@@devookoWell not exactly that because it is still working out something, you can just read it and say no soln exists, because the statement essentially says that find x such that increasing x by 2 is the same as decreasing x by 2, which is not possible. So no solution.
That's how you can tell that this is almost certainly someone yanking our chain. Most of the algebra is just to distract attention from the blatantly ridiculous first step.
@pegasoltaeclair0611 Thanks. Sometimes it's hard to figure out who the comment refers to. But I should have figured out that the OP was referring to the guy who posted the TikTok answer.
I hate how the first step is (x+2)(x-2)=0, which immediately implies x=±2, and then the TikTokker goes on to undo that and expand the quadratic out so that they can solve it by taking square roots instead. That almost bothers me more than the fact that the first step is completely bogus.
@@user-uz4vr7to8c Not sure what you mean by "on both sides of the equality" here. x is only on one side of the equality. When I said "x=±2", I was shorthanding, "either x=2 or x=-2."
@@user-uz4vr7to8c But I never suggested that in the first place. (X+2)(X-2)=0 has exactly two solutions, X=2 and X=-2. You only need one of the terms to equal 0.
I was looking for comment like this. I got micro aneurysm just by looking at it: x+2=x-2 ==> (x+2)(x-2)=0, followed by second step. Yea maybe in some imaginary cosmos where law of mathematics and physics don't exist created by TikTok's influencers.
@@Mike_B-137 TikTok managed to combine additive and multiplicative rules together. Because if you divided one side to the other, the other side would be one. If you multiplied, it would be (x-2)(x+2) but the other side would be (x+2)^2. Not 0. Also, if you subtract the one side, you would get zero, but, the other side would = 4. Which, that is one way to find no solutions just like subtracting just the x on both sides. The problem is they multipled on the left side, but, subtracted everything on the right. And broke Algebra rules. So, they applied multiplication to the left side and then assumed the right side would cancel to 0. Remembering some of Algebra but forgot you have to subtract. And then after that incredibly wrong step, it just gets stranger and more wrong. Lol. Gosh. I think the comment may possibly be a troll. Lol. To mess with TikTok, but, not 100%. Could be someone overconfident in their Math abilities. Either way, they seem to have just kept working until they found a solution somehow. And each step seems reasonable to someone untrained in Math, but, to someone who is trained in Math, it is obviously wrong and insane each step of the way to a misshapen "solution".
I saw some tiktok coders through other platforms and it was down bad, but this is too much, how come as a society we need to explain x+a = x-a if a /= 0 is wrong. System has failed.
I'm sure if the original presenter of this problem and solution had done the check, he would have plugged in the negative value that he obtained on the left side, the positive value on the right, and shown that indeed his solutions work!
I mean they literally could’ve just checked it, there is no way 2+2 = 2-2 (if you use sqrt4 which idk why they didn’t even simplify it down to just 2 but I digress)
So... he solved it the same way a seasoned politician approaches the issue of answering substantial policy questions. Would we call that "politically correct math"? 🤔😇
@@KyrelelWe say “Once TikTok was launched, parents’ nurture f**ked up” (It’s Chinese, I try to translate it, but it still mean TikTok mess up everything)
@whoff59 Exactly, and that's still the case even if X=∞. If you subtract X from both sides of the equation, you are subtracting ∞ from both sides of the equation...and you're still stuck with +2 = -2, which is false. Therefore, no, ∞ is not a valid solution, no matter how much people want to pretend that it is.
Lol yea most people just nope out seeing square roots even though this one is quite simple. But that means they won't catch how the answer doesn't even work if you plug them in which is checking your work 101
I don't think this would happen because someone out there will have the knowledge to challenge them and they risk looking stupid. I think they really were applying maths to the best of their ability, but they have a flawed understanding of algebra
@@SalimShahdiOffWhat my brain did counts. Me: Oh, x+2=x-2 goes into 0=0, so no solution. There are layers to my stupidity, but that’s a pretty good example of “everything here is wrong except the answer you circled on the paper somehow. The math (-2-2=-4), the conclusion (0=0 is infinite solutions not no solutions), all of it.” I somehow survived the calculus series. I don’t understand it either.
It's really unnecessary to do anything past subtracting x on both sides, you got 2 = -2, a likewise always false statement like 0 = -4. What I wanna know is how somebody thought dividing (correction: multiplying) both sides by x - 2 would give 0 on the right side. 🤣🤣🤦🏾♂🤦🏾♂
That was not dividing both sides. That was multiplying the left side by (x-2) while SUBTRACTING (x-2) from the right side to give that zero. What was done as the first step is this : (x+2) = (x-2) → (x+2)*(x-2) = (x-2) - (x-2) → (x+2)(x-2) = 0
@@KualinarNono, it was just dividing both sides by 1/(x-2). I don’t know what composition rules they’re working under where that equals 0 on the right hand side, but technically it was dividing both sides by 1/… just as much as it was multiplying by (x-2)
@@Tristanlj-555 Dividing can never reduce a value to zero. ONLY a subtraction can do that. Then, a division by (x-2) would have made the left side into THIS : (x+2)/(x-2) NOT (x+2)*(x-2)
@@Kualinar I know that. I just finished my last exam, complex analysis for my first year of mathematics at Uni. I alluded to that jokingly by mentioning composition rules.
@@Kualinarinfinity is not a very useful concept in programming, I don't know why one would think about it in that context. It's very useful in abstract math.
@@geometerfpv2804 In mathematics, it's one of the Not a Number entities, a NaN . In programming, ANY logical operation involving a NaN return false, and any mathematical operation with a NaN will return NaN.
You'll run into a problem here, the solution they got is x = 2 or -2. If you plug in -2 on the left and 2 on the right it technically works. Obviously you're supposed to do the same number for both x values but we're past the point of them doing the right thing.
@ I meant graph the left side and right side independently as y=x+2 and z=x-2. Let’s call these the parent functions. Given these functions we can ask a lot of questions, one of which is the question “for which values of X does Y = Z?” Which is the same as asking about the point of intersection of the two functions. This is the ticktock question. Graphing them independently reveals the intuition that the graphs are parallel lines that do not intersect. This technique can be used to build intuition for other forms of functions too.
The real answer is not in finding x. The real answer is that this addition is defined over ring of remainders of division by 4. In which 2 = -2 since 2+2 = 0. Therefore, the equation is true for any x. We are threading in the realms of abstract algebra, where everything is possible.
Yea I’m no calculus major, but I know enough about (X)’s to put all X’s on one side and everything else that you can on the other. And that gets 0X=-4, which is about as wonky as the first question.
@@tundcwe123i mean now that you mention it..... infinity + 2 = infinity - 2. That is if you consider dividing by zero to equal infinity, and not undefined
@@Artleksandr ok but it's a concept If you have infinitely many things (natural numbers), and you simply append 2 numbers (0 and -1) you still have infinitely many numbers Cardinality hasn't changed If you them subsequently remove -1 and 0, you still have infinitely many numbers Sure it's not a "number number" like 5 or 87, but the concept still works Adding or removing finite elements from an infinite set does not change the cardinality of the set nor the number of elements
@xanderlastname3281 *_No, infinity is NOT a valid solution._* Premise 1: A=A Premise 2: A-X ≠ A Conclusion: Therefore, infinity-X ≠ infinity To argue otherwise is to commit a special pleading fallacy. Or, if you prefer: Premise 1: Set {A} includes all real positive numbers (is ∞). Premise 2: Adding any positive number X to Set {A} has no impact because Set {A} _already includes_ X. Premise 3: Subtracting real positive number X _from_ Set {A}, decreases the size of Set {A} by _removing_ something from the set. Conclusion: Therefore, while ∞+X = ∞, ∞-X ≠ ∞. This is _not_ a special pleading because {A} _is defined as_ ∞. To use a more concrete example as an analogy for the second syllogism, let's say {A} equals "all automobiles." When any new year's product line is made available, {A} will remain unchanged because, by definition, it *already includes* all of those automobiles. Conversely, if we subtract "minivans" from {A}, there's a material reduction in the size of {A} that can be observed. If you still disagree, all you have to do just *graph* it. You will end up with two parallel lines. The fact that they are parallel and will never converge proves conclusively that you are wrong. This is all beside the fact that performing the basic algebraic operation of subtracting X from both sides of the equation (this is the subtraction property of equation) yields +2 = -2 which is obviously _false._ And, we can continue the subtraction property of equality to yield 0 = -4, which is also obviously _false._ *_STOP suggesting infinity. It is demonstrably WRONG._*
visually you can take both the x+2 and the x-2 as functions, which is the idea of solving ecuations, you are checking when is it that y1=x+2 intersects with y2=x-2, which you would check intersections by doing y1=y2 and if you graph it, you would see that since there is no intersection, there is no solution. Probably someone already said it but i wanted to say it too :D
you can just plot the (x+2)/(x-2) hyperbole and the asymtotes are obvi x=2 and y=1, so it will only (almost) converge at both positive and negative infinity.
