99% of Math Students Get This STEP CONFUSED! 

Using the null set is both a cheap shot and incorrect. m is not equal to the null set. You could say the set of solutions is the null set. But m is not a set at all.

REBUTTAL, RESPONSE REQUESTED: Okay after watching the rest of the video, I have a rebuttal that I would be very interested to read your response to. Please expand the comment and read it all. When I was initially solving this before checking my answer, I took an extra step that initially you demonstrated as unnecessary by omitting it before squaring both sides. Here's the approach I took initially--same as yours but with the extra step: 3 - sqrt(2m+2) = 8 -sqrt(2m+2) = 5 sqrt(2m+2) = -5 ***

Just re-arrange to sqrt(2m + 2) = -5, and then recognize that the square root does not result in a negative number. Hence, no solution. No need to go further than that.

The problem is that the square function is not invertible. For example both -2 and 2 both yield 4. So when we are just given the answer 4, there is no way of knowing which value was originally plugged in the square function. When "both side of an equation are squared", essentially both sides are sent through the squaring function 1) a = b 2) square(a) = square(b) This is a valid operation since a function when given the same arguments, always yield the same value. However, the information where we are originally were coming from is lost. Hence we have to perform the check. But this is only necessary when we apply non invertiable functions, if we had applied the exponential function for example, which is invertible 1) a = b 2) e^a = e^b we wouldn't have to perform the check.

Definitely learned something on this lesson!

I would recommend on similar problems to show the root and solution set with the quadratic, ie x^2 - 25 = 0 => (x-5)(x+5) = 0 => x = 5 or -5

Well, the problem definition is wrong. You need to specify what is a domain for m. An argument that follows assumes it is R (real numbers) but it is not specified anywhere. And if domain for m is in Z (complex numbers) solution exists

Try a graph. rearrange the eq: subtract 8 from both sides and add to both sides the sqr thus: -5 = sqr(2m+2). Make a graph. (use the graphing program DESMOS if necessary.) graph the left side: y = -5 which is a horizontal line 5 units below the x axis. Graph the right side: x=sqr(2m+2). Notice that there is a left limit of x = -1 and a lower limit of y=0. Therefore, the two graphs do not intercept and there is no solution. QED

I thought for sure this was going to be a trip into imaginary numbers where the variable i is equal to the sqrt of -1. Consequently i^2 = -1. But eventually we are unable to resolve m related to i, so we still end up with no solution.

That's why always encourage students to test their solutions in the original equation and see if it balances.

Thank you for these! I got 23/2 but now when you explained it I’m starting to remember the rules 😃 Awesome!!

For x^2 = 25, that can be solved by x^2 - 25 = 0, which factors into (x-5)(x+5) = 0, which shows the two roots of +5 and -5. At first, I was thinking that 11.5i might be a solution, but quickly realized that didn’t work either. If it had been a quadratic under the square root, an imaginary/complex number might have been a solution.

Per your instructions at 1:08, I came up with m=2-12.5i. It's been 20 years since I've needed to be unreal, due to my current job, so this took me an hour of pondering...on with the video! After video: We can't get there from here! Thanks for the cool puzzle! Fun stuff for sure. A trick I learned long ago is to always estimate what the solution is before starting. With that, I knew that the term with the radical both must be negative and cannot be negative so to pointed me to a non-real solution. What I failed to do is to plug-in my proposed solution as a test to see if it worked.

I noticed that a lot of physicists use this term of ''extraneous solution'' when in quantum mechanics some numbers in their equations don't want to fall in perfectly. Also, they use a lot of number ''i'' which is imaginary number sq rt of minust 1.

This is so much easier to see than that... if you really do your leg work right, you will transform the original problem into sqrt (2m + 2) = -5 and if you now try to express -5 as a square root, you will immediately see that it is sqrt(25), which has two identities 5 and -5 which must both satisfy any condition they appear in; since however it is explicit already that _only_ -5 appears in this context, you will know immediately that this problem has no solution.

I was wondering how he was going to talk for 18 minutes when you can see in the blink of an eye that the equation reduces to sqrt(2m+2)=-5, for which there is no solution, which you can explain in a minute (or 2 minutes if you’re long-winded).

