In this case it would be informative to move the - to the numerical side of the equation. It would yield a sq root equalling a negative number which is an obvious thing to look for.
This is a really important thing to understand. I don't know the exact percent of students who don't get it, but from decades of tutoring, I know it's high. From the comments there are some adults who are also confused by it. (Which is okay.) If you ask the solution of x² = 4, it's x ∈ {-2, 2}. But if you ask what is the solution of √4 = x, then the solution is ONLY x=2. The square root symbol means ONLY the positive square root, or what they call the "principal square root." Start with: x² = 4 Take square root of both sides: ±√x² = ±√4 Simplifying: x = ±2, or x ∈ {-2, 2} This is why in the quadratic equation, they use the ± symbol - because the square root symbol by itself refers to ONLY the positive (principal) square root and since the equation needs to account for BOTH square roots, they have to include the plus or minus sign.
The problem here is that there is a distinction between mathematics meant to have single answers and mathematics used to solve actual real world problems.
The root symbol is for the "principal square root". The the principa square root is a mathematical function which is defined as: "which positive number must be multplied with itself to get the positive number under the root symbol?". This function is only defined for the positive x/positive y-quadrant of the cartesian coordinate system. You as a human know that (2)² and (-2)² both result in 4. But the the principal square root can only reverse the first expression, not the second. That's why you often see ± in front of formulas containing square roots.
I used this rearrangement in my head to come up with a solution but forgot about the possibility of introduction of extraneous solutions. Oops! It's been over 40 years since I've been to Algebra class, perhaps I had a senior moment!
I agree with that method of conviction. However, explain to me how the way the square root equation is written determines whether there is a negative answer.
Your explanation of why you can’t use-5 by demonstrating a quadratic equation is weak for me I think you mean that if there is not a root (unknown) on either side, you cannot introduce 2 solutions (because that would be extraneous) according to wikipedia, by the way, a quadratic equation can have a single solution “double root” please clarify
PEMDAS, but backwards because we are solving for a variable. You need to take care of Multiplication, Division, Addition and Subtraction BEFORE you deal with the square root (which is an exponent (1/2)). If you divide both sides by -1 you see that you need the square root of (2m+2) to equal a -5 (negative 5) we see we have a problem.
What promotes the era is when teachers self contradict when they write on the board something like rad(4) = +/-2. That is never acceptable. In this series of math lessons I’ve seen that done frequently. You cannot ever justify writing that a radical is equal to plus or minus anything. There is no justification for claiming that. Doing that is what creates the problems with radicals.
You are wrong. Two negatives cancel so it is acceptable. -(2)^2 =4. The problem is the new math and it's insane bias against complex numbers at the prealgebra stage when it is required for real world applications using complex analysis.
@@josephmalone253wtf? Who tought you math? First: -(2)^2 = -4 . Put the "-" within the parenteses if you want it to be squared, too...... Second: the radical sign is DEFINED to be the principal root. Always has been. Therefore √4 = +2 and nothing else. If you want to solve x^2 = 4, you HAVE to write x = +- √4 what can be simplified to +-2. Everything else is incorrect notation. And if you knew ANY math at all, you would KNOW that even complex numbers do not solve this. The radical sign is defined the exact same way for complex numbers. Don't need to trust me - try Wolfram Alpha. It'll tell you that there is no solution.
@@wernerviehhauser94I think the problem relies on people not realizing the context of the operation. Yes the radical is defined as the principal root, which is positive but then people get confused in operations like solving for x^2 because then you have 2 roots. If it's just a number with no other context but no relation to anything else then the principal root is the answer. But in this scenario I think the right approach was not to give in the urge to solve for m and just analyze the operations. 3 - (any positive number) =/= 8
@@marilynman I also assume that the culprit here is, apart from "New Math" and the educational system, the use of "root" for two distinct things: roots of a number and solutions of polynomial equations. Here in Germany, you don't find many people making this mistake since we have "Wurzel" as term for roots of a number and "Lösung" for the roots/solutions of polynomial equations. It's much easier to mix up things if they are given the same name.
This is a game of chess and checkers at the same time on the same board. Everyone adopts the rules they want. The square root is defined as non-negative. Not because those who adopted such a definition were stupid, but just to avoid unnecessary ambiguities. The consequences of leaving such ambiguity are much greater than the apparent deprivation of an "alternative solution". If a root can be both positive and negative, then every real number is equal to 0. Proof? Here you go: 2*sqrt(1)=sqrt(1)+sqrt(1), Now, first sqrt(1)= +1, second sqrt(1)= -1 - why not? 2*1 = +1 + (-1) 2 = 0 Now replace "1" with square of any real number - this way it will turn out to be equal to 0. Is this still mathematics?
it works perfectly fine with the principal root. If you think it does not, they you either had a bad math teacher or were a bad math student, since there is obviously something off with your understanding of the problem.