The two statements have the same slope (1) if you look at them as a function of x, but have different y-intercepts (2 and -2). So they are parallel on the same plane, thus never intersect. If you solve you get 0=4 or 0=-4.
@@_JoeVerthe question would be "when if ever does the hyperbola hit y=1?", and this isn't visually obvious from knowledge of asymptotes. Thinking about them as lines is better.
Something drilled into me in school: when dividing by a variable or an expression containing a variable, mind the 0. And, if it is an inequality, mind your negative numbers.
At first I thought "that's impossible". And then I thought "it's been over 20 years since I have done this stuff, I must be wrong and therefore an idiot." I was not wrong but I am still an idiot.
I'm glad it actually has no solution because I was doing the calculation mentally, and I got really confused when I got to the point x-x=-2-2 because those x would certainly result in 0, and I didn't remember what to do at that point. I'll be honest, I watched the whole video to get an explanation, so thank you! ❤❤
In complex numbers, |x - 2| = |x + 2| is fairly obviously just any purely imaginary number i.e. Re(x) = 0. But that's the only way to get anything even approximating a solution.
Another answer could be infinity (minus infinity is still an infinity with the same cardinality). If one would add or substract a number of elements from an infinity it would still be infinity because in the relative size of an infinite set of elements a removal or an addition of some elements would change nothing.
I immediately saw both equations as straight lines with gradient 1 and intercept 2 and -2 (y=mx+c). So two parallel straight lines; therefore no solution. Playing with the equation -> x+2 = x-2 Subtract (x-2) from both sides -> x-x+2+2 = 0 -> 4 = 0, which is categorically false so the original equation can't exist. Now to watch and see what bprp does.
@@kaltaron1284 Not in Euclidean geometry. "Euclid had defined parallel lines to be straight lines in a plane that "being produced indefinitely in both directions" never intersect; and accordingly, will never meet (or "merge") even at infinity. So, if you are in the realm of Euclidean geometry, then parallel lines can never intersect, even at infinity."
@@WombatMan64 The fith axiom actually states: "If a line segment intersects two straight lines forming two interior angles on the same side that are less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles. " That's from The Elements. Where's your definition from? Also Infinity + 2 and Infinity -2 are equal.
@@kaltaron1284 Your definition is specifically talking about non-parallel lines. "less than two right angles" is the key there, for parallel lines, the line segmenting them would be equal to two right angles, not less than. My definition is from every university maths page I could find on the topic. You're treating infinity as if it's a number, which it isn't, it's a concept.
@@WombatMan64 It's not my definition but Euclid's. He doesn't talk about parallels but that non-parallels have to intersect on a certain side. That makes no statement about parallels. Infinity is a concept that can be used in mathematics, what's the problem?
If we allow floating points: x = NaN, +infinity, -infinity, and any number (representable by floating points) so big in absolute value that floating points errors ignore the 4 during addition.
In physics we were taught to look at calculated answers to see if they make sense. It also helps you develop a sense of proportion for real-world values much like a chef knows when a certain amount of salt or other ingredient in an unfamiliar recipe is too much. Like the good teacher concluded @3:00 viewing this from another angle so to speak, once you have a solution, here, ostensibly, x = sqrt(4), i.e. x = 2, put it into the starting equation to see if it works since this is the original goal of the exercise after all. Doing that we find x+2 = x-2 becomes 4 = 0, an impossibility. x=-2 is no better with 0=-4.
To be precise, there is no solution in the real numbers or the complex numbers. But there are solutions in other number systems. For example, both the affine extended real number system and the projective real number system has a value infinity, denoted ∞ (or perhaps +∞ in the affine system). We have ∞+2 = ∞-2 = ∞ so ∞ is a solution. I'm sure there are other solutions in other number systems. Perhaps infinite cardinal numbers?
@@liamernst9626 That is an *excellent* answer! I wish I had thought of it. Of course, modulo 4 also works and has the advantage that "2" is still called 2 in that system.
This is why i had a high school teacher who said he doesn't like calling it bringing to the other side because it causes confusion like that You need to be doing the same thing on both sides, so he emphasizes that point in the equality so none of us do such a bjg mistake
@@rezwhapI agree 100% with cancellation UNTIL every one in the class can say WHY it cancels. Some reasons that they need to articulate are: 1. These cancel because their sum is zero and zero added to rest of the expression doesn’t change it (this is using the idea of additive inverses and addition identity element.) 2. Dividing by the same expression is 1 … 3. Squaring is the inverse operation of square root, but could introduce an extraneous roots.
Looking back, I think my math teachers, all of them, never told me that a math problem can have no answer. It was only in engineering school that I learn that it does happen sometimes.
I legit thought I was an idiot for the first few seconds because I was “this *equation* makes no sense… right? Did I miss something? Oh god what did I forget. Sure I graduated college years ago, but surely I didn’t forget something so basic already… what’s the solution because I’m probably wrong I guess” Swear to god I started doubting my own knowledge so hard
@@be7256even step 2 is an over complication of the supposed solution, you already have a simplified equation and they expand it out which just adds more steps to the totally new problem they wrote in step 1
One nice way to show that there is no solution that fits well with the way kids are taught math: Graph the first line y=x+2. Now graph the second line, y=x-2. Now show that the two lines do not intersect (they have the same slope) and so there is no x that satisfies both sides.
Because he spent so much time going through everything I had an existential crisis where I KNEW that the equation was impossible, and was actively dreading that he was actually going to show a solution that worked somehow and turn my world upside down.
x+2=x-2 x-x=-2-2 0=-4 If we plug in -4 in place of the x, we get -2=-6. If we plug in 0 instead, we get 2=-2. The numbers in the right of the equations are the numbers in the left minus 4 (-4), and for the equations to make sense -4 must be the neutral element of the sum so that -2-4=-6 and 2-4=-2 would be possible. Here is why this doesnt work.
Can you teach me how can I sove this problem, please? sqr(a)+sqr(ab)+sqr(abc)=12 sqr(b)+sqr(bc)+sqr(abc)=21 sqr(c)+sqr(ac)+sqr(abc)=30 Find: (a^2 + b^2 + c^2)
There's probably a better approach but you could solve it by brute force. I suggest substituting √a, √b and √c by u, v, and w just to get rid of the square roots. You'll then get this system of equations: u + uv + uvw = 12 v + vw + uvw = 21 w + uw + uvw = 30 Three equations with three variables should yield solutions. Then, plug the solutions into the last term. (Just remember that it will have to be u⁴ + v⁴ + w⁴.) I had WolframAlpha do the work for me. There are three sets of solutions. the sum of squares that we are supposed to find can be either 2433, 10002, or 312688557441/384160000. Like I said, maybe there's a better way but the old-fashioned way should work. If I had to do it by hand, I'd start by subtracting the second from the first equation: u + uv - v - vw = -9 We can isolate u by factoring it out and bringing the other stuff on the right hand side: u (1+v) - (v+vw) = -9 u = (v+vw+9) / (v+1)
it took me some time and I must have made a mistake in my regular scholarly attempts, but then it jumped at me: it works for a = 1, b = 1, c = 100, so your result is 10002. method: first substitution for those ugly sqrts: A = sqrt(a), B = sqrt(b), C = sqrt(c), so you get A + AB + ABC = 12 B + BC + ABC = 21 C + AC + ABC = 30 now you see every next equation is 9 bigger, so it is as if you subtract 1 and add 10. and this really works if A = 1, B = 1, C = 10 so you get 1 + 1 + 10 = 12 1 + 10 + 10 = 21 10 + 10 + 10 = 30 A^4 + B^4 + C^4 = 1 + 1 + 10000 = 10002
You can also show this visually by graphing y=x-2 and y=x+2. You get two parallel lines meaning they never touch and never give the same output for any x value.