If infinity=-1/12 then we can also say -2=8

When I teach students SAT math I refer to a problem like this as a 3 second problem. That's a problem that can be solved in 3 seconds or less. I teach my students to subtract 3 from both sides and immediately you can see that a negative can never equal a positive so there is no solution. Understanding concepts allows a student to solve problems quickly!

Very clear explanation . Thank you

Outstanding lesson. Thanks.

55 years ago Mr.Greer at my high school drummed into us “square roots are ALWAYS positive” I can hear him say this every time I see a square root. Thanks Mr. Greer!

The square root can never be negative, when working with real number, but with complex number this is ok, so the solution in this case is a complex number (i=sqrt(-1)). Let have this basic equation : sqrt(x)=-1, -1 can be replace by i^2, so x=i^4. In this specific problem, you can never go back to real number, because: sqrt(-1^2)=1, but sqrt(i^4)=-1. So the solution for this problem is: m=25/2 i^4-1

I did the problem in my head. After subtracting three from both sides of the equation, I squared both sides, simplified, and got 23/2. But when you plug it back into the equation, it doesn't work. Then you look carefully at the equation, and realize it can't possibly have any solutions, or at least real ones (I didn't bother to consider complex any ones.) So the answer is the null set.

I get to -5 = sqr(2m+2), and square roots cannot be negative. That's as far as I can get. Unless things have changed. It has been over 50 years since I had any algebra. There was something that seems to have changed in the order of operations.... At least to me. But that is not an issue here. Well, I am relieved to see that there is no answer. But it should be kind of obvious, because there is no number that squared can be a negative number, right?

If you view the problem in the complex numbers, 23/2 works just fine. Call it magnitude 23/2 and direction 2*pi. Double and add 2, and you get 25 with direction 2 pi. Take the square root of that, and you get 5 with direction pi, which is -5. Subtract that from 3 and you get 8.

this is probably incorrect, but I decided to multiply both sides by -1 (but I'm not sure if that is legal with a square root) so ( -1 ) x ( -sqrt (2m+2) ) = ( -1 ) x ( 5 ) result is sqrt ( 2m + 2 ) = -5 then, if I recall my algebra rules (50+ yrs ago), you cannot multiply any number by itself to arrive at a negative number. The only way to get a negative number as the result of multiplication (of 2 numbers) is: Only 1 of them can be negative. Two negatives = a positive, two positives = a positive. Therefore in my little brain, there was no solution, although I remember being confused by irrational numbers with the little i symbol, and hated them.

What if you looking for a solution in the form m=x^2, and in the end you find m from x ?

Fantastic explanation!

I was hoping this would include complex numbers.

I was taught that the SQRT (x^2) = |x| That way you get both positive and negative values when you set |x| = 5. This allows both x = 5 and x = -5 as solutions. It also keeps consistent having the SQRT of 25 be ONLY +5.

Good to see you stayed on task with this problem. Exreaneous explanations often just complicate matteres for studens.

I've never heard of principal square root before. I saw 23÷2 wasn't a solution but kept looking for a solution but couldn't find one. But did not make the final leap to conclude that there is no solution.

Excellent explanation. What I did was to reduce the second term in the equation to a single variable (?). Thus, you say: 3-a = 8. (a is equal to square root of 2m +2). a must equal -5 for this equation to be true, thus there can be no square root. 5 is the square root of 25, but it is ALWAYS positive, thus, there is no solution for -5. The equation is false, according to conventional mathematics. However, I am now wondering about i. The square root of negative one. The imaginary number i would solve this equation, I think. 😊😊

I also spend ages to get students to understand that x=sqrt(4) and x^2 = 4 are not equivalent expressions in the realm of real numbers. At least, we don't have to do the disambiguition between "root" as sqrt(x) and "root" as solution to an equation in german since they are called "Wurzel" and "Lösung / Nullstelle".

