The Algebra Step that EVERYONE Gets WRONG!

[@dogvom]

The first step you have to take is to spell ALGEBRA correctly.

[@1234larry1]

Math skills/language skills....and never the twain shall meet.

[@fredsalter1915]

lol... spotted that straight away

[@luisgq2358]

you so wierd

[@dogvom]

The second step is to spell WEIRD correctly.

[@duanehorton4680]

@@luisgq2358 Weird

[@ProactiveYellow]

3-√(2m+2)=8 √(2m+2)=-5 √x is the primary functional branch of the inverse relation of the function f(x)=x², with branch cuts such that f₁⁻¹(x)≥0 and f₂⁻¹(x)

[@TheFallibleFiend]

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.

[@calicocavalier2254]

Nostalgia here!

[@davidellis1079]

That's a very clear explanation. Your comment deserves to be pinned, as it would help a lot of people.

[@charlesmrader]

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.

[@thomasharding1838]

Except that according to Ozford website, sqrt of X is plus AND minus UNLESS a positive or negative sign is applied to the sqrt symbol.

[@willemakkermans4067]

Except for 'checking the answer with the original problem', you basically explained the whole 17+ minutes video in less than a minute of reading. Thank you, well done.

[@DeanNataro]

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.

[@MikeStallings2023]

Yes, thank you. A teacher pronouncing that "only the principal square root matters in this case" tells the student nothing useful whatsoever. I already knew the answer, and this video almost sucked that knowledge out of my brain.

[@gavindeane3670]

That's true, but it's not really relevant here. Going through a whole process of solving the equation (which involves squaring both sides) only to find that the solution doesn't work when you plug it back in to the original equation is a complete waste of time (not to mention a failure of understanding) when we can see by inspection that the equation is impossible.

[@methatis3013]

This is true whenever you apply a non-injective function to the beginning equality. For example, x=2 If we apply f(x)=x+1/x to both sides, we get x+1/x= 5/2, which has 2 solutions. Mainly, x_1= 2 and x_2 = ½

[@claude77573]

Wow

[@prime423]

The square root indicates a positive real number.Therefore,this equation has no real solutions.

[@nedruss7040]

It does have a complex solution, though. m=12.5 * i^4 - 1. 2m =25*i^4 - 2 2m + 2 = 25 * i^4 - 2 + 2, or simplified to 25 * i^4 The square root of 25 * i^4 = 5i² i² = -1 3 - ((-1) * 5) = 8

[@MarieAnne.]

@@nedruss7040 No, 12.5 * i^4 - 1 is not a valid way to express a complex number. In fact, 12.5 * i^4 - 1 = 11.5, which is a real number. All complex numbers can be expressed in the form *r (cos θ + i sin θ),* where r ≥ 0 and −π < θ ≤ π and where principal square root is *√(r (cos θ + i sin θ)) = √r (cos θ/2 + i sin θ/2)* Since −π < θ ≤ π, then −π/2 < θ/2 ≤ π/2, But cos θ/2 ≥ 0 for values in this interval, then principal square root of any complex number will always a non-negative real component. So for some complex number m, √(2m + 2) = a + bi, where a and b are real and a ≥ 0. Therefore √(2m + 2) = −5 cannot have any solution real or non-real complex.

[@flagmichael]

@@nedruss7040 My front yard is 34 by -19 feet? Domain is always important in any real world math. I have never had any real world use for imaginary numbers, or otherwise out of the domain for the actual answer, except as a flag for "not this solution."

[@flagmichael]

@@MarieAnne. The problem arises when there is one imaginary solution and one real one. That was the case in the only time I used advanced math skills outside a classroom. I was installing an Automatic Direction Finder (ADF) in a small plane and the specifications for the effective height (simple equation) of the wire antenna and capacitance of the antenna to the airframe (another simple equation) could not be met without a series capacitor at the receiver. The solution was the real root of a complex number; the imaginary root was useless.

[@__christopher__]

@@flagmichael The fact that you've not encountered it doesn't mean that it doesn't exist. And BTW, "34 by -19 feet" would still be real, just negative.

[@scottdobson1276]

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.

[@GeoRedtick]

That was basically going to be my comment. You can tell right from the beginning that way that it has no solution.

[@ozviking8052]

Yes. Moving the sign immediately gives the solution (25/2)i - 1

[@illhaveanother8173]

Yeah, it never even occurred to me to square away the negative.