There are actually solutions. Both infinity and negative infinity work. And if you you are in Finite Field of Integers modulo 4 then 2 is a valid solution. The "solution" presented in the video is of course bonkers.
This is why i hate the whole "bring to the other side" terminology. You're not bringing it to the other side, you're dividing by the same amount to both sides of the equation. And this simplification of terminology is fine for most people, but is used way too much in classes where people are trying to learn algebra and then they leave the class not understanding what they actually did. I feel the same way about "cancelling"
I hate people who are bad at math, and think they are good at it. Even worse - people who know SOME math and do wrong things on PURPOSE and then brag about it just to get FREAKING COMMENTS OF PEOPLE WHO GET MAD AT THEM BUT DON'T UNDERSTAND THAT THIS IS EXACTLY WHAT THEY WANT
I was a little bit worried when you started dividing by x-2 and thought I have forgotten everything I ever knew about math. I was relieved when you continued and showed the correct solution, which is also the same solution I got to.
*_No, infinity is NOT a valid solution._* Premise 1: A=A Premise 2: A-X ≠ A Conclusion: Therefore, infinity-X ≠ infinity To argue otherwise is to commit a special pleading fallacy. Or, if you prefer: Premise 1: Set {A} includes all real positive numbers (is ∞). Premise 2: Adding any positive number X to Set {A} has no impact because Set {A} _already includes_ X. Premise 3: Subtracting real positive number X _from_ Set {A}, decreases the size of Set {A} by _removing_ something from the set. Conclusion: Therefore, while ∞+X = ∞, ∞-X ≠ ∞. This is _not_ a special pleading because {A} _is defined as_ ∞. To use a more concrete example as an analogy for the second syllogism, let's say {A} equals "all automobiles." When any new year's product line is made available, {A} will remain unchanged because, by definition, it *already includes* all of those automobiles. Conversely, if we subtract "minivans" from {A}, there's a material reduction in the size of {A} that can be observed. If you still disagree, all you have to do just *graph* it. You will end up with two parallel lines. The fact that they are parallel and will never converge proves conclusively that you are wrong. This is all beside the fact that performing the basic algebraic operation of subtracting X from both sides of the equation (this is the subtraction property of equation) yields +2 = -2 which is obviously _false._ And, we can continue the subtraction property of equality to yield 0 = -4, which is also obviously _false._ *_STOP suggesting infinity. It is demonstrably WRONG._*
You know (just to confuse the chat), there are structures where -4 might equal 0 though. For example, in the GaloisField of order 2 (aka GF(2), or you could call it the Boolean Algebra, if you wanted to; Now basically this is integral math but modulo 2; or in terms of boolean algebra, multiplication is "logical and" and addition is "exclusive or" ). Ok, anyway, so there is not really a symbol 4 there, but you can just interprete it as 1+1+1+1 (ok, -1-1-1-1 for -4), aka 4 times 1, and it does indeed equal 0. So there the equation would actually mean x=x which is always true and in case of GF(2) x can be 0 or 1. Sorry for this knitpicking comment.
We have: x = x + 4 Now substitute x into x: x = (x + 4) + 4 x = x + 8 x = x + 16 … And similarly this can be done by first subtracting 2 over, ie, x = x - 4 The only way adding these finite quantities can work without affecting anything is if x is +/- ♾️
Reminds me of that time I misread "tweezers" as "pliers" and did a ton of unnecessary algebra to calculate the mechanical advantage. Somehow, you remembered how to solve a quadratic, but failed on the most simple part.
*_No, infinity is NOT a valid solution._* Premise 1: A=A Premise 2: A-X ≠ A Conclusion: Therefore, infinity-X ≠ infinity To argue otherwise is to commit a special pleading fallacy. Or, if you prefer: Premise 1: Set {A} includes all real positive numbers (is ∞). Premise 2: Adding any positive number X to Set {A} has no impact because Set {A} _already includes_ X. Premise 3: Subtracting real positive number X _from_ Set {A}, decreases the size of Set {A} by _removing_ something from the set. Conclusion: Therefore, while ∞+X = ∞, ∞-X ≠ ∞. This is _not_ a special pleading because {A} _is defined as_ ∞. To use a more concrete example as an analogy for the second syllogism, let's say {A} equals "all automobiles." When any new year's product line is made available, {A} will remain unchanged because, by definition, it *already includes* all of those automobiles. Conversely, if we subtract "minivans" from {A}, there's a material reduction in the size of {A} that can be observed. If you still disagree, all you have to do just *graph* it. You will end up with two parallel lines. The fact that they are parallel and will never converge proves conclusively that you are wrong. This is all beside the fact that performing the basic algebraic operation of subtracting X from both sides of the equation (this is the subtraction property of equation) yields +2 = -2 which is obviously _false._ And, we can continue the subtraction property of equality to yield 0 = -4, which is also obviously _false._ *_STOP suggesting infinity. It is demonstrably WRONG._*
x+2=x-2 x=x-4 x/x=x/x-4/x 1=1-4/x 0=-4/x x=-4/0 x=unsigned infinity Now let's see if solution correct. unsigned infinity + 2 = unsigned infinity - 2 unsigned infinity = unsigned infinity Any real number added to unsigned infinity doesn't change it. Solution is correct.
Just in case this isn't a meme, infinity isn't a number. You can never have x equals infinity, only x aproaches infinity or lim x = inf. There's a reason why R = (-inf, inf) and not [-inf, inf]
*_No, infinity is NOT a valid solution._* Premise 1: A=A Premise 2: A-X ≠ A Conclusion: Therefore, infinity-X ≠ infinity To argue otherwise is to commit a special pleading fallacy. Or, if you prefer: Premise 1: Set {A} includes all real positive numbers (is ∞). Premise 2: Adding any positive number X to Set {A} has no impact because Set {A} _already includes_ X. Premise 3: Subtracting real positive number X _from_ Set {A}, decreases the size of Set {A} by _removing_ something from the set. Conclusion: Therefore, while ∞+X = ∞, ∞-X ≠ ∞. This is _not_ a special pleading because {A} _is defined as_ ∞. To use a more concrete example as an analogy for the second syllogism, let's say {A} equals "all automobiles." When any new year's product line is made available, {A} will remain unchanged because, by definition, it *already includes* all of those automobiles. Conversely, if we subtract "minivans" from {A}, there's a material reduction in the size of {A} that can be observed. If you still disagree, all you have to do just *graph* it. You will end up with two parallel lines. The fact that they are parallel and will never converge proves conclusively that you are wrong. This is all beside the fact that performing the basic algebraic operation of subtracting X from both sides of the equation (this is the subtraction property of equation) yields +2 = -2 which is obviously _false._ And, we can continue the subtraction property of equality to yield 0 = -4, which is also obviously _false._ *_STOP suggesting infinity. It is demonstrably WRONG._*
Compulsively reducing things just leads to connections being lost. For example, the values of sin and cos on the unit circle from 0 to 90 degrees are so nice, unless you reduce them, then they get a lot harder to remember.
On the worst case scenario, if this was a variable question, it's still (probably) unsolvable because we weren't even explicitly told what X is supposed to be...
Common sense tells you that any finite real number for x won’t make sense because anything plus something other than 0 is no longer the same thing, or addition would be meaningless as a term and concept. Idk if maybe infinity would fit for x, since infinity breaks a bunch of rules. What is infinity plus 2, but just infinity still. It seems like the whole thing is either simply unequal or if you seek a bs answers then it’s a trick question. When I looked it up to see if I had the right idea regarding infinity it does appear that the conventions used by mathematicians is such that infinity plus 2 is considered equal to infinity. With that in mind, if x is even allowed to be infinity in the scope of the problem then that’d be an answer, and if the scope of allowed or relevant answers doesn’t allow for anything but real numbers than it would be simply an inaccurate mathematical equation as the two sides are not equal.
