is -√x. And if you need both square roots (like, for example, in the formula for solving a quadratic equation) you write ±√x.
There is no sign indicator in front of the radical symbol in this question so therefore the question explicitly requires the principal square root and only the principal square root.
You may not have been taught this. It does seem to be a point that isn't always taught very well - as is evident whenever this topic comes up. But that doesn't change the fact that it's true.
Undoubtedly, the symbol has been defined this way so that it can be used clearly and precisely. Clarity and precision are important in mathematical notation.
Yes, if someone's goal is to pass a test then they should answer questions in the way they were told to answer by the teacher whose job it was to prepare them for that test. But if someone is being taught incorrectly then explaining to them how it really works is a good thing. They might need the incorrect understanding on test day, but it's the correct understanding that matters on every other day.
The range for a square-root function w/o shifting up or down is y ≥0.
-7
There is no real solution but there is a complex solution
Another way to say this is that you assume a positive square root unless the problem says otherwise.
I have solved the underlying problem in general - it's very tricky and this video can't even scratch the surface (question underlying: why and when do extra solution show up?)
Solution in general: check all the solutions you get in the original equation (and cross your fingers or better check that every step you made was allowed for all x you consider)
So basically the square root of 3X does not equal -7. Would it actually be an equation if the two sides are not equal. for example I would say that 2+2=7 is not an equation.
It is a false statement. The solutions for it are included in the empty set (a set without any element).
Be good to know the grade level of each question
I wasn’t confused until I watched this. Way too long
It's really intuitive actually in its penultimate form. You would never assume the sq rt of 49 to be -7. If you did that while within a larger equation you will always get the wrong answer.
Again you took 14 min when you should have taken 5
Wish my math teacher was this slow.
He probably also doesn't know what it means if we can see a daytime crescent moon. And the sun is well above the horizon. 😂😂😂
Took like 5s to workout from the thumbnail no way it should have taken that long.
Yes, on top i guess pupil would be completely lost with such unsynthetic, unorganized and monopace flow of explanations.
😢🎉😢🎉
I ALWAYS test my answers in the original equations as I certainly am fallable in my math manipulations. So I would have caught the impossibility of a real number solution to this equation.
✓(x^2) = |x| = the absolute value of x
The equation ✓(3x) = -7 has no solution
On the other hand, there is one solution to the equation
(-7)^2 = 3x
x = 49/3
I do not get it. When you resolve this equation you upgrade to square the left side of the equation and the right side. So -7 becomes +49 and the solution is x=49/3 But is true the checking is not working so what is wrong? What is the correct answer?
Only a quadratic squared root has two outcomes, being both the negative and the positive number. He explained, that there exists a RULE, that if a squared root is NOT quadratic, you may not alternate it into a quadratic one to get both answers. When you have a simple squared root, the answer MUST be a positive number. Since the number here is a negative, namely - 7, that alone tells you, that the answer cannot be given. For with the simple squared root the answer ALWAYS has to be positive. You don't even have to complete the math here to know that. (But I bet math teachers still prefer you to write everything out anyway...)
Silvia: the checking IS working though. If you plug in “49/3” as the value of “X”, you get 2 possible answers, one of which is “-7”.
It is true that “-7” is not the only answer, but it is not wrong.
i guess my not so much mathematical schooled brain helped me because i immediately thought that the squareroot of something cannot be a negative number as a negative number multiplied by itself becomes positive. and of course a positive number stays positive. So i got to the right solution by only considering the mathematical basics i learned in primary school.
Is there some convention baring imaginary roots? Why can we not use 49/3 i^4 as an answer??
i^4=1 though so that wouldn’t work.
There's no convention barring imaginary and complex numbers. But that doesn't help - there is no solution, real or complex, to this equation.
(49/3)i⁴ is not an imaginary number. It's a real number. It's just a fancy way of writing 49/3.
If x = (49/3)i⁴ the 3x = 49i⁴ = 49.
And √49 does not equal -7. It equals 7.
The « principal » square root is a subjective concept not a mathematical one.
We see the problem so is there an answer? After watching this I was still waiting for an answer or is it impossible to square root a number to make a negative answer?
Per Definition is sqrt( a) never negative.sqrt(16) is only 4 and never - 4
If someone gave me that on a test (I’m a teacher so that would probably be another teacher, or the people who put together standardised tests) I’d school them on how to use notation. The square-root symbol written like that is by definition the positive root, and the whole equation is just nonsense; it’s not merely a thing where “no solutions” is an appropriate response, but “that’s not a question” works too. 👍🏻😂
√66 3^2 3^2 (x+2x-3) (x+2x-3)
You cannot have AC current with just principle roots.
sq rt (3x) = 7*(i^2); where i^2 = -1
(sq rt(3x))^2 = 49*(i^2)*(i^2)
3x = 49*(-1)*(-1)
3x = 49
So, x = 49/3 or 16(1/3)
Can't this be the solution for x? If not, why? Please explain. Thank you.