[@hasan7275]

that’s how i looked at the problem from the get go

[@gavindeane3670]

@@ozviking8052 That's not a solution. That gives 2m+2 = 25i. But √(25i) does not equal -5. I think you have the relationship backwards. It looks like you're thinking √i = -1, but that's not correct. The relationship is √(-1) = i.

[@brianwilson14]

It is possible if -1 is replaced wIth i^2 and is carried over to m as i^4: 3 - sqrt( (2* (i^4*23/2) +2) = 3 - sqrt( (2* (i^4*23/2) +2i^4) = [added i^4 multiplier to 2 as i^4 is 1 and 2*1=2 3 - sqrt( (2* (23/2) +2) *i^4) = 3 - sqrt( (23+2) *i^4) = 3- sqrt( 25 * i^4 )= 3 - (5*i^2)= 3 - (-5)= 3+5=8 This solution is a bit long-winded and relies heavily on i, but it does work.

[@penguincute3564]

Smart! But… you did the thing sqrt(-1) * sqrt(-1) = sqrt(-1 * -1) = sqrt(1) = 1 Which you cannot do as that.

[@claude77573]

My hangup with this approach is i^4 = i^2 x i^2 = -1 * -1 = 1. So 3 - sqrt (25*i^4) can just as easily be simplified to 3 - sqrt(25), or 3 - 5, which does not equal 8.

[@oldestpi]

@@penguincute3564 sqrt(-1)*sqrt(-1) = sqrt(i^2)*sqrt(i^2) = i*i = i^2 = -1

[@PeteTechGuy]

It’s ABSOLUTELY TRUE that the square root of 25 is plus or minus 5 HOWEVER The definition of the mathematical symbol √ is NOT simply the square root. The MATHEMATICAL DEFINITION of the symbol √ is “the principal square root”. ie. the positive square root. That’s why mathematical formulas like the solution to the quadratic equation include the symbols ±√ Those two symbols used together are referring to both the positive AND negative values of the square root.

[@PhillipRhodes]

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.

[@alexaneals8194]

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.

[@evgtro8727]

@@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.

[@alexaneals8194]

@@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).

[@evgtro8727]

@@alexaneals8194 Ironically I was the 8th grade student exactly at the same years, 1978-1979 🙂

[@raoulduke5690]

I was a math major in a university in 1970. I have NEVER heard of " principal square root." Mathmatics is centuries old, this is a new invented term.

[@CharlesBallowe]

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.

[@-danR]

If I had to grade this video, and most of the comments, on a gray-scale of right to wrong, I'd give it a dark gray. For the greenhouse-plant-algebra of high school, you can go ahead and only allow principal roots (and if you want to pass exams, you better beware), but that may not give you _working_ solutions in engineering and physics. In another reply here, I mentioned a 1899 Algebra (school) textbook that was a little more nuanced, where the telling phrase appears on page 241 "... _if other than principal square roots _*_be admitted_* ..."

[@danv2888]

Exactly what I was thinking. This all comes down to what level of math your studying and what crazy rules your teacher wants to add in.

[@misterroboto1]

​@@-danRNah, man. You're just confusing two different notations: x^0.5 and √x . The former refers to both roots while the latter really just refers to the principal root. Otherwise, why is it that the quadratic formula has a ± in front of the √? It would be redundant if, as you claim, √ refers to both roots in "more advanced mathematics".

[@benprice3586]

I'm sorry you went through that.

[@ianw6352]

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.

[@flagmichael]

Not the Magic Decoder Ring of PEMDAS/BOMDAS again! That should be trashed in favor of using nested parentheses because it is unreliable.

[@penguincute3564]

Technically this is HOW you evaluate something but abiding by this way… isn’t how it works… everything is solved in the same order but we must add stuff to find it.

[@rodrodrigues5402]

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.

[@josephmalone253]

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.

[@bernhardbauer5301]

We see more and more this kind of nonsense.

[@wernerviehhauser94]

​​​@@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.

[@marilynman]

@@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

[@wernerviehhauser94]

@@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.

[@WestPicoBlvd]

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

[@jim2376]

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.

[@mathepunk]

If he says everybody this must include himself by definition.

[@vrmerc]

but √(-5)√(-5) = i√(5)i√(5) = i^2√(25) = -1 *5 = -5

[@michaelragusa5138]

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!

[@ailblentyn]

I haven’t done math since high school, and even I saw the same thing!