@@harrietjameson X+2=x-2 0/0+(2*0/0)=0/0-(2*0/0) 0/0+0/0=0/0-0/0 this Is true. Inf+2=inf-2 this is also true, as you can't add or take away from infinity
@@SeleverEnjoyer not true, 0/0 is indeterminate form for a reason, doesnt matter if its common denominator. Plus you're still adding on one side and subtracting from the other however, as x approaches ±infinity the equation does get more and more correct, but its not because you "cant take away from infinity", its because the tiny 2 has literally nothing in comparison to an unbounded sum but at that point you arent solving, just finding some limit property
5 หลายเดือนก่อน +1
Actually, I can solve this one. If you think about the problem you realise that we talking about a point (x like treasure) and when we move oposite directions the same ammount (+2; -2) we get the same place. I can only think about a circle (maybe part of a sphere or something). Now we know x is a point on a circle and the opposit "end" 2 units from this point both sides and we also know the circle circumference is 4. From these informations we can deduce that the question is the radius of the circle. r = ?; K = 4; K = 2 * r * Pi 4 = 2 * r * Pi r = 2 / Pi r = 0,6366... Solved! xD
another way to think of it is that x must occupy a unique place on the number line, and thus cannot be translated in different directions simultaneously and still retain a unique position
A simple way to illustrate the solution is with a graph, showing the two parallel lines that do not converge hence can not have the same value at any point.
Once in a while I have nightmare of being back in middle school and having to do stuff without the use of a calculator, or being back in advanced math in highschool and not remembering ANYTHING... But I'm pretty sure I'd be okay if that was the level of equations I got.
There is one place where you can do x = x - 4, not as an equality but an instruction. In some programming languages, this would mean define a new variable x, then set the value as 4 less than what x was originally.
I could see a teacher in an elementary algebra class giving the equation for the students to solve, as an object lesson that not all equations actually have solutions.
If you want the graphical solution: x+2 is a line parallel to the x-axis, but two units up. x-2 is a line parallel to the x-axis but two units down. The solution is where the two parallel lines meet.
1 divided by 0 (a 3rd grade teacher & principal both got it wrong)
th-cam.com/video/WI_qPBQhJSM/w-d-xo.html
You wasted your time with this video.
If you apply limits and assume x is going to negative or positive infinity you get an answer.
*Reddit* isn't any better.
@@ToguMrewuku Hi #bot
@@Real-Name..Maqavoy at least reddit has downvotes and a place for math, so people know if they're wrong
"x+2=x-2"
"No it doesn't"
Nuh uh
"Time of day" + 12 hours = "Time of day" - 12 hours
@@khaitomretro you missed a (mod 24) at the end
@@kebien6020 "x+2 = x-2... in ℤ₄; I'm fine!"
@@khaitomretro no. That cannot be used to solve this problem.
-Remembers the negative square root
-Forgets division exists
-remembers the negative square root
-doesnt realise a number minus two would never be equal to the same number plus two
@@kraquinette2430 unless the numbers are mod 4. Then x=2 would satisfy it.
@@bobbobert9379 Sure, but then you'd expect that to be explicitly mentioned.
Let's be honest, you can already see just by looking at the question that this will have no solution...
Yes, I can.
This stupid comment has more likes than the comment below this, which has actual valuable information unlike this junk
(X+4) does not equal x
Who came up with it was for sure just writing down random numbers, the statement "a number plus two is equal itself minus two" is a paradox
@@devookoWell not exactly that because it is still working out something, you can just read it and say no soln exists, because the statement essentially says that find x such that increasing x by 2 is the same as decreasing x by 2, which is not possible. So no solution.
Computer Scientist: "x = x+1"
Mathematician: "Nuh uh"
Computer science square root of -1... because computers don't work without "I"
well its an assignment operator, not checking for equal. Checkng for equal is ==.
@@user-vs1mn8ig8wwoosh
@@user-vs1mn8ig8wThat’s part of the joke bro
That's becausewe don't have proper notation for iterable variables. In computers x = x + 1 means. (new/next x) = (current/previous x) + 1
Guy had the mental prowess to apply difference between two squares, but not enough to do the first step right💀💀
That's how you can tell that this is almost certainly someone yanking our chain. Most of the algebra is just to distract attention from the blatantly ridiculous first step.
Bros probably solve the 1st line, then ask chatgpt to solve the rest 💀
What he is showing first is the solution from the TikTok video. Then he proceeds to show how it is wrong.
@@cybore213 OC wasn't talking about bprp
@pegasoltaeclair0611 Thanks. Sometimes it's hard to figure out who the comment refers to. But I should have figured out that the OP was referring to the guy who posted the TikTok answer.
I hate how the first step is (x+2)(x-2)=0, which immediately implies x=±2, and then the TikTokker goes on to undo that and expand the quadratic out so that they can solve it by taking square roots instead. That almost bothers me more than the fact that the first step is completely bogus.
The average tiktokers knowledge of math:
it’s probably a 12 year old who was just really eager to use the new identity he learnt
@LumiaFenrir-nn2pz and you’d be surprised by the number of Asian fetuses that know how to solve quadratic equations in microseconds
@@user-uz4vr7to8c Not sure what you mean by "on both sides of the equality" here. x is only on one side of the equality. When I said "x=±2", I was shorthanding, "either x=2 or x=-2."
@@user-uz4vr7to8c But I never suggested that in the first place. (X+2)(X-2)=0 has exactly two solutions, X=2 and X=-2. You only need one of the terms to equal 0.
This is not math this is meth
😂
I was looking for comment like this. I got micro aneurysm just by looking at it: x+2=x-2 ==> (x+2)(x-2)=0, followed by second step.
Yea maybe in some imaginary cosmos where law of mathematics and physics don't exist created by TikTok's influencers.
Meth is herd
@@Mike_B-137
TikTok managed to combine additive and multiplicative rules together.
Because if you divided one side to the other, the other side would be one. If you multiplied, it would be (x-2)(x+2) but the other side would be (x+2)^2. Not 0.
Also, if you subtract the one side, you would get zero, but, the other side would = 4. Which, that is one way to find no solutions just like subtracting just the x on both sides. The problem is they multipled on the left side, but, subtracted everything on the right. And broke Algebra rules.
So, they applied multiplication to the left side and then assumed the right side would cancel to 0. Remembering some of Algebra but forgot you have to subtract.
And then after that incredibly wrong step, it just gets stranger and more wrong. Lol.
Gosh. I think the comment may possibly be a troll. Lol.
To mess with TikTok, but, not 100%. Could be someone overconfident in their Math abilities.
Either way, they seem to have just kept working until they found a solution somehow. And each step seems reasonable to someone untrained in Math, but, to someone who is trained in Math, it is obviously wrong and insane each step of the way to a misshapen "solution".
I saw some tiktok coders through other platforms and it was down bad, but this is too much, how come as a society we need to explain x+a = x-a if a /= 0 is wrong. System has failed.
The guy didn’t even check his solutions back by plugging them in
That’s rule #1 for anything you want to know you’re reasonably correct on
I'm sure if the original presenter of this problem and solution had done the check, he would have plugged in the negative value that he obtained on the left side, the positive value on the right, and shown that indeed his solutions work!
My eyes are bleeding from the proposed solution
I mean they literally could’ve just checked it, there is no way 2+2 = 2-2 (if you use sqrt4 which idk why they didn’t even simplify it down to just 2 but I digress)
There's nothing wrong with the way they've written it because √ is always positive.
I think the thing is that -2+2=2-2.
@@error_6o6but you're saying x=-2=2
Quadratics can have multiple solutions, x=2, x=-2 indicates that x could be either, and not that it is both
The first step is basically "forget about the equation and let's just solve another one" 😂
So... he solved it the same way a seasoned politician approaches the issue of answering substantial policy questions. Would we call that "politically correct math"? 🤔😇
@@clefsan no they would give a wrong solution to another question.
The fact that he had zero on the right, not 1 implies he was mixing up subtracting with multiplication, not division.
No, he simply copied what tiktok gave as the answer and then explained why it is complete bullcrap.
The tiktok was a joke, literally.