For reference, I asked the following question to the AI, which seemed to be unreliable, and received the following answer. "What problems arise when using the definition of √1=±1?" "First, if we define √1=±1, the properties of the square root will be contradictory. Typically, √a is a function that returns a non-negative number. For example, √4=2, but not -2. Adopting such a definition makes the properties of the square root inconsistent. Adopting such a definition may also undermine consistency with other areas of mathematics. For example, in areas of mathematics such as calculus and complex numbers, √a usually returns a non-negative value. Using such definitions can lead to incoherent mathematical theories. Furthermore, using such definitions can also complicate the handling of solutions to equations. For example, the equation x^2=1 usually has two solutions: x=±1. However, if we define √1=±1, there are an infinite number of solutions to this equation. Due to the above problems, the definition √1=±1 is not generally adopted. In a mathematical context, √a is usually defined as a function that returns a non-negative value."

What if you allow an imaginary solution? We would need sqrt(2m+2) to be -5, yielding 3 - (-5) = 8. If we try to solve for sqrt(2m+2) = -5, we can try setting 2m+1 = 25i^4. Solving for m = (25/2) i^4 - 1. (Yes, I^4 is 1, but want to keep it in this form for now). Checking sqrt(2m+2) = sqrt [ 2 ((25/2)i^4 -1) +2 ] = sqrt(25i^4 - 2 + 2) = sqrt(25i^4). Taking sqrt(i^4) = i^2 we obtain: sqrt(2m+2) = i^2 sqrt(25) = -5 as required.

Thank you for the helpful videos I passed my math test thank you Lord thank you Lord at the age of 34 I am class of 2023 I got my GED

It answers my confusion in math. To make it clearer: If x=sqrt(25) then x is the positive value of sqrt(25); as the sign of x as well as sqrt are positive. However if x²=25 --> x=±sqrt(25), as by taking square root there are two value of x satisfying x²=25. A lesson here is when squaring is involved in solving math problem always plugging back the solution to the original equation to check whether a solution is correct or not.

Very good video…thanks!

But at 16:45, can’t you just move the +5 across so that it becomes 3 = 8 - 5?

Pretty hard to subtract anything from 3 that makes it greater than 8 no reason to look farther. Take a quart of water and poor enough out of it to make it equal a gallon.

What I usually do in these cases is to simplify the equation and then find the root of the equation (P(x)=0). Then try to figure out by inspection if there are any values for the variable(s) that can result in 0. For this specific case we get -sqrt(2m+2)-5 = 0. That means that sqrt(2m+2) has to be equal to -5 so that -(-5) = 5, then 5-5 = 0. This is impossible because there is no way the squareroot is going to be negative, as the negative of a square root is undefined (unless we go into imaginary numbers), so the equation has no solution. There are a lot of problems in algebra that can be solved easily by inspection like this, and it's actually done a lot later in higher courses.

I have a associate in mathematics and I was cringing at first because I like the negitive five answer at first then I had to look up the principal square root or rules of square root after watching this and then I can see I am rusty at math. THIS IS CORRECT!!! Thanks for the reminder 😁😁😁

Can we just change square root to power of ½ and solve as a quadratic?