[@freefall9832]

I have one semester of college algebra. How do you subtract from 3 and get a bigger number. It doesn't make sense to me.

[@gregsmith4102]

You are limiting yourself to a real number solution. There is no real number solution, just as there are no integer solutions nor rational number solutions. However there are complex number (real plus imaginary) number solutions.

[@methatis3013]

​@@gregsmith4102 no, there aren't. Please provide an example of a complex solution

[@divermike8943]

​. So what is the complex solution?

[@hitest8925]

Should start by checking the radicand for non-permissible values: ie 2m + 2 must be zero or greater or m must be greater than or equal to -1 which identifies -5 as an extraneous root. Cheers.

[@OleJoe]

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.

[@omarjette3859]

Well, for x=0 then f(x) = sqrt(2)-5 = -3.585... so maybe infinitely close to x=0. Could we say "at the limit"?

[@bjornfeuerbacher5514]

@@omarjette3859 "Well, for x=0 then f(x) = sqrt(2)-5 = -3.585" Where did you get the -5 from? The comment your replied to explicitly wrote +5- "so maybe infinitely close to x=0" ??? Where did you get that from?!?

[@divermike8943]

Quick question in terms of symbology: does (25)^1/2 mean the same as square root of 25? I don't don't know how to type a square root symbol on my phone but that's what I meant. In other words does using the 1/2 exponent mean both +5 and -5 are solutions because you are not using the square root symbol?

[@StevenCarr-rh6dd]

I would expect the solution is m=(25/2)i - 1, because 2m+2 = 2x((25/2)i-1)+2 = (25i -2)+2 = 25i. With sprt(25i) = sprt (25) x sqrt(i) = 5 * (-1)= -5 the equation is fullfilled.

[@MarieAnne.]

√(i) ≠ −1. You're confusing it with √(−1) = i In fact i = cos(π/2) + i sin(π/2), so √(i) = cos(π/4) + i sin(π/4) = 1/√2 + i/√2 There are indeed no complex solutions to this problem. Complex solutions occur when we have an equation of the form (ax+b)² = negative value but not when we have an equation of the form √(ax+b) = negative value ² ³ − √ ° ∠ △ × → ~ ≅ π ≈ αβθ ± ≤ ≥ ⁻¹ ⇒ ₁ ₂ ≠ ╪ ·

[@gavindeane3670]

You've got the identity backwards. You're trying to do √i = -1 but it's √(-1) = i.

[@AJHyland63]

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.

[@andrewclifton9772]

Absolutely agree. It is foolish to ignore the negative value of square roots. To say 11.5 is not a solution is just plain wrong.

[@ChuckCreagerJr]

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.

[@misterroboto1]

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?

[@ChuckCreagerJr]

@@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.

[@bjornfeuerbacher5514]

You are totally wrong. This has nothing at all to do with being a real-world problem or not. If you wrote down an equation like 3-(2m+2)^0.5 =8 and claimed that this described a real-world problem with an actual solution, you are simply describing the real-world problem in a wrong way! This equation can _not_ describe a real-world problem with a solution, because this equation has no solution.

[@wardebert8341]

In practical situations this equation would arise while solving a problem with +/- in front of the radical and the steps you illustrate would eliminate the case with the - sign.

[@SaiyanMaster95]

Another way to really emphasize the principle square root as a positive value is using functions. In order to pass the vertical line test, you have to have separate functions, f(x) = sqrt(x) and g(x) = -sqrt(x).

[@scottguitar8168]

It's been a while since taking algebra and I think I vaguely remember this lesson. I think spending a lot of time with quadratic equations can cause one to forget this more basic lesson. When squaring a number, you have to consider both the positive and negative values that work. When taking the square root, unless otherwise denoted, such as in the quadratic equation with + - square root, it is only the positive root and in this case the positive root doesn't work as a solution and the equation doesn't work as a quadratic.

[@flintcoat2596]

Yes, I was one of those doing algebra in the '60's! Retired from Power Industry, but now have Parkinson's Disease and was looking for a channel like yours to try to keep sharp! I saw the null set right away! However, I do not recall having ever heard of the "Principal Square Root" before. Somehow, I learned the concept because -5 for the answer DID NOT occur to me. I remember that quadratic equations result in the answer being plus or minus, but I don't remember how to check those results. Unless you substitute in the original equation and discard the extraneous result.? Not sure. I will be sure to follow up with your other episodes! Semper Fi Bob

[@macfrankist]

I write the square root of x^2 as |x|=5

[@flagmichael]

Valid, but requires splitting the path in two from there on - one for 5 and one for -5. Remember the outcome is one OR the other.