@@Kyrelel talking about the commenter, not the guy explaining
@@KyrelelWe say “Once TikTok was launched, parents’ nurture f**ked up”
(It’s Chinese, I try to translate it, but it still mean TikTok mess up everything)
I'm acting like that wouldn't happen to me, but it happened even on tests (the worst part is that I'm a physics undergrad 😅)
Wich would make sense
Since this type of mixing crap never made me like math
The first line is actually saying
+2 = -2
as you can subtract x on both sides.
That's it.
Exactly
@whoff59
Exactly, and that's still the case even if X=∞. If you subtract X from both sides of the equation, you are subtracting ∞ from both sides of the equation...and you're still stuck with +2 = -2, which is false. Therefore, no, ∞ is not a valid solution, no matter how much people want to pretend that it is.
@@Bardineerwhy +2 is Not -2?
@@bobbychess5652
No.
@@bobbychess5652 that's like saying hands = no hands
That proposed solution is got to be a ragebait ,
How did you do that?
@@blacklight683 there is no sol.
Definitely something I'd land on
During college I helped my roommate in chemistry...the guy didn't know how to solve x+2=-2
@@zebefreod871 They had to be joking. I refuse to believe they were genuine.
The question goes like:
I love Math = I hate Math
It's a love-hate relationship
I love Math = I love Math
More like
I love Math = the sky is cucumber
I love Math = I love Meth
@thomashobbes8786 That's funny but no it has to be two contradicting things
Not simplifying into 2 is probably intentional so that people won't just mentally check their solution and find out how garbage it is.
Bro got a degree in psychology but failed math 💀
@@Бобреккакойтане пиши сюда больше никогда
Lol yea most people just nope out seeing square roots even though this one is quite simple. But that means they won't catch how the answer doesn't even work if you plug them in which is checking your work 101
@@Бобреккакойта garbage - мусор
I don't think this would happen because someone out there will have the knowledge to challenge them and they risk looking stupid. I think they really were applying maths to the best of their ability, but they have a flawed understanding of algebra
My favorite subgenre of this is when there are multiple fundamental errors, but the result happens to be correct.
I yearn for an exemple, if anyone has any
@@SalimShahdiOffWhat my brain did counts.
Me: Oh, x+2=x-2 goes into 0=0, so no solution.
There are layers to my stupidity, but that’s a pretty good example of “everything here is wrong except the answer you circled on the paper somehow. The math (-2-2=-4), the conclusion (0=0 is infinite solutions not no solutions), all of it.”
I somehow survived the calculus series. I don’t understand it either.
It's really unnecessary to do anything past subtracting x on both sides, you got 2 = -2, a likewise always false statement like 0 = -4. What I wanna know is how somebody thought dividing (correction: multiplying) both sides by x - 2 would give 0 on the right side. 🤣🤣🤦🏾♂🤦🏾♂
Even worse, this was obviously multiplying both sides with (x-2).
That was not dividing both sides. That was multiplying the left side by (x-2) while SUBTRACTING (x-2) from the right side to give that zero.
What was done as the first step is this : (x+2) = (x-2) → (x+2)*(x-2) = (x-2) - (x-2) → (x+2)(x-2) = 0
@@KualinarNono, it was just dividing both sides by 1/(x-2). I don’t know what composition rules they’re working under where that equals 0 on the right hand side, but technically it was dividing both sides by 1/… just as much as it was multiplying by (x-2)
@@Tristanlj-555 Dividing can never reduce a value to zero. ONLY a subtraction can do that.
Then, a division by (x-2) would have made the left side into THIS : (x+2)/(x-2) NOT (x+2)*(x-2)
@@Kualinar I know that. I just finished my last exam, complex analysis for my first year of mathematics at Uni. I alluded to that jokingly by mentioning composition rules.
0:16 I’m sorry but exactly WHERE did that come from
Meth
Fr, I can't even follow the thought process, i'm so confused
It was revealed to me in a dream
Isn't it supossed to be
x+2=x-2
x-x=2+2
0x=4
meaning it has no solution?
@@Trap-chan750 little mistake, you forgot the - on the first two, meaning 0x = 0
The very FIRST step of the proposed «solution» is totally wrong.
There are NO solution as this define two parallel lines.
So geometrically we can say that the solution is x->infinity, since all parallel lines meet at infinity. /smart-alec
@@professorhaystacks6606 Well... Infinity is NOT a number, it is a NaN.
∞ = ∞ return false.
∞ > ∞ return false.
∞ < ∞ return false.
∞ ≥ ∞ return false.
∞ ≤ ∞ return false.
-∞ < ∞ also return false.
EVEN ∞ ≠ ∞ returns false.
@@Kualinarinfinity is not a very useful concept in programming, I don't know why one would think about it in that context. It's very useful in abstract math.
@@geometerfpv2804 In mathematics, it's one of the Not a Number entities, a NaN . In programming, ANY logical operation involving a NaN return false, and any mathematical operation with a NaN will return NaN.
x-x=-4 0=-4 not true. End
This is why you should always plug your solution into the original equation to make sure it's correct.
You'll run into a problem here, the solution they got is x = 2 or -2. If you plug in -2 on the left and 2 on the right it technically works. Obviously you're supposed to do the same number for both x values but we're past the point of them doing the right thing.
@@jacobisbell9388 Yeah to be fair that makes sense.
@@jacobisbell9388yea but even just looking at the equation, how would x plus n be equal to x minus n.
no calculations needed.
@@Tealen
Unless n is 0, which it is not.
@@Pingwn yes, in this case its 2. I should have mentioned it
I have a solution, change = to ≠ 😅
Surely,
The equation
x + 2 = x - 2
should be replaced with
x + 2 ≠ x - 2
another solution is changing = to >
x+2=x-2 (false)
x+2>x-2 (true)
Another solution, the equation is in Z/2Z
I love this comment. 😂
@@wqrwtrue dat.
The graph also helps. It’s y=x graphed with two different offsets. Intuitively they are parallel and will never cross.
Thank you, there are no x-intercepts so finding x is not possible but this is still a function that can exist.
@ I meant graph the left side and right side independently as y=x+2 and z=x-2. Let’s call these the parent functions. Given these functions we can ask a lot of questions, one of which is the question “for which values of X does Y = Z?” Which is the same as asking about the point of intersection of the two functions. This is the ticktock question. Graphing them independently reveals the intuition that the graphs are parallel lines that do not intersect. This technique can be used to build intuition for other forms of functions too.
Hes getting stronger.
He can manipulate the board by sinply tapping it with the back of his marker.
We must stop him before its too late
😂
One day he shall no longer have a need for markers, his mind is enough
The real answer is not in finding x.
The real answer is that this addition is defined over ring of remainders of division by 4.
In which 2 = -2 since 2+2 = 0.
Therefore, the equation is true for any x.
We are threading in the realms of abstract algebra, where everything is possible.
Lmao! 😂😅
He edited it out
Black magic 1:14
Just seeing that on the comment preview i was wondering what you meant but i see clearly now it was indeed black magic.
I love the tap erase. Super tech white board ;)
Thanks!!
I got it from Amazon haha
@@adityagoyal7110
0-2=-2
0+2=2
-2=2
So no, its not correct
@@adityagoyal7110 aah yes 0 - 2 = 0+2 which would imply or -2 = 2
@@adityagoyal7110 Since when adding or subtracting to 0 gives back 0? Are you implying 0 is infinite?
Using Tiktok is already a signal for the lack of common sense
Yea I’m no calculus major, but I know enough about (X)’s to put all X’s on one side and everything else that you can on the other. And that gets 0X=-4, which is about as wonky as the first question.
I am sure that if you would stop at 0X=-4, someone would say X = -4/0
@@tundcwe123i mean now that you mention it..... infinity + 2 = infinity - 2.
That is if you consider dividing by zero to equal infinity, and not undefined
@@xanderlastname3281infinity is not a number. Can't work with it like that outside of a limit or other special conditions.