BRB, more complicated than I originally thought, 1 sec I'M BACK! First off I'm no mathematician, I am a factory worker who happens to have a associates degree in math. This is a thought experiment, just trying out a strange idea. So I had a think on this. I hate empty sets and in mathematics he is right, empty set....however math is also a playground for cheating. Lets be weird and see where it goes, I played with this idea and got a strange property of a ghost negative sign. Let the number j = a number in some reality where j=1 with the property that the sqrt(j)=-1 Note: sqrt(4j) =-4 sqrt(4/j)=-4 thus we can postulate that while j*j should result in j^2, j = its reciprocal of 1/j, thus j*j can in fact = 1, kind of like a ghost - sigh that just cant interact with other numbers from our reality, which if true can be very useful, I'm not going to try to be rigorous, this is just for fun, but if some egghead can prove this is wrong, be my guest, i would be very interested in the way it fails. So on to the problem, in "j space" 3-sqrt(2m+2) = 8 sqrt(2m+2)=-5 (2m+2)j=25 2m+2=25/j (note 1/j = j, it at least functions as such here, so for ease of writing its just 25j) 2m+2=25j 2m=25j-2 m=12.5j-1 back substitute (going to do both 12.5/j and 12.5j) for 12.5/j 3-sqrt(2(12.5/j-1)+2)=8 3-sqrt((25/j)-2+2)=8 -sqrt(25/j)=5 -sqrt(25)/sqrt(j)=5 -(25/-1)=5 5=5 for 12.5j 3-sqrt(2(12.5j-1)+2)=8 3-sqrt(25j-2+2)=8 -sqrt(25j)=5 -sqrt(25)/sqrt(j)=5 -(-25)=5 5=5 The thought process behind this is how did the 'i' we know and love come into being in the first place, I'm sure I'm not 100% accurate on how this all works, and some quirks exist but preliminary testing on how it functions seems interesting, if largely useless. The whole goal of "imaginary numbers" is to find 2 of the same impossibilities and have them cancel each other out, j in this case is just a placeholder for "this is an impossible thing, find a way to delete me in a larger set of equations" but a more useful (by useful i mean descriptive) answer in the real world than "NO!" Basically the takeaway of this thought experiment, is is thus, you can encapsulate an impossibility under a flag of a variable as long as its attributes remain respected and in turn can solve otherwise impossible problems. The usefulness of this is when you hit a wall in a set of calculations because something is 'impossible' you may encounter another 'impossible' problem of the same caliber later in the equation, and you can effectively multiply 2 impossible problems together to get a possible solution even though the second problem might happen far down the road. that's why 'i' works....if this is not a field of study, it really should be. The only problem I see is how to procedurally introduce j, from what I've tested it can only come into being if the person solving the problem acknowledges that squaring a root with a negative result is impossible in the first place. It's less cut and dry obvious than "i". It's a legal move in math proofs, it's little more than marking and tracking a result so you can return to its source accuratly. So here's the best procedural use I can figure out that does not break any rules. 3-sqrt(2m+2)=8 Sqrt(2m+2)=-5 Ah there we go, brainwave. Let j equal a number in another reality such that j=1 and sqrt(j)= -1 Rewrite Sqrt(2m+2)=-1×5 Sqrt(2m+2)=sqrt(j)×5 (sqrtj = -1) (Sqrt(2m+2))/sqrtj=5 (2m+2)/j=25 2m+2=25j m=12.5j-1 Perfectly legal QED, j exists.

I passed Algebra, Trig, Geometry, and Calculus...... 50 years ago. I also taught all but calculus to my own kids and other home schooled kids. I also co-coached a Math Counts Team, which won the reagion and placed 10th in my state (my kid was one of them, bragging here) . But I was stumped. Of course, now that you say the answer it makes sense.

Then what would the result be if you had (1 - i)^0.5?

Can spmeone explain why (25i - 2) / 2 is no solution? because if I plug this into m it results in (25i)^(1/2) which I can write as 25^(1/2) * i^(1/2) = 5 * (-1).

y = x^2 is a quadratic and has 2 solutions only one of which works here so the correct answer is 11.5* where * means only the -root works. To say there is no answer is wrong, there is an answer but it is qualified.

The square root of a number gives the nonnegative result that, when multiplied by itself, equals the original number. The reason for this is to make the square root function a well-defined function (i.e., each input gives exactly one output). Therefore, when you ask for the square root of 25, the result is 5. On the other hand, when you are solving an equation like x^2 = 25, you're looking for all possible solutions, not just the nonnegative one. Since both 5^2 and (-5)^2 equal 25, the solutions to the equation x^2 = 25 are x = 5 and x = -5. This is due to the property of real numbers that states both a positive and a negative number squared will result in a positive number. So, the square root function and the equation x^2 = 25 are asking slightly different things: the former asks for the nonnegative number that squares to 25, and the latter asks for all numbers that square to 25. That's why the square root of 25 is just 5, but the solutions to x^2 = 25 are x = 5 and x = -5.

went to 23/2 but then i plugged it in and noticed it gives -2=8. then started trying to consider answers using i but got weird stuff that didn't work and concluded there was no answer

How long does it take to cover 100 feet at 70 miles per hour?

