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.
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).
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.
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
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.
I disagree. It appears that you believe there is no solution to the original equation? But surely every equation has a solution? I believe that the solution to the original equation IS indeed 49/3. Substitute 49/3 into the original equation gives sq rt49 = - 7, +7. Hence we satisfied the solution for -7, but +7 is not required. I believe that your error lies in the fact in the fact that every number actually does have two square roots.
Absolutely correct. The video contradicts itself when it claims that every equation has at least one solution, then claims no solution for what is clearly an equation. Furthermore every number, including imaginaries, does have 2 square roots.
Yes, 49 has two square roots. But the expression √49 refers to exactly one of those square roots - the positive one. That's the definition of the √ symbol. If the equation was ±√(3x) = -7 then x=49/3 world be the solution, because ±√ tells us that we need to consider both square roots.
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.
@@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.
So not really an error, just schoolteacher-level hair-splitting pedantry. Professional pure mathematicians would see no error. It is like the 'trick' question, "what is the differential of log( log[sin x])? One cannot really mark the answer as wrong simply because the function does not exist at some points.
Well when I went to school, a square root had both, positive and negative numbers as solution. Must be a very special odd definition in American schools to say otherwise.
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).
This guy cannot possibly be a mathematician .... the square of a sqrt ALWAYS has two solutions unless stated otherwise. In this problem, x can be -ve or +ve ..... so the solution is x= +49/3 or -49/3 ......Even though the value of x is negative, it is not a complex number..... the sqrt of -x IS a complex number (a very subtle but important difference)
I am a physicist and have a PhD in chemical engineering. I always thought as the way as in this video for 60 years. However recently a mathematics professor corrected me. You are right. Many people think they know the answer and even making videos to show others. However they are not correct. I eat my humble pie.
Another poor example. This problem sets up the student for failure. Every square root has a possible outcome of +/- the resultant value. However, in the case of this problem, in order to be algebraically balanced, the positive 7 value has to be discarded in only the negative 7 (-7) value will satisfy the equality requirement of the problem and written.
I see a pattern in a lot of your videos, trickery with no solution. I don't remember teachers doing this type of stuff when I went to school and I never had issues not going the school of trickery. You seen to overcomplicate things to prop up your intelligence.
If you have a radical with a negative solution, then the negative root is MUST be assumed. Don't trust this guy with his misunderstanding of basic mathematics
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…
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! 🙄
✓(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
Astonishingly incoherent, muddled and generally poor explanation. Write down what you have to say in advance, say it slowly and clearly giving your audience the opportunity to follow, don't repeat yourself unnecessarily, and reduce the distracting movement on the screen. Watch 3 Blue 1 Brown on how to do it.
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!
10:24 I disagree. Square root of 16 is plus or minus 4, in pure math. It is just a CONVENTION to keep only the positive values for cases where the negative value is out of the domain (such as what is the height of the water in a river, it can't be zero under THIS context, but pure math has no context), and so, square root of 16 IS plus or minus 4 and THAT is the right answer. Furthermore, your assertion that square root has just one (positive value) solution is totally wrong too: Counter example: within complex number a cubic root has 3 solutions. And that argument or sinus equals to zero has an infinite number of solutions (2n*pi, n = an integer, positive, negative or nul).
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.
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.
This video is totally wrong. Yes the square root of 49 is both 7 and -7, An nth root of a non-zero number has n distinct complex values. That's a trivial corollary of the Fundamental Theorem of Algebra. A principal square root exists, but so does a non-principal root. The point of the principal root is that if one restricts to the non-negative reals the nth root function is unique, where otherwise it would not be. When a question is explicitly asking for negative number it isn't asking for the principal root.
You appear to assume that 10th/11th graders (or so ) know nothing about complex numbers. But the better ones will, and will be terribly confused by the title and the first 9 minutes of your video (until you mention a bearably audible caveat). I know I would have been. You have to cater to all students, such as to not to inflict lasting harm.
It is a nonsense only using the principal sqrt. Consider 4-8 but someone says you can't use negative numbers then the answer is null. If you think there is only one root for a sqrt please state your nationality. If you can pls reference a text book or paper.
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.
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
In the middle of your analysis you casually mention that you want to confine solutions to real numbers and toss off something about makng that clearer at the beginning of the problem. To which I respond, ""Well, YEAH, that woild have saved me wasting 10 minutes listening to your puerile explanation." My answer, of course, was 49/3 i. If you had made your parameters clear, anyone should immediately have seen that there's no real number solution. What is sad is that this problem is supposed to illustrate the convention of "principal square roots"; instead, you post a problem that actually DOES have a solution. Technically, your null set answer is incorrect for the very reason that you failed to state parameters. As a teacher, you should appreciate what happens when you rely upon unstated assumptions. Since you seem attracted to using smiley faces and to judging students efforts, I will do the same. I assign you 😖 and a D for wasting my time. I suggest you strive to be clearer in future.