@@Artleksandr ok but it's a concept
If you have infinitely many things (natural numbers), and you simply append 2 numbers (0 and -1) you still have infinitely many numbers
Cardinality hasn't changed
If you them subsequently remove -1 and 0, you still have infinitely many numbers
Sure it's not a "number number" like 5 or 87, but the concept still works
Adding or removing finite elements from an infinite set does not change the cardinality of the set nor the number of elements
@xanderlastname3281
*_No, infinity is NOT a valid solution._*
Premise 1: A=A
Premise 2: A-X ≠ A
Conclusion: Therefore, infinity-X ≠ infinity
To argue otherwise is to commit a special pleading fallacy.
Or, if you prefer:
Premise 1: Set {A} includes all real positive numbers (is ∞).
Premise 2: Adding any positive number X to Set {A} has no impact because Set {A} _already includes_ X.
Premise 3: Subtracting real positive number X _from_ Set {A}, decreases the size of Set {A} by _removing_ something from the set.
Conclusion: Therefore, while ∞+X = ∞, ∞-X ≠ ∞.
This is _not_ a special pleading because {A} _is defined as_ ∞.
To use a more concrete example as an analogy for the second syllogism, let's say {A} equals "all automobiles." When any new year's product line is made available, {A} will remain unchanged because, by definition, it *already includes* all of those automobiles. Conversely, if we subtract "minivans" from {A}, there's a material reduction in the size of {A} that can be observed.
If you still disagree, all you have to do just *graph* it. You will end up with two parallel lines. The fact that they are parallel and will never converge proves conclusively that you are wrong.
This is all beside the fact that performing the basic algebraic operation of subtracting X from both sides of the equation (this is the subtraction property of equation) yields +2 = -2 which is obviously _false._ And, we can continue the subtraction property of equality to yield 0 = -4, which is also obviously _false._
*_STOP suggesting infinity. It is demonstrably WRONG._*
visually you can take both the x+2 and the x-2 as functions, which is the idea of solving ecuations, you are checking when is it that y1=x+2 intersects with y2=x-2, which you would check intersections by doing y1=y2 and if you graph it, you would see that since there is no intersection, there is no solution. Probably someone already said it but i wanted to say it too :D
you can just plot the (x+2)/(x-2) hyperbole and the asymtotes are obvi x=2 and y=1, so it will only (almost) converge at both positive and negative infinity.
The two statements have the same slope (1) if you look at them as a function of x, but have different y-intercepts (2 and -2). So they are parallel on the same plane, thus never intersect. If you solve you get 0=4 or 0=-4.
@@_JoeVerthe question would be "when if ever does the hyperbola hit y=1?", and this isn't visually obvious from knowledge of asymptotes. Thinking about them as lines is better.
the solution to this problem is the friends we made along the way
Something drilled into me in school: when dividing by a variable or an expression containing a variable, mind the 0. And, if it is an inequality, mind your negative numbers.
At first I thought "that's impossible". And then I thought "it's been over 20 years since I have done this stuff, I must be wrong and therefore an idiot." I was not wrong but I am still an idiot.
It's not impossible though. The answer shown is of course.
I'm glad it actually has no solution because I was doing the calculation mentally, and I got really confused when I got to the point x-x=-2-2 because those x would certainly result in 0, and I didn't remember what to do at that point. I'll be honest, I watched the whole video to get an explanation, so thank you! ❤❤
Tic toc math on 4:00: now we have proof for 4=0 for any x.
My calc teacher in high school was known for saying "your calculus would be fine if your algebra wasn't horrible horrible"
In complex numbers, |x - 2| = |x + 2| is fairly obviously just any purely imaginary number i.e. Re(x) = 0. But that's the only way to get anything even approximating a solution.
Another answer could be infinity (minus infinity is still an infinity with the same cardinality). If one would add or substract a number of elements from an infinity it would still be infinity because in the relative size of an infinite set of elements a removal or an addition of some elements would change nothing.
@@jaaguar13 Another solution is to stipulate that we are searching for x in a Finite Field. Both Integers modulo 2 and 4 allow solution(s).
This transition 2:59 is amazing
I agree with the silly cat
0:44 "What's wrong with this?"
EVERYTHING
This honestly feels like something I would genuinely mess up with at some point because I tend to overcomplicate everything with math
You can't just multiply one side by (x-2) and not the other side.
I immediately saw both equations as straight lines with gradient 1 and intercept 2 and -2 (y=mx+c).
So two parallel straight lines; therefore no solution.
Playing with the equation -> x+2 = x-2
Subtract (x-2) from both sides -> x-x+2+2 = 0 -> 4 = 0, which is categorically false so the original equation can't exist.
Now to watch and see what bprp does.
Parallels do meet in infinity though.
@@kaltaron1284 Not in Euclidean geometry.
"Euclid had defined parallel lines to be straight lines in a plane that "being produced indefinitely in both directions" never intersect; and accordingly, will never meet (or "merge") even at infinity. So, if you are in the realm of Euclidean geometry, then parallel lines can never intersect, even at infinity."
@@WombatMan64 The fith axiom actually states: "If a line segment intersects two straight lines forming two interior angles on the same side that are less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles. "
That's from The Elements. Where's your definition from?
Also Infinity + 2 and Infinity -2 are equal.
@@kaltaron1284 Your definition is specifically talking about non-parallel lines. "less than two right angles" is the key there, for parallel lines, the line segmenting them would be equal to two right angles, not less than.
My definition is from every university maths page I could find on the topic.
You're treating infinity as if it's a number, which it isn't, it's a concept.
@@WombatMan64 It's not my definition but Euclid's. He doesn't talk about parallels but that non-parallels have to intersect on a certain side. That makes no statement about parallels.
Infinity is a concept that can be used in mathematics, what's the problem?
If we allow floating points:
x = NaN, +infinity, -infinity, and any number (representable by floating points) so big in absolute value that floating points errors ignore the 4 during addition.
The fact that in his mind he found a way to pass to the other side the x-2 as a multiplication is crazy 💀
In physics we were taught to look at calculated answers to see if they make sense. It also helps you develop a sense of proportion for real-world values much like a chef knows when a certain amount of salt or other ingredient in an unfamiliar recipe is too much.
Like the good teacher concluded @3:00 viewing this from another angle so to speak, once you have a solution, here, ostensibly, x = sqrt(4), i.e. x = 2, put it into the starting equation to see if it works since this is the original goal of the exercise after all. Doing that we find x+2 = x-2 becomes 4 = 0, an impossibility. x=-2 is no better with 0=-4.
To be precise, there is no solution in the real numbers or the complex numbers. But there are solutions in other number systems. For example, both the affine extended real number system and the projective real number system has a value infinity, denoted ∞ (or perhaps +∞ in the affine system). We have ∞+2 = ∞-2 = ∞ so ∞ is a solution. I'm sure there are other solutions in other number systems. Perhaps infinite cardinal numbers?
Integers mod 2 has infinite solutions :)
@@liamernst9626 That is an *excellent* answer! I wish I had thought of it. Of course, modulo 4 also works and has the advantage that "2" is still called 2 in that system.
@@rorydaulton6858🤔 this is essentially a question of whether two parallel lines can intersect at one point
@@vdm942no
@@vdm942 Agreed.
Remember what Sal Khan always said in the middle school math courses; What you do to one side, you also have to do to the other side
This is why i had a high school teacher who said he doesn't like calling it bringing to the other side because it causes confusion like that
You need to be doing the same thing on both sides, so he emphasizes that point in the equality so none of us do such a bjg mistake
My teacher also hated ‘cancel’ for the same reason. There’s no need to invoke any magic!
@@rezwhapDoes that also apply to Twitter?
@@rezwhapI agree 100% with cancellation UNTIL every one in the class can say WHY it cancels. Some reasons that they need to articulate are:
1. These cancel because their sum is zero and zero added to rest of the expression doesn’t change it (this is using the idea of additive inverses and addition identity element.)
2. Dividing by the same expression is 1 …
3. Squaring is the inverse operation of square root, but could introduce an extraneous roots.
Looking back, I think my math teachers, all of them, never told me that a math problem can have no answer. It was only in engineering school that I learn that it does happen sometimes.
Interesting. You didn't have definition sets and solution sets?
Why not remove x immediately and you got 2 = -2. No fuzz, one step.