I think, perhaps we should not exclude -5 as one proper solution of sqrt of 25 completely. Mathematics often shows us, that things, which look weird at first glance, turn out to be useful in several cases. At least we can keep it in mind...

I got a solution that I only saw one comment here has also got. My solution is m = (25i^4/2) -1, because when you plug that into the original equation, the sqrt(i^4) will give us i^2, which is equal to -1, making it possible for a square root to result in a negative number. Although you can say that i^4 = 1, I suggest leaving this number notated as i^4, so the equation can be solved.

Before the squaring operation, I multiplie both sides by -1 to get the left side positive. It is true that the interim result for the right side will be a -6. this will ot matter as the squaring process gets ride of the -5 problem. If the slight modification of the initial equation by -1 is allowed,, whicy would better agree with the intent, then the solution would be valid.

What is the solution if we allow imaginary numbers? solving sqrt(2m+2) = -5

Interesting. I've had math through ODEs, and never encountered this! But, SQRT(--25) = j5. So, there's a solution, but it's complex.

Why is that no one bothers to explain to students that the square root symbol ( √ ) represents the power of 1/2 ( √x = x^½ ) which makes more sense hence squaring it rids the problem of the square root symbol since it ends up becoming a power of 1 🙃 I wasn't made aware of this till my first year of college Calculus. I guess most teachers just ASSUME their students realize this huh 😮‍💨 Also, if you graph x^½ you will see it's half a parabola turned on its side with no negatives hence it can never have more than one solution unlike x^² which can have up to two possible solutions.

About 03:00f This is a bit sloppily written since m is meant to be a number rather than a set of numbers, right? Actually, {m∈ℕ | 3 − √(2m + 2) = 8} = ∅.

Fascinating! Real food for thought!

No, I disagree. Taking the so called principal square root is a convention. You can take either. It depends on the context.

This not a mathematical issue. It's a semantics issue, because math is simply a tool we use to find solutions to numerical problems. So it doesn't really matter how someone might want to classically define the proper function of an operator like say a square root symbol so long as it results in a solution that is practical. As to this problem, whatever real world situation that may have produced the equation, the fact that it has but one unknown intuitively suggests it has a solution. Therefore in noting that 11.5 might be a solution, the real test is to apply that solution to the orignal real world situation and see if it works there. If so, that is where your happy face A++ belongs. And who cares if it runs afoul with how someone wants to define the square root operator.

Thanks for the clarification.❤

Can someone explain why you can't square both sides as the first step (so, the whole numbers 3 and 8 would be squared, instead of subtracting 3 from 8 and squaring 5)? Doing this gives an answer of -53/2, which is obviously wrong, but what is the rule that says you cannot square everything first?

The square root (s) of any number (x) can have the value -s or +s. If you are going to teach math you need to know this and teach it. The solution is 23/2. You should not be teaching math. Example: the square root of 25 is +5 or -5 because if I square either one of those I get 25. Are all your math books from junior grade school?

3 - √((2m) + 2) = 8 To solve this equation, we can follow the following steps: 1. Subtract 3 from both sides of the equation: - √((2m) + 2) = 8 - 3 - √((2m) + 2) = 5 2. Square both sides of the equation to eliminate the square root: ((2m) + 2) = 5^2 (2m) + 2 = 25 3. Subtract 2 from both sides of the equation: 2m = 25 - 2 2m = 23 4. Divide both sides by 2 to solve for m: m = 23 / 2 Therefore, the value of m is 23/2.

I think, you get the problem clear, when you look at thr two underlying geometrical functions y=x^2 and y=x^(1/2) plot this geometrical and you will find out, that y=x^2 is defined for negativ values of x, for 0 and for positiv values BUT this is not the case for y=x^(1/2) . This function is only defined from 0 to positiv values of x up to,positiv infinity When you want to have the negativ part of the horizontal parabel, you have to use another function. y = - x^(1/2) But again also for this function, their are no solutions for y , when you put in negativ values for x. That helps me a lot to understand the mathematical rule: there are no solutions, when you want to have the root of a negativ number.

What about imaginary roots? Maybe, m = (i^2)*23/2, to force the negative square root solution. This problem is the collision of the rules department and the solution department. At first glance I thought there would be no solution but then I said to self to check the imaginary roots. Again, this may violate the rules?