So, your answer is, you only take the positive Square root cos I say so? Terrible explanation my friend. You took 14 minutes to say " the answer you solve is wrong because, randomly Square roots only have one solution except when they're solutions to quadratic, no reason given. Trust me"
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.
You are wrong. The answer to the squareroot of any positive number has always (mathematically speaking) two solution, one positive and one negative. When I was studying Mathematics there was a joke: Ask a mathematician, a physicist and an engineer what is the square root of 4. The engineer will say 2, the physicist it is +2 or -2, and the mathematician will say, the problem has at least one solution but no a unique solution. So you are saying only the engineer is correct? Well, that is nonsense without stating beforehand you are looking for a positive solution only. BTW if you are of the opinion that the squareroot of any positive real number can never be negative, why make any transformation on the equation? No need for that. Just say at the beginning that in your, very personal opinion, a square root always has to be posivite and you are finished with the problem. (well of course not in the mathematical world, because there, your opinion is just wrong)
I think that you have a right point... it is semântics... And if n a way that sqrt(x) is defined as a fonction il think thé maker of thé qu'estion Brings... só strictly said he may ben right... but it dépends on what for you need it ... and in that case even one van enhance thé mathematical definion of squarerrot depending ik n thé system or context... nothing with physics or engineering, just thé domains where you except a solution
On thé otherhand in général or mostly thé symbol means principal sqrt... it is also confusing... when it comes tô Powers that van alsof be fractions... in a lot of problems one should look at thé context... but je is right as it is defined as thé principal squarerrot... not all littérature is consistent on this...
@@josleurs4345 yes, if the solution has to be in a certain range, and that range are positive numbers, than the squartroot of a positive real number has only one solution. If the solution has to be an integer, there is no solution. BUT He did not talked about a specific range for the solution. But that was not the problem here. The only way to claim there is not solution to this problem was to claim that the squareroot of a positive number has to be a positive number, and there was especially mentioned "in the mathematical sense". Sorry, but mathematicallly the squareroot of any positive number (greater zero) can be both, a positive and a negative number. And ... why make the effort of writing the problem in another way? If the squareroot has to be positive just looking at the problem says, there is no solution.
I have never used "sqrt". Well English is not my language, but even in mathematical books at University level I have never seen this function, alway seen the symbol for the squareroot. And the answer to the question what is is the squareroot of 4 is it is both +2 and -2, at least as long as one does use negative real numbers at all. And as there was a negative number in the problem from the beginning, they are included. @@MrSummitville
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.
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.
Why it doesnt have solution: The square root Index is a Even Number (this means result is going to be always positive) You cant use imaginary numbers because theres a difference between 7i and -7 (one is fake number and other is real number lol)
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.
You said put what you think is the correct answer in the comments - so before watching the video I'm going to go out on a limb and say there are no correct answers - that is to say, the "correct answer" is that there is no solution. This is because the square root is a function whose range is all non-negative numbers and the right hand side is less than zero. Therefore the question itself is a contradiction.
sqrt(3x)=-7 sqrt(3x)+7=0 (sqrt(3x)+7)*(sqrt(3x)-7)=0*(sqrt(3x)-7)=0 sqrt(3x)*sqrt(3x) - 7*7 = 0 3x-49=0 3x=49 x=49/3 You got here without squaring across '='. Ans is correct, but you seem to not like the question (1st statement)... so skip and move on :)
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.
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.
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.
Another word salad. No wonder so many people fail at mathematics. You are not really explaining math TERMINOLOGY or THEORY. For example, what is a "quadratic" equation? Merriam Webster: "involving or consisting of terms in which no variable is raised to a power higher than 2" I've heard you say quadratic equation 100 times but you never have broken your explanation down to the definition of terms level. John, you are falling short of teaching math. I'm giving you a D-. Please bring a revised lesson to your next math video. It will help your grade. Who knows? You may become a professional expert in math. 😕
Let's make the equation to : ((3X)^2)*2=( -7)*2 When X is big more than 0, we have 3X=49 X= 49/3 = 16 +1/3 or X=( -7/3^2)*2 When X =0 ,the equation is wrong. When X less than 0 , the equation is nonsense.
There's nothing 'fundamental' here. The principal square root is merely a convention, to be understood when a problem like this is given. Mathematically, the square root with no such convention is both positive and negative. Since square root function does not provide an unambiguous answer, it is not a true function - by the definition of such. The 'principal square root' is a function with a restriction on its range, y=square root(x) belongs to R/x
Yeah. I do "see" but my anxiety is magnified, by this explanation, because I see the path forward, to Engineering Calculus (Advance Physics, et. al.) is a MINEFIRLD of radicals, each of which, with an attached pedigree, which must be, discreetly, tested and verified. How does Albert Einstein look at that LONG blackboard, LOADED with radicals, at every scale and order, and *know, how to "treat" each instance, in the moment, within his "stream of thought" ?