I legit thought I was an idiot for the first few seconds because I was “this *equation* makes no sense… right? Did I miss something? Oh god what did I forget. Sure I graduated college years ago, but surely I didn’t forget something so basic already… what’s the solution because I’m probably wrong I guess”
Swear to god I started doubting my own knowledge so hard
I love how literally every step they take is incorrect in some way
I mean only the first step really is
@@be7256even step 2 is an over complication of the supposed solution, you already have a simplified equation and they expand it out which just adds more steps to the totally new problem they wrote in step 1
One nice way to show that there is no solution that fits well with the way kids are taught math: Graph the first line y=x+2. Now graph the second line, y=x-2. Now show that the two lines do not intersect (they have the same slope) and so there is no x that satisfies both sides.
Lol I can really relate your sadness at 1:51
2:41 divide by x-2
😊
Because he spent so much time going through everything I had an existential crisis where I KNEW that the equation was impossible, and was actively dreading that he was actually going to show a solution that worked somehow and turn my world upside down.
1:10 Gesture erasing? Nice feature 😂
x+2=x-2
x-x=-2-2
0=-4
If we plug in -4 in place of the x, we get -2=-6. If we plug in 0 instead, we get 2=-2.
The numbers in the right of the equations are the numbers in the left minus 4 (-4), and for the equations to make sense -4 must be the neutral element of the sum so that -2-4=-6 and 2-4=-2 would be possible.
Here is why this doesnt work.
Can you teach me how can I sove this problem, please?
sqr(a)+sqr(ab)+sqr(abc)=12
sqr(b)+sqr(bc)+sqr(abc)=21
sqr(c)+sqr(ac)+sqr(abc)=30
Find: (a^2 + b^2 + c^2)
Tried but couldn’t solve it. I’d love a video on this problem
(a^2 + b^2 + c^2) is below the first three equations and on the right side of 'Find:'. Thanks me later.
There's probably a better approach but you could solve it by brute force. I suggest substituting √a, √b and √c by u, v, and w just to get rid of the square roots.
You'll then get this system of equations:
u + uv + uvw = 12
v + vw + uvw = 21
w + uw + uvw = 30
Three equations with three variables should yield solutions.
Then, plug the solutions into the last term. (Just remember that it will have to be u⁴ + v⁴ + w⁴.)
I had WolframAlpha do the work for me. There are three sets of solutions. the sum of squares that we are supposed to find can be either 2433, 10002, or 312688557441/384160000.
Like I said, maybe there's a better way but the old-fashioned way should work.
If I had to do it by hand, I'd start by subtracting the second from the first equation:
u + uv - v - vw = -9
We can isolate u by factoring it out and bringing the other stuff on the right hand side:
u (1+v) - (v+vw) = -9
u = (v+vw+9) / (v+1)
it took me some time and I must have made a mistake in my regular scholarly attempts, but then it jumped at me: it works for a = 1, b = 1, c = 100, so your result is 10002.
method: first substitution for those ugly sqrts: A = sqrt(a), B = sqrt(b), C = sqrt(c), so you get
A + AB + ABC = 12
B + BC + ABC = 21
C + AC + ABC = 30
now you see every next equation is 9 bigger, so it is as if you subtract 1 and add 10. and this really works if A = 1, B = 1, C = 10 so you get
1 + 1 + 10 = 12
1 + 10 + 10 = 21
10 + 10 + 10 = 30
A^4 + B^4 + C^4 = 1 + 1 + 10000 = 10002
@@popularmisconception1 Thank you, I liked your idea, but you found the answer by guessing.. Can we solve it by mathematical steps?
You can also show this visually by graphing y=x-2 and y=x+2. You get two parallel lines meaning they never touch and never give the same output for any x value.
Before I watch your video I'm going to say NO SOLUTION.
How can there be ?
There are actually solutions. Both infinity and negative infinity work.
And if you you are in Finite Field of Integers modulo 4 then 2 is a valid solution.
The "solution" presented in the video is of course bonkers.
There is a solution but a graph is more useful as this is a function that has no x intercepts and is parallel.
@@kaltaron1284 Well ...
You can subtract 2 from infinity but if you can add 2 then it wasn't actually infinity to start with 😛😜😝🤪
@@MrMousley I don't think you understand the concept. Search for the Hilbert Hotel for a visuaisation.
This is why i hate the whole "bring to the other side" terminology. You're not bringing it to the other side, you're dividing by the same amount to both sides of the equation. And this simplification of terminology is fine for most people, but is used way too much in classes where people are trying to learn algebra and then they leave the class not understanding what they actually did.
I feel the same way about "cancelling"
I will be born tomorrow and i solved this,how could tiktokers not
Happy birthday
W. I was enthralled. Told in a comedic but informative way. Gud content. Gud teacher.
This is wrong. x is obviously {0, 1} in Z_2 (mod 2).
Or 2 in Z_4.
I got of the Z_4 answer myself, but I would have thought the problem couldn't exist in Z_2 because 2 isn't an element of the Z_2 group.
@@willempye73 2 can't be a solution because it's not part of the field but can be part of an operation.
It has no solution, yes. But it can be solved in certain condition.
For an example, in modulus 4 any natural number can works for x.
I hate people who are bad at math, and think they are good at it. Even worse - people who know SOME math and do wrong things on PURPOSE and then brag about it just to get FREAKING COMMENTS OF PEOPLE WHO GET MAD AT THEM BUT DON'T UNDERSTAND THAT THIS IS EXACTLY WHAT THEY WANT
I was a little bit worried when you started dividing by x-2 and thought I have forgotten everything I ever knew about math. I was relieved when you continued and showed the correct solution, which is also the same solution I got to.
Actually, allowing for sufficient inaccuracy, x=infinity.
As
x 》infinity
Then
(infinity+2)/(infinity-2) 》1
Exactly. And -infinity as well.
@@vasiliynkudryavtsev True. Well spotted. 😊
Tis not how infinity works
@@treeNash Ah, yes, and neither is it how infinity DOESN'T work.🙃
*_No, infinity is NOT a valid solution._*
Premise 1: A=A
Premise 2: A-X ≠ A
Conclusion: Therefore, infinity-X ≠ infinity
To argue otherwise is to commit a special pleading fallacy.
Or, if you prefer:
Premise 1: Set {A} includes all real positive numbers (is ∞).
Premise 2: Adding any positive number X to Set {A} has no impact because Set {A} _already includes_ X.
Premise 3: Subtracting real positive number X _from_ Set {A}, decreases the size of Set {A} by _removing_ something from the set.
Conclusion: Therefore, while ∞+X = ∞, ∞-X ≠ ∞.
This is _not_ a special pleading because {A} _is defined as_ ∞.
To use a more concrete example as an analogy for the second syllogism, let's say {A} equals "all automobiles." When any new year's product line is made available, {A} will remain unchanged because, by definition, it *already includes* all of those automobiles. Conversely, if we subtract "minivans" from {A}, there's a material reduction in the size of {A} that can be observed.
If you still disagree, all you have to do just *graph* it. You will end up with two parallel lines. The fact that they are parallel and will never converge proves conclusively that you are wrong.
This is all beside the fact that performing the basic algebraic operation of subtracting X from both sides of the equation (this is the subtraction property of equation) yields +2 = -2 which is obviously _false._ And, we can continue the subtraction property of equality to yield 0 = -4, which is also obviously _false._
*_STOP suggesting infinity. It is demonstrably WRONG._*
You know (just to confuse the chat), there are structures where -4 might equal 0 though.
For example, in the GaloisField of order 2 (aka GF(2), or you could call it the Boolean Algebra, if you wanted to; Now basically this is integral math but modulo 2; or in terms of boolean algebra, multiplication is "logical and" and addition is "exclusive or" ). Ok, anyway, so there is not really a symbol 4 there, but you can just interprete it as 1+1+1+1 (ok, -1-1-1-1 for -4), aka 4 times 1, and it does indeed equal 0. So there the equation would actually mean x=x which is always true and in case of GF(2) x can be 0 or 1.
Sorry for this knitpicking comment.
You are SOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO right!
TikTok should be banned!!
x = infinity might be a solution, since if we put that value for x, it balances both equation.
