Everything is possible | A tricky math question

[@snigdhasingh5682]

This problem is complex only in the sense that it uses the concept of "complex" numbers.

[@exyile3323]

so it isn't really all that complex if u just think that the complex numbers are just algebra lol?

[@erikhicks6184]

Did you assume something else?

[@TheREALDocRabbit]

"complex" numbers aka numbers that the real rules of math prove aren't possible being used to "prove" a concept are as useful as eating the "air" sandwich in your hand.

[@Segalmed]

y = 1-x; insert in 2nd equation; x(1-x)=1 => x² - x + 1 = 0 ; standard formula yields x1,2 = 0.5 +/- 0.5 * SQRT(-3) = 0.5 +/- 0.5*i*SQRT(3) => y1,2 = 0,5 -/+ 0,5*I*SQRT(3)

[@PatrickL91-l1g]

moi à l'iut je faisais de tete lorsque on a a + b = S ab = P a et b sont les solutions de l'equation t² - St + P Delta = S² - 4P x,y =( S +/- racine (Delta) ) / 2 x = e^+/-ipi/3 en 4eme !!!! | x².y + x.y² = -540 | x.y = -45

[@therongjr]

Everything is possible if you don't limit yourself to reality.

[@bananacorn4375]

You just have to use your imagination

[@litrehead]

Jokes aside, why should call them complex not imaginary. Everything is possible, but probably not because it's complex. Is eipi transcendental?

[@maxpolaris99]

🤣 I've suspected that for a long time, thanx for verifying.😁

[@UserSams-ve2mj]

There's a much faster way to get x² -x + 1 =0 The sum S and the product P of x and y are directly given; S = 1 and P = 1. You can apply the formula x² - Sx + P = 0 x² - x + 1 = 0

[@ferreolduboiscoli]

That's what I wanted to say lol

[@БелАлекс]

Это ведь теорема Виета.

[@UserSams-ve2mj]

@@БелАлекс No translate option, what are you trying to say?

[@TenToTu]

@@UserSams-ve2mj he said "This is Vieta's theorem."

[@UserSams-ve2mj]

@@TenToTu Ok ty

[@warrengibson7898]

This dude’s singsong delivery has me LMAO

[@sankadubinin]

(x + y)^2 = x^2 + y^2 + 2xy = x^2 + y^2 +2 = 1; x^2 + y^2 = -1; NO ROOTS. THATS ALL! If you want unreal roots, you MUST make a note about it!

[@GiovanniMascellaro]

The square of (-1) is (1), because ((-1) \times (-1) = 1). However, when we talk about (i), the imaginary unit, the situation is different. By definition, (i) is such that (i^2 = -1). By definition i^2 = −1

[@CebGIN]

The square root of -1 isn't 1, 1*1≠-1

[@NightHawkGameplays]

He said square, not square root

[@CebGIN]

​@@NightHawkGameplaysAnd was has to do this with the problem?

[@NightHawkGameplays]

@@CebGIN it has to do with you wrongly correcting his comment

[@CebGIN]

@@NightHawkGameplays I mean with the math problem In the video

[@octobrot7591]

I just got here, gotta say I love the handwriting

[@wes9627]

The simple approach. Substitute x=1/2+z and y=1/2-z into the second equation: (1/2+z)(1/2-z)=1 or z^2=-3/4. Thus, z=±i√3/2, x=(1±i√3)/2, and y=(1∓i√3)/2

[@Mosentsev_Danielle]

nice to know I am smart enough for that :3

[@matthewlloyd3255]

Why can't you just graph y = 1 - x and y = 1 /x and find the intereception points between the two curves on the graph. It might not be all the solutions but it will be some of them.

[@RikiFaridoke]

You can using complex analysis method sir.

[@Mustafa_Shahzad]

You can use* not using

[@AIOGaming07w]

​@@Mustafa_ShahzadI was about to write the same, lol

[@RikiFaridoke]

@@Mustafa_Shahzad where is my mistake located?

[@Stuytz]

​@@RikiFaridoke in your comment

[@RikiFaridoke]

@@Stuytz try by yourself solving it via complex analysis sir.

[@anatolykhmelnitsky2841]

Let's substitute x for x₁ and y for x₂. Now we have x₁ + x₂ = 1 and x₁ * x₂ = 1. For x² + bx + c = 0 Vieta's formula is x₁ + x₂ = -b and x₁ * x₂ = c. All we need now is to solve quadratic equation where b is -1 and c is 1. It's just x² - x + 1 = 0. Quadratic formula will do the rest of the job. That's all folks!