TI 89 yields "false". Multiplying both sides of -√(2m + 2) = 5 by -1 gives √(2m +2) = -5. A square root will not give a negative real solution. Negative 5 a solution to √25? If so, that would mean √(-5)√(-5). Bzzt! Wrong.
I got "no solution" a slightly different way. I don't consider "-(sqrt(2m+2)) to be fully isolated, since there's a -1 being multiplied by (sqrt(2m+2)) there. So I divided through by the -1 to get sqrt(2m+2)=-5. At which point you already know there is no solution, since a sqrt can't give a negative answer.
Actually a sqrt can give a negative number. Without complex numbers you can't take the sqrt of negative number. - 5 * - 5 = 25 and 5 * 5 = 25 so the sqrt of 25 is +/- 5.
@@alexaneals8194 The problem here is you both are talking about different sqrt functions. You need to agree first which sqrt to use. One of them can give values in the form of negative real numbers and the other one is not allowed to do that.
@@evgtro8727 It's actually not which sqrt to use. It's the fact that they have introduced the concept of principal sqrt after I attended school. I took Algebra I in the 8th grade in 1978-1979 and Algebra II the following year 1979-1980. At that time there was no principal sqrt. The sqrt equation was assumed to include both positive and negative values. So, the roots to 25 would be +5 and -5. Even the math and engineering handbook that I have from before 2000 does not mention the principal sqrt. So, the concept is new to me. It makes sense since most people use calculators in math class and the calculator like the programming languages treat the square root as a function and not as an equation (which can return more that one value for a given input).
The easiest way to see that there is no real solution is to move everything to the rt and graph it. Y = sqrt(2x+2) + 5. (Have to change the m to an x for the calculator) Then graph. Notice that the graph never touches or crosses the x axis.
Writing this before watching the video so I can have my thought process out there and see how it compares. 3 - sqrt(2m + 2) = 8. Subtract 3. - sqrt(2m + 2) = 5. Multiply by -1. sqrt(2m + 2) = -5. Square to get rid of sqrt. 2m + 2 = (-5)² = 25. Subtract 2. 2m = 23. Divide by 2. m = 23/2 = 11.5
If I was to apply this to an engineering problem such as “the difference of pressure of a pipe from a large diameter to a small diameter is 3-(2m+2)^0.5 =8 (Excuse the nomenclature, my phone doesn’t have a square root symbol) where m is the pressure difference in psi, then using (25)^0.5 =-5 is definitely valid as a negative pressure difference shows a drop across the pipe restriction. In fact, that there is just one answer of 11.5 is correct.
You don't even have to go to engineering just a pure mathematics perspective. The fact is that negative 5 is a square root of 25 it may not be the Principal square root but it is a square root it is a solution.
X^0.5 is not the same as √x . The √ symbol denotes the principal root while the exponentiation notation denotes all the roots. That's why there's a ± symbol in front of the √ in the quadratic formula. After all, if √ really did refer to all the roots, then the ± would be redundant, right?
@@misterroboto1 if √ does not include all roots why does the quadratic formula bother with ± Why not just do the quadratic equation like this: x = (-b + (b^2 - 4ac)^0.5)/2a It would be the same thing as x = (-b ± √(b^2 - 4ac))/2a The result is the same thing, including the fact that you come up with only two unique values. What the ± does is clarify the fact that there are two unique values for x.
I have a master's in math. I saw that was no solution immediately because the negative square root of (2m + 2) is not 5. When you put something inside a square root radical with no sign in front of the radical, it is understood "Take the positive square root of that something". I learned that in high school. When dealing with radical equations like the ones above, always test the answers!
This feels like one of those things that comes up differently depending on the level of math that has been taught. The entry point is people only think about the principal square root, but as things like polynomial equations are introduced, you start to get "oh yeah... theres a secondary root". Then again, i was always getting into trouble on tests for having read ahead and applying theorems that hadn't been introduced in class yet.
So, basically, the square root of (2m+2) must equal -5 for the original equation to work. This is what I was thinking… looks like my algebra is quite rusty. Thanks for the explanation. Makes total sense that there is no answer.
Forget the square root part, just looking at it and the problem is getting a bigger number than 3 while subtracting from 3. Nothing about it makes sense to my college algebra semester understanding.