We have: x = x + 4
Now substitute x into x:
x = (x + 4) + 4
x = x + 8
x = x + 16
…
And similarly this can be done by first subtracting 2 over,
ie, x = x - 4
The only way adding these finite quantities can work without affecting anything is if x is +/- ♾️
Reminds me of that time I misread "tweezers" as "pliers" and did a ton of unnecessary algebra to calculate the mechanical advantage. Somehow, you remembered how to solve a quadratic, but failed on the most simple part.
x+2=x-2 -> No solution
x+2=-x-2 -> Solution = -2
^
|
As a math major, that thumbnail caused me actual mental pain.
The ans is infinity
Or negative infinity -♾️
*_No, infinity is NOT a valid solution._*
Premise 1: A=A
Premise 2: A-X ≠ A
Conclusion: Therefore, infinity-X ≠ infinity
To argue otherwise is to commit a special pleading fallacy.
Or, if you prefer:
Premise 1: Set {A} includes all real positive numbers (is ∞).
Premise 2: Adding any positive number X to Set {A} has no impact because Set {A} _already includes_ X.
Premise 3: Subtracting real positive number X _from_ Set {A}, decreases the size of Set {A} by _removing_ something from the set.
Conclusion: Therefore, while ∞+X = ∞, ∞-X ≠ ∞.
This is _not_ a special pleading because {A} _is defined as_ ∞.
To use a more concrete example as an analogy for the second syllogism, let's say {A} equals "all automobiles." When any new year's product line is made available, {A} will remain unchanged because, by definition, it *already includes* all of those automobiles. Conversely, if we subtract "minivans" from {A}, there's a material reduction in the size of {A} that can be observed.
If you still disagree, all you have to do just *graph* it. You will end up with two parallel lines. The fact that they are parallel and will never converge proves conclusively that you are wrong.
This is all beside the fact that performing the basic algebraic operation of subtracting X from both sides of the equation (this is the subtraction property of equation) yields +2 = -2 which is obviously _false._ And, we can continue the subtraction property of equality to yield 0 = -4, which is also obviously _false._
*_STOP suggesting infinity. It is demonstrably WRONG._*
There is no solution because the slopes are the same but the y-intercept is different. It makes parallel lines that will NEVER intersect.
x+2=x-2
x=x-4
x/x=x/x-4/x
1=1-4/x
0=-4/x
x=-4/0
x=unsigned infinity
Now let's see if solution correct.
unsigned infinity + 2 = unsigned infinity - 2
unsigned infinity = unsigned infinity
Any real number added to unsigned infinity doesn't change it. Solution is correct.
Plus or minus infinity also works as a solution btw
Just in case this isn't a meme, infinity isn't a number. You can never have x equals infinity, only x aproaches infinity or lim x = inf. There's a reason why R = (-inf, inf) and not [-inf, inf]
this is what i thought right away even without actually solving idk if it's correct tho
*_No, infinity is NOT a valid solution._*
Premise 1: A=A
Premise 2: A-X ≠ A
Conclusion: Therefore, infinity-X ≠ infinity
To argue otherwise is to commit a special pleading fallacy.
Or, if you prefer:
Premise 1: Set {A} includes all real positive numbers (is ∞).
Premise 2: Adding any positive number X to Set {A} has no impact because Set {A} _already includes_ X.
Premise 3: Subtracting real positive number X _from_ Set {A}, decreases the size of Set {A} by _removing_ something from the set.
Conclusion: Therefore, while ∞+X = ∞, ∞-X ≠ ∞.
This is _not_ a special pleading because {A} _is defined as_ ∞.
To use a more concrete example as an analogy for the second syllogism, let's say {A} equals "all automobiles." When any new year's product line is made available, {A} will remain unchanged because, by definition, it *already includes* all of those automobiles. Conversely, if we subtract "minivans" from {A}, there's a material reduction in the size of {A} that can be observed.
If you still disagree, all you have to do just *graph* it. You will end up with two parallel lines. The fact that they are parallel and will never converge proves conclusively that you are wrong.
This is all beside the fact that performing the basic algebraic operation of subtracting X from both sides of the equation (this is the subtraction property of equation) yields +2 = -2 which is obviously _false._ And, we can continue the subtraction property of equality to yield 0 = -4, which is also obviously _false._
*_STOP suggesting infinity. It is demonstrably WRONG._*
If you are going to TikTok to learn math, you have already failed.
dawg why did bro do subtraction but multiplication 😭
One solution (over the space of functions, not the reals nor the complex numbers) is that x is a periodic function of period 4, such as sin(pi x/2).
Compulsively reducing things just leads to connections being lost. For example, the values of sin and cos on the unit circle from 0 to 90 degrees are so nice, unless you reduce them, then they get a lot harder to remember.
On the worst case scenario, if this was a variable question, it's still (probably) unsolvable because we weren't even explicitly told what X is supposed to be...
Common sense tells you that any finite real number for x won’t make sense because anything plus something other than 0 is no longer the same thing, or addition would be meaningless as a term and concept.
Idk if maybe infinity would fit for x, since infinity breaks a bunch of rules. What is infinity plus 2, but just infinity still.
It seems like the whole thing is either simply unequal or if you seek a bs answers then it’s a trick question.
When I looked it up to see if I had the right idea regarding infinity it does appear that the conventions used by mathematicians is such that infinity plus 2 is considered equal to infinity.
With that in mind, if x is even allowed to be infinity in the scope of the problem then that’d be an answer, and if the scope of allowed or relevant answers doesn’t allow for anything but real numbers than it would be simply an inaccurate mathematical equation as the two sides are not equal.
Technically, you COULD make x a division by zero, that way the "common denominator is zero" and the equation works (also, infinity)
no, you cant
@@harrietjameson
X+2=x-2
0/0+(2*0/0)=0/0-(2*0/0)
0/0+0/0=0/0-0/0 this Is true.
Inf+2=inf-2 this is also true, as you can't add or take away from infinity
@@SeleverEnjoyer not true, 0/0 is indeterminate form for a reason, doesnt matter if its common denominator. Plus you're still adding on one side and subtracting from the other
however, as x approaches ±infinity the equation does get more and more correct, but its not because you "cant take away from infinity", its because the tiny 2 has literally nothing in comparison to an unbounded sum
but at that point you arent solving, just finding some limit property
Actually, I can solve this one.
If you think about the problem you realise that we talking about a point (x like treasure) and when we move oposite directions the same ammount (+2; -2) we get the same place.
I can only think about a circle (maybe part of a sphere or something).
Now we know x is a point on a circle and the opposit "end" 2 units from this point both sides and we also know the circle circumference is 4.
From these informations we can deduce that the question is the radius of the circle.
r = ?; K = 4;
K = 2 * r * Pi
4 = 2 * r * Pi
r = 2 / Pi
r = 0,6366...
Solved!
xD
3:56
"Of course, its bull..."
My brain autocompleted your sentence that way with your voice.
Was quite the fun surprise
This has the same energy as
"I have twins, how many months should they take to arrive ?"
"9+9=19 😊"
another way to think of it is that x must occupy a unique place on the number line, and thus cannot be translated in different directions simultaneously and still retain a unique position
A simple way to illustrate the solution is with a graph, showing the two parallel lines that do not converge hence can not have the same value at any point.
Once in a while I have nightmare of being back in middle school and having to do stuff without the use of a calculator, or being back in advanced math in highschool and not remembering ANYTHING... But I'm pretty sure I'd be okay if that was the level of equations I got.
I appreciate your commitment to making a short video with the perfect duration. You da best!
There is one place where you can do x = x - 4, not as an equality but an instruction.
In some programming languages, this would mean define a new variable x, then set the value as 4 less than what x was originally.
the first thing that I learned for solving equations like this: plug your result into the original equation and see if it checks out. its so simple...
I love how he keeps doing false steps just to mess with us, pretty funny
I put in in a graph just to avoid the hassle and nothing appeared, so it checks out.
I genuinely didn’t see any issues about the tik tok solution at first, but then I saw the question
I could see a teacher in an elementary algebra class giving the equation for the students to solve, as an object lesson that not all equations actually have solutions.
Laughed uncontrollably from 2:30 on
If you want the graphical solution:
x+2 is a line parallel to the x-axis, but two units up.
x-2 is a line parallel to the x-axis but two units down.
The solution is where the two parallel lines meet.