[@matthewalan59]

Wow, what a huge number of comments from people who have an extremely limited background in mathematics. Just because we have used the term "imaginary" does not mean that such numbers are any less "real" than any other kind of number. I recall grade eight students trying to argue that you could not have less than zero and that negative numbers were not real. Many years ago, as an exercise, I used the general solution for a cubic equation to solve a specific cubic equation. What was fascinating to me is that the three roots were all real but the details of the general solution involved complex numbers. In the end the imaginary parts all canceled out. If, during the solution, you were to conclude that there were no real solutions because square roots of negative numbers were appearing, you would be completely mistaken.

[@KrytenKoro]

x2-x+1=0 X=[1+-(1-4)]/2 1/2+-(3)i/2

[@rainerzufall42]

Another Vieta problem => Immediately clear, that x, y are the two z solutions of z^2 - bz + c [-b = 1, c = 1] = z^2 - z + 1 = 0 => z_1/2 = b/2 +/- sqrt(b^2/4 - c) = 1/2 +/- sqrt(1/4 - 1) = 1/2 +/- 1/2 sqrt(3) i => x = z1, y = z2 OR x = z2, y = z1. Use the symmetry for God's sake! x and y are totally exchangeable!

[@anonymouscheesepie3768]

y= 1/x and y = 1 - x 1 - x = 1/x x - x^2 = 1 -(x^2 - x + 1) = 0 x = (1 +- sqrt(1 - 4))/2 x = (1 +- sqrt(3)i)/2

[@ahojukka5]

Raise the first equation to the power of two, get x^2 + y^2 + 2xy = 1, substitute the second equation to get x^2 + y^2 = -1 = i^2.

[@pjaj43]

Obviously x & y are interchangeable by inspection so x1 = y2 and x2 = y1 therefore no need to calculate one from the other, just use the two results of the well known binomial solution. This is almost mental arithmetic.

[@eyalbryan9708]

Better way to find y: Xy=1--> y=1/x--> y is the conjugate of x

[@maxpolaris99]

I've done problems that required finding all roots to a quadratic before, but I'm still befuddled. It still makes me scratch and shake my head. 😁

[@lijath]

Why don't you just apply both of them to an array? It seems like the long way around.

[@MrCool-fm5ty]

The first root x1 =e^ipi/6

[@TransportGeekery]

Love it! Especially the additional steps to work out 1/4 + 3/4 😂

[@Rafachus]

Y=1-x x^2-x=1 x^2-x-1=0 ∆=1+4 √∆=√5 x1=(1-√5)/2 x2=(1+√5)/2 y1=1-(1/2-√5/2) y1=(1+√5)/2 y2=1+(1+√5)/2 y2=1,5+√5/2

[@katsiarynadubovik4213]

x+y=1 xy=1 x^2+xy=1 xy=1 x^2+1=1 xy=1 x^2=0 xy=1 x=0 xy=1 x=0 0*y=1 0*a=0 answer: equation doesn't have roots

[@n_kliesow]

I prefer completing the square, but I've never seen this way of solving. In Germany we are told either pq-formular, which is similar to this you used or completing the square (Quadratische Ergänzung). Most teachers prefer pq-formular because it's just learning a series of terms with no explaination. I was in luck learning both, my teacher hat 2 examples where the formular failed, and I am not a fan of learning too much stuff. I rather like to understand how it works. 😅 Under the hood they are both as same as fast. But I'm a bit confused there is another formular.

[@istvange11ai]

I just want to ask what kind of pen you use. Because your writing style looks very nice.

[@martinphipps2]

xy=1 and x+y=1 so x+1/x=1 so x^2+1=x so x^2 -x +1=0 so x=[1+-sqrt(1-4)]/2 which means x is either [1-isqrt(3)]2 or [1+isqrt(3)]/2. Yeah, sure. Real hard.

[@remy131]

i achieved the same solution setting xy=x+y, it just took more steps! :)

[@kalki4163]

11 minutes and it took me 3 seconds. Speed running champion

[@gregorymorse8423]

I could solve it in my head without seeing the video. So "tricky" is a pretty relative term.

[@RealQinnMalloryu4]

(0.5)+(0.5)=1 (y ➖ 0.5x+0.5). (xy ➖ 1xy+1). .

[@WyzZ_03]

Excuse me, what?

[@knucklehead83]

No 2 numbers less than 1, when multiplied can equal 1. So x or y has to be negative. But not both. I haven't watched yet.

[@bosssnurp5912]

Linear graph and a hyperbola

[@SALogics]

Very nice problem! ❤❤

[@wolfmanoh]

Is the D=(b^2)-4ac thing just to find out if the answer is a complex answer? Cause I was just taught to use the Quadratic Formula, which is what the x=[-b±SQRT(D)]/2a is.