Unless otherwise stated, the PSR is the only correct answer for sqrt( ). The sqrt(49) = 7. Not -7. Type it into your calculator!
x = [-7(sq rt of 3x)] / 3
I like it. It's crazy wrong, but fun.
So...
(3x)^2 = 9x^2
[ sq rt (3x) ] X [ sq rt (3x) ] = -7^2
OR [ sq rt (3x) ] X [ sq rt (3x) ] = -7[ sq rt (3x) ]
....
hmmmm....
9x^2 = 49
x^2 = 49/9
some crazy number...
or maybe...
x might be negative... oh sheeeeeit....
LOL
LMAO!!!
These are THE BEST equations!!!!!
Though a sqrt is never negative, for x=49/3, -7² would still lead to (3*49/3).
Don't need 15 min for that
So -7 is not a valid root of 49? Who re-wrote maths in the 50-odd years since I took my A-levels? We'd have lost marks, or even been given zero, if we only gave the positive root as the answer!
Is the full answer to the problem X= (49/3) which is not true? What is the full answer ie. how would the answer be fully verbalized? I can follow the math, but it seems at the end when you plug in the answer it is not correct.
OK, you said what is not true. What is the correct solution?
But what if: (-square root of (3x))squared = 7squared? You will get 3x=49, since one is allowed to multiply both sides by -1!
(− √ (3 x))²=7²
Sure, but the answer still does not work for the original equation. That's kinda what this is about.
I'm FAR from a math genius, but if the right side of the equation is negative, the left side can not be all positive, right?
Yes, that is correct. But as a square root of any positive real number can be a positive and a negative real number the equation does have a solution.
I only came here because there is no answer and I was ready to fight. LOL
I understand why -49/3 wouldn’t work, because would lead to an imaginary answer of + or - 7i. But I don’t understand why +49/3 would not work. That would yield the answers +7 and -7. Granted, when you think of square root of 49, you usually think of the positive square root of 7, but I don’t see how -7 would be incorrect because (-7) squared yields 49.
Unless otherwise stated, the PSR is the only correct answer for sqrt( ).
did I not see the answer, you said sq rt of 49 IS NOT -7, what is the answer? anyone?
Ok so whats the answer?
= 2.1 2.1 (x+1x-2) (x+1x-2)
Another way to prove this is to square ROOT both sides. Now you have 4th-root of 3x=sqrt(-7), and we know we can't use complex numbers here.
How about x=49/3i^4
Yeah, that would be my solution as well.
Yep. Or (49/3)e^(i*2*pi). But both of those are equal to 49/3 anyway, though hinting that the solution requires the complex plane.
You can, of course, have a square root of -49 which is 7i where (i) is the square root of -1, an imaginary number. This, however, is not the point of this video.
Even so if the equation was sq rt (x) = - 7 the answer would not be x = 49i
couldn't you actually say (49/3) times imaginary i ? Or does that only work if you are taking a square root of a negative number.
No (49/3)i doesn't work.
That would work if √i equalled -1. But that's backwards. It's not √i = -1, it's √(-1) = i.
So bringing i into this doesn't help. There's still no solution to this equation.
Wish i had your videos when i eas in college 😂
1. The variable x domain has not been established.
2. The squaring operation on equations introducing unexpecting values to the variables and should be avoided.
3. No square root can be taken of a negative number within the system of real numbers, because squares of all real numbers are non-negative.
The square root of a positive number can't be negative. That was my first reaction.
Ya got me pretty good there guy, or did I just get myself, hard to say. Anyway I was blasting verbosity at my monitor as I was watching this video.
"Of course there is a solution you dolt and it's not the nul set, How styuupid can you be?"
So to prove how smart I am and how dumb you are I wrote the equation down and solved it. Then solved it. Then solved it. Then solved it. Then solved it. Then solved it. Then solved it.
Tail appropriately tucked, thank you as much for the leason in humility than the math leason. Great video.
You can always tell this type of problem has no solution by just looking at it, because if you square root any real number it will never be negative.