[@macfrankist]

@@flagmichael That is what the absolute value implies no? Guess what. I don't leave it as |x|.

[@tensor131]

A lot of emphasis there on correct maths, which is fine .. but wouldn't it be easier simply to take the original equation, remind the student that the square root symbol means (by definition, e.g. the requirement to append a +- in the QE formula), the POSITIVE root and so the equation is impossible.

[@annelarrybrunelle3570]

When you evaluate a term, you have to resolve a radical before applying a sign to it; there are implicit parentheses around the radical expression. In this example, you can subtract 8 from both sides, giving -5 - (radical expression) = 0 upon which you can add the radical expression to both sides, giving -5 = (radical expression), and you know immediately that won't fly for real numbers. This is somewhat similar to the kinds of multiple-root fallacies you can get in calculating an internal rate of return (IRR) for a business project. It's a pitfall people too frequently fall into in trying to get numbers to sell a project to management, and, too often, neither those who present the numbers nor the management receiving them understand how such unrealistic predictions occurred until the unrealism comes jome to roost. Moral: don't use IRR to sell something; instead, calculate the range of credible NPVs.

[@dono42]

Rearrange the terms and you get √(2m + 2) = -5. The square root is always positive, so this has no solutions.

[@richardvoogd705]

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!

[@benprice3586]

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.

[@freefall9832]

But they are trying to subtract from 3 and trying to get a bigger number. Am I missing something?

[@zweisteinya]

This is equivalent to Gödel respecting “This statement is false” .Both are lies, the work of The Deceiver

[@flagmichael]

@@benprice3586 sqrt 2m = -7 (i.e. -5-2), or we can wrote it as sqrt m = -7/2. No real root.

[@Bunny99s]

I think the main issue is that people seem to remember that when resolving a square root you introduce a "+-" but that's not the case. The "+-" does not come from solving the quare root on the right side but it comes from the left side. x² = 25 means x is a variable and when taking the quare root of x², this is where we introduce the two solutions, not when resolving the sqrt(25). So in some sense it would make more sense to write x² = 25 sqrt(x²) = sqrt(25) +-x = sqrt(25) +-x = 5 => x1 = +5, x2 = -5

[@KimberlyCharles-x2j]

This content was a fantastic addition to my learning!

[@claude77573]

For me, what helps me understand what's going on is to hypothesize a simpler example. What is the value of x where √x = -1? Well, let's substitute the imaginary number i so that i² for -1, giving √x = i². Squaring both sides would give x = i^4, or x = i² * i², or x = -1 * -1 = 1. But substituting back into the original equation, √1 does not equal -1, it equals 1. So, there is no solution.

[@grzegorzkondracki4630]

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?

Thank you for explaining in 10 seconds what the video failed to explain in 17 long minutes!!

[@oinkinollie4792]

Can't this be "solved" by making all +/- agree?

[@felipe-n9t]

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.

[@sparty94]

i'm an adult that's brushing up on my math skills. i find your classes interesting and helpful. thanks.

[@flagmichael]

Please don't rely on the Magic Decoder Ring of PEMDAS/BODMAS. If you are working on somebody else's problem you never know whether it was written correctly (if the expression does not parenthesize enough to be clear it was sloppy from the outset. (Aced my math ACT test, scored second in the school in the MAA competition in my senior year of high school. I have tutored friends and family in math off and on since John F Kennedy was still the US president.

[@ffggddss]

At first glance, there can be no (real!) solution to this equation, because a quick rearrangement of it yields: 3 - 8 = -5 = √(2m+2) which is impossible, because a radical must always be ≥ 0. Bottom line: *No Solution.* Fred

[@mathyeti]

Tidying up, I get -5 = sqrt(2m+2). Since the radical sign implies a positive square root, there is no solution (in real numbers) because the left side is negative. [ Final answer! ] When I check for imaginary roots, I square both sides, and get (-5)^2 = absval(2m+2). Split this up into two possibilities, 25 = 2m+2 and 25 = -(2m+2). This ends up with m1 = 23/2 and m2 = -27/2. Neither of these is equal to 8, so there is no complex solution either. P.S. I also taught high school algebra for 30 years, but I retired 20 years ago, so I'm a little rusty!

[@michaelsanders2655]

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.