No solution to this problem

Your teaching method is why students come to me so ill prepared. You can try letting your students think! Your students do know that 3 - some number = 8 will result in that some number equaling a negative number and the square root of no number is ever negative. Done! Even if they isolated the radical, as you did, they should know that there is no solution at that point.

Actually m=phi is not correct, because the Greek letter phi refers to the empty set, which is not a real number. It would be more correct to say that the set of solutions is empty.

by putting the value of m=23/2 we get 3 -√25 = 8 3 - (5 or -5) = 8 when 3 - 5 is not equal to 8 we reject the value. when we take m= -5 we get 3 - (-5)= 3+5=8 so m = -5 satisfy the equation.

Is there an answer that involves a complex number (imaginary + real)?

If we let m=12.5*e^(2pi*i)-1=23/2 since e^(2pi*i)=1, the radical becomes [25e^(2pi*i)]^(1/2) and using exponent rules, this becomes 5e^(2pi*i*[1/2])=5e^(pi*i)=5*(- 1) = (- 5) so then we have 3- -5= 8. It requires a trip to the complex numbers, but concludes sqrt has a valid positive and neative solution in the complex world, and the question didnt assume real numbers.

What if we multiply both terms by the conjugate of the first term, are we still come to the same empty set solution?!!!!؟!!!!?

When I reached -√(2m-2)=5, I stopped there, because a square root shouldn’t give a negative physical number, even with a complex number under the square root as far as I remember.

So √ -1 = i and i is not positive nor is it null.....although +i is pos as opposed to -i

@5:00 you correctly state that sqrt()=N. But at @6:00 you fail to rearrange fully to that point. Multiplying through by -1 returns sqrt(2m+2)=-5, then squaring both sides returns 2m+2=25, and this simplifies to m=23/2.

Why is m not equal to 25i/2 -1 ? Thanks

what if 2m+2=25exp(2πι) is that permissible

Square root is always a positive number, or 0. So there is no way to get a solution of 8 when you subtract a positive number from 3. No calculation necessary in this case. You see immediately there is no solution.

Empty set? Isn't that mixing apples and oranges? If for example real numbers are used I don't think that the empty set is a member of the set of real numbers.

But WHY is PLUS preferred over MINUS?

m should be equal to (25/2*j^4)+1 where i^2=-1. In this way the resulting number is equal to - 5

Paused at 1:14 to follow instructions. I did this in my head from the thumbnail before clicking the video and I got 11.5 (or 23/2). Paused at 2:59 after what you said prompted me to check my answer. Sure enough, 23/2 doesn't yield equivalence between the two sides of the equation, as -2 does not equal 8.

OK so you arrive at m=11.5 and you check by plugging the numerical value back into the original equation. You get 3 - ±5 = 8. How do you deal with that? 3 - - 5 = 8? A truism which takes you no further. What about the alternative 3 - + 5 = -2 So - 2 = 8. Now plug that back into the original equation by substituting -2 for 8 and we arrive at 2m + 2 = 5² = 25, so 2m = 23, so m = 11.5.

Is it possible to have a, 110% or should a 100% be final?

Another issue with how math is taught is at least for me i dont recall much if any instrcution on when an equation doesnt have a solution as they all did. That symbol looks like a zero to me because i was never taught that.

Whenever I get a minus sigh and a square, I keep having to image the solution.

I found this very confusing. What convention are you using that only the "principle square root" is the ONLY square root? Doesn't matter if an equation is quadratic or not.

A square root cannot be negative. 3 - X = 0.