If you misspelled "ALGEBRA", there is a high probability that you also omitted the negative state sign that was to precede the square root sign. We are sure that the problem was intended to be 3 minus the negative sqrt(2m + 2) = 8 which results in 0 minus the negative sqrt(2m+2) = 5 and that means that sqrt(2m+2) = negative 5 and squaring that results in 2m+2 = 25, 2m = 23, m =23/2. Or, maybe you just slipped up a little in writing "ALGEBRA". Either way, I enjoy your classes here and this one gets another Thumbs UP!!
Just a sidenote: At 12:37, the resolution to the equation shown at the right of the screen yields the correct result but there's a problem on the second step. We know that √x²=|x|. Yet you wrote √x²=±√25 which is wrong. It should be: √x²=√25 Then |x|=5 And finally x=±5. By saying that √x²=±√25 you would be saying that √x² can be a negative number, which it cannot be. Just a small inaccuracy that somewhat contradicts the main point of the video. Now, that resolution to the equation may very well be there to demonstrate how some people may get that step wrong while solving for x, but as it is not stated anywhere on the video i suppose it's just a small mistake that made it's way into the video.
That’s something I can’t understand. If someone thinks that sqrt(16)=+/-4 and sqrt(9)=+/-3, what will be the result for sqrt(16)+sqrt(9)? Will it be either 7 or -7 or 1 or -1?
I didn't see how there could be a solution, but since you said to go ahead and use your calculator to get one, I came up with 23/2 and joined the majority.
Be trend site it sound like don't follow back I recognize the presonilty you want to be I guess word would use heared. Not precisely related I am thinking of trendsiter
Squaring roots can introduce extranous solutions which is why you always have to check after solving an equation this way. This is exactly what he is showing here.
Why would you grade the solution 23/2 a C+ or 5/10 ??. I see at least 2 reasons it should have been graded F or 0 - Domain of validity of m not defined: if solving this in R, then the radical does not exist if m
"When you use -5 you get the right answer"... he said that! Then said it was wrong. Change the rules to suit the problem.. how do you then justify (-5)^2=25??
I can remember from maths lectures during my uni studies of chemistry we should ALWAYS keep in mind that squaring is not an equivalent transformation. Bad things can happen...! 😉
It is a CONVENTION that when solving for square roots, only the principal square root is used. However, the secondary square root still exists. So think otherwise is as foolish as believing that negative square roots don’t exist simply because they are referred to as “imaginary,” and do not show up on the Cartesian graph of real numbers. But imaginary roots still exist, and you ignore them at your peril.
Just curious why you didn't use PEMDAS this time. Because then you could have dealt with the sqrt first. I am not an expert so any response is acceptable.
I didn’t quite get why you consider both solutions in a quadratic equation but only consider the primary root of 25 in this equation. Your explanation didn’t do it for me 😬
"We just need to use the value of 5 and not -5, because when we do use -5 the answer turns out to be right, but this is not the case..." 😯😑😤 If really that's your explanation, I'll think twice before I let my son/daughter be in your class!
Let’s assume that the equation you started with, p, is true. You asserted. You gave us an equation. You TOLD us they were equal. That statement was FALSE. Statement p is false. If p then q is true. My advice: don’t write false statements. If p is true then q. The above statement is true for any q since p is false. So if q is the solution you don’t like it matters not.
An important general point is this: when you square both sides of an equation, your new equation can have more solutions than the one you started with. Ex: x = 2 has one solution, x^2 = 4 has two solutions. When you solve an equation after having squared both sides of the original, your new equation may have one or more "false roots". Thus, it is necessary to take your bottom line "possible solutions" back to the original to see which ones don't work.
so what you teell us is that m equals the empty set is the solution? That means that we replace m in the formula by the empty set we have an equality? If there are no solutions that means that the set of solutions is empty, but it does not mean that the variable we are looking for is the empty set. We are lookinf for a value for m, so for a number. And because that number does not excist (ithe real values!) the set of solutions is empty. But you give a "value" for m (the empty set) and that is very wrong. In fact you say: 3-sqrt(⌀ + 2) = 8 Please stop teaching the wrong things about math, you still make too many errors.
From the open, my brain said their is no solution... then wasted 16 minutes to confirm. There must be a better title to not fool people into watching. Unsubscribe.
If everyone gets it wrong maybe you should do something about it as its obviously not being taught correctly. I can tell you 100% i have never heard the term principle sqrt ever when learning about sqrt.
The red flag should happen as soon as you see a square root of something equating to something less than zero. That statement can't be true with real numbers. In any case, it should be apparent to the student that they can't just decide that the square root of 25 *is* -5. Why must it be -5? Solving for m means that you are telling me definitively that m is a solution, not a solution under certain made up circumstances.