[@AvoidsPikes-]

This math problem should only exist to explain null answers (or the existence of crappy math problems.)

[@freefall9832]

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.

[@ArtVanAuggie]

3 minus -5 = 8 Given that a square root cannot produce a negative result, the solution is null. Or you could just graph both sides; the graph stops as m approaches -3 and never intersects.

[@BMGBOX]

Thanks for this great video !!!

[@reedchaber8239]

I simplified question to 3-x=8 Seeing x had to be a negative and intuitively knowing you can't get a negative from square root knew something was wrong. First thought it might be imaginary but that didn't work out either so knew there was no solution. Did I just get lucky or was logic correct..

[@junkdubious]

Why wouldn't this, (1+5/2i)=m be a solution? Does it need to be extended to be continuous to the complex set?

[@mouradbelkas598]

I agree, sqrt(25) = 5, and therefore, there is no solution. 25^(1/2) is not =-5

[@matthewkendall5235]

By observation the square root must be negative five and the square roots domain only covers positive and imaginary numbers - never negative ones - so their can be no solution - real or complex...

[@fizisistguy]

So does the equation really have any solution?

[@tormkg9879]

Good video, thank you. A couple comments. When showing the algebra that leads to x=+/-5 you should include the step in between with the absolute value being the result of the square root of a squared variable. I take that opportunity to explain that we have to consider that if an unknown number was chosen, we cannot be certain what the sign was, and ABS(x) is how we admit that, while the sqrt(25) is 5, always and forever, due to PRT as you noted clearly. The other thing I would have mentioned is that you should avoid squaring negatives away by first moving it to the other side of the equation and observing that allegedly Sqrt(2m+2)=-5 which can't be true, because again the result of a sqrt is always positive.

[@MarieAnne.]

YES. When showing someone how to solve by completing the square, I also show the "missing" step with absolute value. As you say √(a²) = |a| so we can get solution to a quadratic equation as follows: x² − 2x − 3 = 0 x² − 2x + 1 = 4 (x − 1)² = 4 √((x − 1)²) = √4 |x − 1| = 2 x − 1 = ±2 x = 1 ± 2 x = 3 or −1

[@martinbennett2228]

Mathematics needs a symbol for the operation that returns a positive and a negative value for the square root, in other words for the reverse of the square function. There needs to be an operation where the value for -5 (or some other number) squared can then be reversed to return the original value.

[@gavindeane3670]

We have a symbol if you want to refer to both the square roots: it's ±√25.

[@hdfoster5507]

When I was in school, back in the ice age, we called the 0 (null) answer as "undefined''; does that answer still apply today?

[@davidbroadfoot1864]

"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} = ∅".

[@knutholt3486]

Well, the equation has no solution if you stick to a rigid definition of a square root as something always positive, but if you allow yourself to let a negative number also be a square root, it indeed has a solution. What it also shows is that mathematics is not entirely without faults.

[@WK-5775]

No. It just shows you didn't understand square roots. Apart from that, in mathematics one should always stick to the definition (call it "rigid" if you want) and not allow yourself to let other things be something which they are not.

[@knutholt3486]

@@WK-5775 I understand that thing very well. But this implies that I also understand that what is wrong in the first place, is math itself, as it is formulated.

[@flagmichael]

Not a fault, but one of many real world quirks. I would go outside and take a picture to demonstrate, but I can't get through my -8 foot tall door.

[@davidnelson4960]

I see. Isn't there a subtle difference further down the road which provides a subset of quadratic which recognized nul sets with negative solution?

[@gregsmith3406]

This was informative. Thanks!

[@MrMousley]

3 - square root of (2m + 2) = 8 subtract 3 from both sides square root of (2m + 2) = 5 square both sides 2m + 2 = 25 subtract 2 from both sides 2m = 23 divide both sides by 2 m = 23/2 = 11 1/2 Is that it ? I'll check now.

[@christopherellis2663]

Square BOTH sides

[@MrMousley]

@@christopherellis2663 Yes .. square BOTH sides of the equation The square root of (2m + 2) SQUARED is 2m + 2 and .. on the other side of the equation .. 5 SQUARED is 25

[@tom-qj6uw]

3 - square root of (2m + 2) = 8 subtract 3 from both sides square root of (2m + 2) = 5 square both sides This unfortunately is wrong. You get - square root of (2m + 2) = 5 square both sides Don't simply lose the '-'!!!