(@2:03) Well, it’s clear that neither 0 nor 23/2 are solutions to the original equation, 3-sqrt(2m+2)=8. Plugging in 0, we get: 3-sqrt(2(0)+2)=8 3-sqrt(0+2)=8 3-sqrt(2)=8 and sqrt(2) not = -5, so 0 cannot be a solution. What about 23/2? 3-sqrt(2(23/2)+2)=8 3-sqrt(23+2)=8 3-sqrt(25)=8 3-5=8. This is close, but the sign is incorrect; we need -5, not positive 5. But to get sqrt() to give us a negative value, we have to wander into the complex numbers; those of the form a+bi, where i=sqrt(-1). Now i has some interesting properties if you start raising it to integer powers. i^1=i, i^2=-1, i^3=-i, and i^4=1. Here, we want the 4th power, as x^4=(x^2)^2. Now, the square root will get rid of that pesky ^2 at the end, so sqrt(25((i^2)^2)) = sqrt(5x5 x ((i^2)^2), or 5x(i^2), which is 5(-1)=-5. But this isn’t the end; we need to reverse the variable substitution for ‘m’. So 2m+2 has to = 25(i^4). First, divide both sides by 2. m+1=12.5(i^4). Now, subtract 1 from each side, or m=12.5(i^4)-1, or m = -1+12.5(i^4). Will this work? To figure it out, plug it back into the original equation. 3-sqrt(2(-1+12.5(i^4))+2)=8. 3-sqrt(2(-1+12.5((i^2)^2)+2)=8 3-sqrt(-2+25((i^2)^2)+2)=8 3-sqrt(-2+2+25((i^2)^2))=8 3-sqrt(25((i^2)^2))=8 3-sqrt(25 x ((i^2)^2))=8 3-sqrt(25) x sqrt((i^2)^2))=8 3-5(i^2)=8 3-5(-1)=8 3-(-5)=8 3+5=8 is that the solution? Yes. And no. Recall that we substituted (i^2)^2 for i^4 so that the square root in the original equation would get rid of the extra squaring operation for us, but if we immediately substitute i^4=1, then sqrt(25(i^4)) simplifies to the original incorrect value of 3-5, or -2 which is not = 8! So which is it. My only response is, ask the Mathologer! I’ll leave that as an exercise for the viewer.

SQRT (i*i) what is ? 3 - SQRT (25i*i) ? in int no solution (SQRT=>0), but complex

12:49 Just a tiny thing to clarify: √(x²) = ±√25 is actually wrong, because √(x²) = |x|, so, it's like saying |x| = ±√25, which is incorrect because it means absolute values are both positive and negative at the same time (which contradicts the definition of an absolute value). That ± comes from resolving the absolute value, so it's better just to write √(x²) = √25, then |x| = √25, *and just then* x = ±√25

The solution does in fact exist if you consider imaginary axis does it not? This may in fact may indicate 11.5 is a valid answer for some electrical engineering or other field in which the complex numbers are valid. So the Algebra depends on the context of the problem to determine whether it is valid . The method to solve must take into account what domain problem we are trying to solve it in. Throwing the eq out there with the assumption that PSR is all that is wanted is probably unwarranted.

Did you really say m=empty set? What is 2x a set. What is 2x a set plus 2. And finally how is a square root of a set defined?

You can also see that the equation has no solution if you subtract the 8 at the right side: -5 - (sqrt. 2m+2) = 0. Then multiply both sides with -1: 5 + (sqrt. 2m+2) = 0. This means that (sqrt. 2m+2) has to be -5 and that is impossible because of the rule that the result of a square root always must be a positive number (the so-called P.S.R. rule: only the positive answer of a square root is valid). For example: sqrt. 25 = (always) 5 and never -5 (even thought -5exp.2 = -5 x -5 = 25. That is the essence of this video.

I got 11.5i, which is neither of us the provided answers provided at the beginning. As you didn't limit the domain to Real numbers, the Empty set is incorrect.

Probably missed something simple as one tends to do, but why do I get a different answer if I square everything before moving the constant to the right side? 3 - sqrt(2x+2) = 8 3^2 + (-sqrt(2x+2))^2 = 8^2 9 + 2x + 2 = 64 2x + 11 = 64 2x = 53 x = 53/2

3-SQRT(2m+2)=8 SQRT(2m+2)=-5 Substitute SQRT((5^2)(i^4)) for -5 Giving SQRT(2m+2)=SQRT((5^2)(i^4)) Square giving 2m+2=(5^2)(i^4) 2m=(5^2)(i^4)-2 m=((5^2)(i^4)-2)/2