Great video! LOL, you made me feel bad and then good again:-) I could simply not find a solution, but I thought there had to be one, so I tried and tried again. When you then answered that there were no solution, hen did I feel good:-)
The square of a negative number is positive. To say that the square root of a number cannot be negative is a meaningless, although usually helpful, convention. The problem is only unsolvable if you stick to this meaningless convention. The question is, does it make sense in a real situation or is it a problem that remains forever in the realm of theory? We might equally decide that negative numbers cannot exist but it is quite useful using them. Sorry, I am now having an existential moment, realising that the whole capitalist system is based on the misconception of negative numbers, the idea that something can be owed to another and be in deficit.
Good math texts should make a big deal of teaching the principle square root so that when a student studies quadratic equations they won't default to this misunderstanding. This is an important issue in teaching math.👍🏾
there is a much easier way to explain this. The algebra we learn to do has a hidden assumption in its methodology which is that the steps must be reversible. This is why you plug in the answer to check. extraneous solutions are asymmetrical which is why you cant use them.
I have an intuition that the +5 solution would draw a different shape geometrically than the -5 solution. I have no intuition that sqrt(25) does not include -5 as an answer.
NO because the square root encases the 2m+2 as follows [(2m+2)^(1/2)]. As usually this guy's problems are poorly defined with lots of ambiguous aspects.
I studied math back in the dark ages when there was no such thing as a principal square root. Root 25 = +/- 5 . For many purposes the positive square root is the one you use but they both exist. m = 11.5 can be put into the original equation and works fine so it is a solution. Thinking that root 5 is + / - 5 is NOT an error. Definitions vary - do you ignore negative solutions to a simple quadratic equation? To ignore negative square roots is just plain bogus BS.
This isnt a real application so it doesnt require a real number for the radical. Just do what Al Kharizimi would do and add the radical to both sides. Then further simplify till m =11.5. Plug 11.5 back into your equation and see that it checks out. It does so we are done. We were not asked to debate complex numbers only to solve for m. People are over thinking the problem and that is bizarre. They are ignoring what algebra is. There is no multiplication of negative 1, there is only a positive radical being subtracted from both sides. I am appalled by the fragrant attempts in the comment section to say there is no answer when there is.
Does anyone know if the “Principle Square Root” convention is solely an American thing, or is it employed in the UK and other developed countries as well? It’s just that it seems crazy to me. There doesn’t seem to be any logical reason for it.
The Principal Square Root idea is good because it lets you quickly see that there is no solution. if you don't like that idea, its fine; You can do the problem the hard way - that is, search for a value of M that makes the equation work - and there is no value that works. you can't use 23/2, or -23/2, or any other value.
The principle square root is *a* solution, it is not the operation. The only reasons I can think of for it are computers or lazy instructors. Mathematical functions can only return one value (one value of y for any value of x for y=func(x)). Square roots are an operation (not technically a function) and actually actually usually have two solutions. A function can only return one, so the function returns the positive solution, or zero (or an error if x
@@pmw3839 It's all about notation. The question reads as "3 minus the principal square root of (2m plus 2) equals 8". So although -5 is a square root of 25, it doesn't apply to this equation.
@@davidellis1079 How are we supposed to know that? I mean you say yourself that -5 is a square root of 25. I see no explanation for why we should not use it here!
That the sqr is only positive is only an arbitrary definition, not something fundamental. In the real world you should take it as a possibility that the writer intended +- sqr. Especially if they state that there is a solution.
sorry to break this to you, but this has nothing to do with "real world applications" and all with "didn't really understand how numbers work". If the author of the question intended a +- square root, he was in error.
You forgot a step. You forgot to tell us the answer had to be a principle square root. You mentioned it later, but not before I solved the problem. i = sqrt(-1). So, i^2 = -1. And, i^4 = i^2 * i^2, which =1. But, sqrt(i^4) can be either 1, or -1. So, a valid answer is m = 25/2 i^4 -1.
The result of the square root of 25 is the value of 5. If you square a positive 5 or a negative 5 you still get the same result of positive 25. That is why the square root of a number technically is +/- the magnitude of the result of the square root operation.
@@GaryBricaultLive Yes, I got confused. The square root of a number can be negative. I was thinking of the square root of a negative number, which is a different thing.
"undefined" and "no solution" are the correct ways to express it. What he wrote ("m = ∅") is incorrect. If you wanted to use set notation, you'd have to write "{m} = ∅".