[@harrymatabal8448]

There's other ways of becoming famous

[@HerrFinsternis]

Hey this is interesting. Without having watched the videos, here's some thoughts. √(2m + 2) = -5 can't be solved on the real plane, so it must be a complex something. It looks like the modulus of a complex number [Z_modulus = √(a^2 + b^2)] Z_modulus = √(2m + 2) = -5 Write that as the conjugate [Z_conjugate = a - ib] and you get the following I guess? Z_conjugate = √(2m) - i√2 = ??? Don't really know how to continue, or if these even remotely made sense or is useful.

[@BalthasarCarduelis]

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.

[@JacquesLafont]

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?

[@aek03030731]

It looks like there is a solution set.

[@user-gr5tx6rd4h]

7 is the ONLY answer

[@JacquesLafont]

@@user-gr5tx6rd4h Of course, this also my opinion, because I think one can only write sqrt(16)=4 and not +/-4.

[@davidbroadfoot1864]

IF you defined the sqrt() function that way, then the expression in question has four solutions. The set of solutions is {-7, -1, 1, 7).

[@flagmichael]

@@JacquesLafont There is a workable solution if the domain of the answer is complex numbers or the numerical solution is in the realm of imaginary numbers (sometimes that is true).

[@maxvangulik1988]

sqrt(2m+2)=-5 but since the principal branch of the square root function is strictly positive, there are no solutions.

[@tulliusagrippa5752]

I’ve heard of algebra. But what is algeba? Is it non-rhotic mathematics from which the letter “r” is expelled?

[@generaclesdey4622]

There is no way to subtract the positive radical from 3 and get a positive value 8!!

[@HimitsuYami]

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

[@bjornfeuerbacher5514]

And you made exactly the error the video warns about.

[@tinogabriel5096]

The step "square both sides" is fatal ... because if that was allowed as a proof then 7 would be equal to -7 : ... 7 = -7 square , so: 49 = 49 ...

[@derekloukes6225]

You should have moved the "-" to the right, instead of squaring both sides. You would seen square root of something = -5. The only answer to that is with complex numbers! The square root of -5 = 25i which then requires 2m + 2 = 25i but that is only possible if m = 12.5i - 1 and I don't think that algebra will allow m to be a combination of real and imaginary components.

[@yetidynamics]

that was really informative, if you rewrote it as 3-(2m+2)^(1/2) = 8 would there be a solution? i'm guessing no

[@GaryBricaultLive]

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.

[@geeache1891]

Got the error directly: the r is missing!

[@TheFallibleFiend]

Step two is either sqrt(2m+2) = -5 or you can use your step two and make this step three. Once you get to this point, you have a square root equal to a negative number, but square root always means ONLY the positive square root. There is no square root that can equal -5; therefore, there is no solution.

[@Magnus_Loov]

Isn't (-5)*(-5)=25? And that means the square root for 25 is +/-5. I was always tought by all teachers that there can be two square roots, a positive and a negative, if not stated explicitly before the square root sign with a plus-sign. I have never ever heard of "principal square root" and I don't get why it is used here. Still see it is two possible answers with either 23/2 OR no solution. Maybe this is a new convention that wasn't taught in the 80:s? (or maybe no in Sweden?). I mean many others writing in here who were taught it decades ago seems to think the same thing.

[@davidbroadfoot1864]

@@Magnus_Loov It is not new. It has been that way ever since the radical sign was invented in the year 1450.

[@anastasiyad6038]

All you need to do is to multiply each side of the original equation by (-1) in order to get rid of the negative sign in front of the radical. You will get an identical equation but will not have to deal with the confusing negative sign. Thus, you will get [-3 + radical = - 8], which leads to [radical = - 5], which is impossible for real numbers (unlike for imaginary ones).

[@gavindeane3670]

It's impossible for imaginary and complex numbers too. There's no value for m, real or complex, that gives √(2m + 2) = -5

[@pennstatefan]

The first step is move the 3 to the other side - (2m + 2)^1/2 = 8 - 3. square both sides and one gets 2m + 2 = (5)^2. The final solution is m = 23/2

[@flagmichael]

The problem arises in "square both sides." For example, take the equation 10 = -10. Square both sides: 100=100.

[@Raulitango]

However, if one multiples by -1 both sides and subtracts -3 on both sides and powers 2 both sids one ends up with 25=25 and 23/2 then seems to be the answer!

[@tomlake2732]

I multiplied both sides by -1 before squaring both sides.