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.
I disagree completely with everything in your commentary. Completely false. The square root symbol is exactly the same as raising the number to the power of 1/2. So all square roots have 2 values. We just keep thing simple when we supply the positive value. Root 49 is equal to -7, this is true.
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.
This video fails to explain why the square root of 16 (for example) cannot be -4. I watched through the video hoping for an explanation, all I got was an assertion that it is not the negative number. In mathematics there have to be logical reasons, if not there have to be a few basic axioms, even so there still needs to be some kind of justification. What is the mathematical problem with a solution of -4 for the square root of 16? Unless this question can be properly answered the claims of this video are useless.
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. 👍🏻😂
OMG, in France we consider that sqrt(a) is the positive solution of x²=a. This is so simple, why make it so complicated? You can tell in less than a second that there is no solution. What are you trying to achieve?
It's a common fallacy to suppose that if you say something multiple times, the listener will understand better. In fact the listener gets bored and stops listening. Americans are particularly bad at repeating themselves. This was a really boring video as a result of this long-winded and repetitive waffle.
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.
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.
Common error or philosophical difference? What if we object to concept of "i" on the basis that it is only imaginary? Mostly being a perverse Devil's Advocate here, but maybe has a speck of truth to it.
Colossal waste of time. Just say that when you solve a radical equation you only consider the positive (principal) square root. When you solve a quadratic you solve for both roots. Period.
It looks like too much of explanation for a simple thing. If we are looking for solutions in real numbers, then the square root is a real valued function and must have one output, a nonnegative one. If complex numbers can be used for solutions, then the square root must be a complex valued function which is multivalued and has two outputs as its two branches.
There's no negative value when square-rooting any sum. That's the golden rule. It's as simple as that. It took me 1 sec instead of 5, not to mention 14. Silly demo.
"There is no answer" is your solution? Then the error is in the question, not in attempts to solve it. Don't give up. Maths and science developed out of attempts to explain the inexplicable.
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.
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. 😫
Wrong- the answer is 49/3. Negative answers are perfectly legitimate for square roots as every textbook will tell you if you read it. You should be ashamed telling people that negatives are not legitimate for square roots- I am so glad I didn’t rely on you for my college mathematics classes as you would have cost me my honors ranking
If you've got a textbook that says √49 = -7 then put it in the bin. The author doesn't know what they're talking about. Who knows what other nonsense might be in that book. By definition (the definition of the √ symbol) √49 equals 7. If you want the other square root of 49 then you need to write -√49. And if you want both square roots it's ±√49.
@@stonerdave None of my textbooks failed to use the radical symbol correctly. 49 has two square roots. They are 7 and -7. The expression √49 refers to exactly one of those square roots: the positive one. That is the definition of the √ symbol. The other square root is -√49, and if you want to refer to both square roots (for example, the formula for solving a quadratic equation) you write ±√49
@@gavindeane3670 in all my textbooks the square root has 2 solutions- typically written with the plus/minus next to the answer - nothing says you need to put any symbol before the radical to indicate the desired solutions- you use the solutions that work that’s how math works
@@stonerdaveImprecise, ambiguous notation where the reader gets to choose the interpretation that's most convenient for them is not how math works. It's exactly the opposite of how math works. The definition of the radical symbol gives a clear and precise way to refer to specifically and only the principal square root, a clear and precise way to refer to specifically and only the other square root, and a clear and precise way to refer to specifically both square roots. It's not defined that way by accident. How did your textbooks tell you to refer to just one of the square roots, if not like this? How did your textbooks set out the formula for solving a quadratic equation if not with a ± before the √ symbol?
First of all you have to state that x is an element of R. Then there is no need for 15 minutes of crap and you can immediately say that there is no solution. Alternatively you could try to teach something and solve fir x as an element of I...
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 if -7 is not the solution, which is the solution? You focus your whole video on what is not the answer without giving what you believe is the answer
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?
Because he didn't gave a real explanation, he just jumped around to invalidate the fact that if the equation stated a negative result then you use the negative root and NOT the "principle"... x=49/3 because the square root of 49 is ±7.
By definition, the square root symbol leads to a value >=0 so from the beginning there is NO SOLUTION. You don't need to go through all this work. From the outset the notation leads to no solution. Period. If you want to denote "any number whose square is 3x" the proper notation is (3x)^1/2 Simple rule to remember. No need for a long confusing video.