Ans: (-7)^2/3=(49/3).i^4
3x=49 -> x=49/3=16 and 1/3
If I can be honest about this, the explanation confused me. This is definitely "NEW" Math for me. When I went to High School, way back in the dark ages, it was always the case that a positive number could have both positive and negative square roots; a negative number could NOT have a square root! Seemed unfair for the negative number, but there you go! Nobody in those days said that Math had to be fair! I really do not understand this unique explanation. I do like the term "unsolvable" which we were encouraged to use back in the "dark ages" as an answer to a negative square root problem. Though I suspect true mathematicians dislike the word unsolvable!
You are good when you do not accept his explanation, because it is just wrong. The square root of 49 has two solutions + 7 and - 7. And has for a very long time now, in like for centuries.
"Unsolvable" was expressed as "doesn't belong to the field of x" where x is integer, real, rational, etc. When I was in high school. I think there's no "unsolvable" in mathematics, just undetermined like when there's a division by 0.
Of course, the calculator can't show you "undetermined" in a 8 digits display, so they just went with "error" for anything the limited programming (and possibly understanding of the programmer) couldn't solve.
There is a square root for a negative number, but it is an imaginary number, and is generally not taught in Algebra 1, but may be taught in Algebra 2.
How do you explain (-4)(-4)=16 ?
(-4)(-4) = 16 because -4 is one of the square roots of 16. It's not the principal square root of 16 though, which is the issue here.
So, it takes over 14 minutes to say that the square root sign means "the positive square root of." The example shown says "the positive square root of (3x) = -7" which clearly has no solution.
Yep, all of these videos take sooooooooooooo long to get to the point.
@@billk9856 Perhaps you would find Organic Chem Tutor more amenable
The square root sign always means the principal square root. . .is not a universally-agreed rule, it is a convention used by some for what purpose. For example, if I have the quadratic equation x^2=64, and the directions are "solve by taking the square root of both sides" (which in that particular problem is possible method to solve), if I do that, are you saying to solve that, you would write "x=8, x=-8, can't have a negative square root, so x=8?" My math teachers (and my brother with his undergrad degree in mathematics) would fail that answer, saying you did not understand the concept of a quadratic equation ALWAYS has two solutions.
So the principal square root rule does not work everywhere in mathematics.
It also does not work is if you have something like "provide the zeroes of x^2=√16"
So for that because x is squared, anything squared will be greater than or equal to zero (negative times negative always makes positive)
So the first step we simplify the radical, the square root of 16 is 4, -4 but only 4 will work in the equation NOT due to any principal square root convention, but SIMPLY because the SQUARE of a number can never be negative.
Take the SQUARE ROOT of both sides
x=2 or x=-2
Now. . .one last thing. A square rootcof a number x is any number that when multiplied by itself equals x.
For example, (-2)*(-2)=4 and 2*2=4
Conventions are one thing. Proper math is another.
LHS of the equation is nothing more than exp((1/2)log(3x)) = exp((1/2)Log(3x) + kπi), k∈ℤ; RHS is exp(log(-7)) = exp(Log(-7) + 2lπi) = exp(ln(7) + Arg(-7)i + 2lπi) = exp(ln(7) + (2l+1)πi), l∈ℤ and you may proceed from here...
When you square both sides you are introducing extraneous roots
Try this sqrt(9)
Sqrt(9)=3. From your logic.
However, sqrt(9)= sqrt(9*i^4)= sqrt((3*i^2)^2) = 3*i^2= -3. Therefore both 3 or -3 are correct.
So are you saying you can't have a square root of a negative number.
No, that's not what he's saying. You can have a square root of a negative number, but that's irrelevant here.
What he's saying is that the principle square root of any number can't be a negative number.
I got (49/3)*i as the answer. Is that NOT correct?
Shouldn't it be times i^4, not just i?
Wouldn't -i work? I can't remember that one because I haven't used complex numbers in about 2 decades...😅
Yeah it’s i^4. The square root of i is also complex.
Phew, I got that right.
(x=16 1/3 I) but since I is not a real number means it is it is unsolvable I= the square root of -1
( 16 1/3 i ) is not a solution.
Note that there are no complex solutions either. The same logic applies. Any complex number, z, whose square root was -1 would satisfy the equation z = (-1)^2 = 1. That's the only possibility and it's not a solution.
May I suggest it's a convoluted description? More simply: Square roots are all positive if not zero. Negatives of square roots are all negative if not zero.
√16 =4 is true.
-√16=-4 is true
±√16=±4 is true
√16=±4 is false
(Hence, the quadratic formula, for example, necessitates the ± symbol, not merely the +, to produce 2 solutions. If anyone thinks square roots themselves can be negative, remove the ± from all your quadratic formula statements please. And draw y=√x and y=~√x as the same thing as y=±√x).
The initial equation here in this video has no solution simply because it states that a positive number (left) equals negative seven (right).
OK, I'm learning. Thanks.