[@EinSofQuester]

3 - +-5 = 8 The solution works for the minus option of the +-5 If we say there is no solution, that's only true if we adopt the convention that the root if x is only the positive option. But mathematics transcends human convention. So I disagree that there is no solution.

[@knutritter461]

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...! 😉

[@Astrobrant2]

What I noticed right off the bat is that √(2m+2) has to be -5. A real square root can't be negative.

[@GaryBricaultLive]

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.

[@Astrobrant2]

@@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.

[@davidbroadfoot1864]

@@GaryBricaultLive Your first and last sentences contradict each other.

[@wernerviehhauser94]

Not even complex numbers can have negative principal roots.

[@flagmichael]

@@wernerviehhauser94 There is sometimes value in the negative aspect of a square root. If we are going to fall short, it may be helpful to know whether we can come up winners next time. It won't work this time, but may define a fix next time.

[@ganterpv]

Isn't the square root of 25 plus or minus 5?

[@petertiger1134]

Right!!

[@whoff59]

no.

[@ganterpv]

Please explain.🙂@@whoff59

[@arnothar8035]

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.

[@davidbroadfoot1864]

no

[@stevenpace892]

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.

[@wernerviehhauser94]

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.

[@flagmichael]

Yes and no. The only time I used my advanced math skills in the real world was to calculate the value of a series capacitor I had to put in line with a wire antenna to bring the capacitance down to spec. If the value had been imaginary I could have looked at shortening the wire antenna because it would not work this time.

[@davea136]

There are no real roots because the radical indicates only the positive square root. Which is why in things like the quadratic equation they explicitly state +/- the radical. You are welcome.

[@boringusernamelmfao]

this video was extremely helpful, thanks bro

[@christophernuzzi2780]

Title should be: "The Spelling Step that WE Got Wrong!"

[@Man19699]

Great,many thanks !

[@bjornfeuerbacher5514]

You could have made the point clearer by using a simpler equation, i. e. sqrt(x) = -1.

[@russelllomando8460]

great lesson. negative radical stops everything. i got the 'null' as unsolveable.

[@stevelc777]

what is Algeba?

[@johnburgess2084]

Although there is no REAL-number answer, is it legitimate to answer by introducing the i = √(-1) and getting an answer with an imaginary number?

[@feathersfeathers2022]

Subtract 8 from both sides U.tiply -1 on both sides Results. 5 + sqroot (2m+2) = 0

[@markmurto]

Retake 10th grade. Sqrt [a] can not equal -b. No solution by simple observation.

[@RealMesaMike]

OK, since 5 is already greater than 0, you can see there is nothing you can add to it to bring the answer down to 0, since SQRT(X) cannot be < 0.

[@i_am_a_gugugu]

√(2m+2) = a 3 - a = 8 , -a = 5 , a = -5 √(2m+2) = -5 2m + 2 = 25 2m = 23, ==> m = 23/2, is m = 23/2 really the correct answer? No, there is no answer to this problem. Because √ only gives us +, not -. Maybe.. to summarize, it would be this. " a-√x ≯ a " ( Even if it sounds weird.. please understand..!! ) - someone used Machine translator

[@wes9627]

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.

[@StephenRayWesley]

You have became follower not lead. You have that type presonilty be different do thing on way

[@StephenRayWesley]

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

[@markmurto]

You need to go back to kiddy math. SqrtA can not equal -B No solution. Public school has so failed the majority!

[@whoff59]

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.

[@flagmichael]

@@markmurto That gets into splitting hairs in the real world, particularly in AC electrical situations and in electronics. Half a century ago tunnel diodes were a darling of radio electronics; if biased properly they exhibited *negative* resistance over a small operating range and the speed was effectively at the speed of light. They would still be a major factor in some electronics today if they didn't have a problem of short life (months or less, depending on the service). In the domain of real numbers (most of the world around us) you are right, In some of the darker paths of exotic tech, you are left behind.

[@bhaktabachalabehera2654]

To eliminate that - on left side can't we just multiply -1 on both sides to get (-5) on RHS and solve further to get 23/2 or 11.5?😅😅

[@gamer122333444455555]

How do you show that you want the negative square root? There is a standardized way to do so right?

[@davidbroadfoot1864]

Put a minus sign in front of it. Or put a +/- sign in front of it if you want both.

[@gamer122333444455555]

@@davidbroadfoot1864 thanks

[@migsy1]

What software do you use to make these?

[@raynewport9395]

Not just everyone, everyone in CAPS. Shock. Horror. You won't believe what happened next.

[@flagmichael]

The answer will amaze you!

[@devonwilson5776]

Greetings. The answer is M=23/2 This is we do it. We rewrite the expression as-(2M+2)^1/2=5, and (2M+2^1/2)=-5 after dividing both sides by negative. Moving on, we square both sides to get (2M+2) =25, and 2M=25-2 for 2M=23, and M=23/2.

[@MrSeezero]

... but you still have to use the original equation to check your answer. Even if you divided both sides by negative one after pushing all the non-radical stuff to the right side of the equation, you still would end up with 5 -5 to re-indicate that there is no solution. Whenever you take a radical, and put it to whatever integral power (In this case, the integer power is 2.) is needed to get rid of the radical, then you need to check your answer(s) if you get at least one using the original equation.

[@1234larry1]

This equation is an inverse function of a quadratic whose y intercept is 23/2.

[@markwarner327]

yeah, and I get the answer of null and 23/2 for the function

[@såshäÖ2024]

​@@markwarner327🤓um actually 23 divided by 2 is 11.5

[@ralvarado2764]

A square root always has a positive and negative valued solution. The problem can be solved by squaring the equation Z = -5 where Z = SQRT(2m + 2) and then converting this equation into the quadratic equation: ZZ-6Z-55 = 0. This can be factored yielding the solutions m = 59.5 and m = 11.5

[@philippebrillault1140]

It seems to be a contradiction in your video at around 13:37 According to the principle of the square root (PSQR); SQRO(25) = 5 [1] (writing SQRO(25) - the SQuare Root Operator - and not sqrt(25) - the square root function -) Later, you solved the quadratic equation X² = 25 [2] SQRO(X²) = +/- SQRO(25) [3] SQRO(X²) = +/- 5 [3] As, according to [2], X²=25, replacing X² by its value in [3] gives: SQRO(25) = +/- 5 [3] which contradicts with [1] saying SQRO(25) = 5 or … your equation [3] is ill-stated and [2] should be resolved otherwise: X² - 25 = 0 coud be factorized as (X-5)(X+5)=0 hence the 2 roots R1=+5 and R2=-5. There is no need to solve it by using the square root operator.

[@Jahwobbly]

What's algeba?

[@harrymatabal8448]

Mr michael intelligent people dont boast. A high school learner will be able to give the answer

[@wmfield152]

Why is the answer that works not the correct anwer? When considering functions to determine Domain & Range values, there are some possible answers that are calcuated from the same algebraic expession. Some answers are workable and some answer may not be. Those that aren't workable are simply rejected for further use. This is the case with the expression considered here. One answer works and the other does not. The value in this lesson is that there are two possible answers to be considered. However, being consistent in mathematical principal requires using the workable answer. Therefore, the correct answer is m=23/2.

[@daemeonp]

That's amazing handwriting for paint

[@nobleneckbeard7356]

Why do you have to plug it in to "check" if it's a solution. Why isn't the answer guaranteed from doing the algebraic steps?

[@stevenjohnson1143]

First step is to get the radical alone by subtracting 3 getting -√2m+2=5 dividing by -1 gets ✓2m+2=-5 squaring both side creates 2m+2=25 subtracting 2 from sides 2m=23 dividing by 2 creates m=11.5 or 23/2

[@kee1zhang769]

I assume you are dealing only with real numbers. Thus sqrt(2m+2) if exists is >=0 sot So 3-sqrt(2m+2)

[@disraelidemon]

The algeba step everyone gets wrong - missing out the "r" in "algebra" :-)

[@wachu8522]

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.

[@bjornfeuerbacher5514]

How does PEMDAS help with the square root? I don't understand what you mean.

[@flagmichael]

For many moons I have tried to get this channel to stop pushing PEMDAS/BODMAS. Math is universal; PEMDAS and BODMAS are English only; the acronyms are for English words. A quick google search shows english is only used in 1/6 of the world.

[@bjornfeuerbacher5514]

@@flagmichael Yes, English is used as the _main_ language in only 1/6 of the world. So what? Nevertheless, it's a language which is _understood_ by most people on the internet. Myself, I am German. But nevertheless I have no problems at all when an _English_ KZbin channel uses _English_ abbreviations like PEMDAS.

[@TheArchitect316]

Shouldn’t the correct answer be 59.5i ?