[@UserSams-ve2mj]

If the discriminant is positive >0, then there's 2 real different roots. If it is = 0, there's 1 real double root. If it is negative, there's no real roots

[@ml09394]

Yeah, the idea is the discriminant D determines what the answers will be like. All the other parts of the quadratic formula are normal numbers but the discriminant is under a square root. That could make it imaginary, irrational, or rational.

[@shailenderrai]

the discriminant (sqrt of b^2-4ac)determines the number of solutions. if it is =0, there is 1 solution. if it is greater than 1, then there are 2 solutions. if less than one then there are complex solutions.

[@t_c5266]

What I did was type it into Matlab and hit "solve"

[@Misiok89]

It would be easier to solve geometricly. Just draw unit circle where the real part will add to one.

[@douglaswolfen7820]

I'm sorry, how does that help? If you're drawing the unit circle on the complex plane, then that gives you all the numbers with magnitude equal to 1 But how do you know that one of those numbers is going to be the answer? And how do you know which one?

[@Misiok89]

​@@douglaswolfen7820Both. I assumed without any solid basis for it that for both roots of this equations we have absolute value equal to 1 so we are allow to look for solution on unit circle. In order for the imaginary part to cancel out, we need a certain value and its conjugate (base on x+y = 1 part) - it could lead us to the fact that we can look for solutions on the unit circle based on x*y = 1 part real part can be found based on x+y =1 and imaginary part from Pythagorean theorem will give us square root of 3 imaginary part.

[@douglaswolfen7820]

@@Misiok89 Huh, interesting. If you want two numbers that add together to make a real number, then they don't _have_ to be complex conjugates. But I suppose it is an easy way to find pairs of numbers that fit the requirements. It guarantees that their product and their sum will both be real And putting them on the unit circle guarantees that their product will have a magnitude of 1. So then you just have to pick the pair that has a real part of ½, so that the sum is also 1. (And then Pythagoras gives you the imaginary part) I like it!

[@Misiok89]

@@douglaswolfen7820 You are right... it is all about angles... (2 + i) * (4 - 2i) gives 10 so they do not have to be complex conjugates for result of multiplication to be real number... But at least have to lay on lines that are symmetrical about the real axis... So the number of potential solution would explode because i do not know how to narrow it down to initial assumption about unit circles. Their absolute value would have to just multiply to 1 and real sum add to 1. We will have to find that if we get something different then unit circles that lay on those lines and its multiplication is equal to 1 then real part sum would be greater then 1... but i think that it wont be possible to get it from drawing... you can show multiplication by drawing line that cross unit, but from drawing you do not get grounds to assume that the real part must be equal to each other... "It guarantees that [...] their sum will [...] be real" - that was maybe that missing part. We can know that they lay on the symmetrical lines to real axis (passing through the point 0 + 0i), and in equal distance from it and it will give us the right to assume that they are conjugated.

[@douglaswolfen7820]

So after some thought, it turns out that they really do have to be complex conjugates When two complex numbers are multiplied, their arguments are added. So if the product is going to be a real number (argument 0), then the two numbers need to have equal and opposite arguments When two complex numbers are added, the imaginary parts add. So if the sum is going to be real (imaginary part zero) then the imaginary parts have to be equal and opposite If the original numbers aren't both real numbers, then they have to have equal and opposite arguments _and_ equal and opposite imaginary components. That means they're guaranteed to be complex conjugates

[@michallesz2]

With the system of equations x+y=1 and xy=1 we calculate y from the second equation y=1/x. Now we substitute y into the first equation and get x+1/x=1. In this way we have obtained an incorrect equation, which means that the result of the equation with the value 1 cannot exist. If we substitute any number for x into this equation, it turns out that the result cannot be numbers greater than -2 or less than 2, because such a result does not satisfy the conditions of the equation. To prove that the equation is incorrect, it is enough to substitute the variable a/b for x, then the variable 1/x will be b/a. So let's solve the equation with the new variables. a/b+b/a. The result will be (a^2+b^2)/ab. As we notice that a^2+b^2 is equal to c^2 from the Pythagorean formula, therefore the equation looks like c^2/ab. This means that the square of the hypotenuse is divided by the product of its remaining sides. From such division we get that only with equal legs the result c^2/ab will be the number 2. For example, in a right triangle with sides 1,1 the third side will be V2. So V2^2/1*1=2. In other cases the result will be greater than 2 and I will give an example with sides 3,4,5 where 5 is the side opposite the right angle. So 5^2/3*4 = 25/12 = 2.0833333.. And since in your equation we have x+1/x=1, it means that the result 1 is wrong. If you do not believe that this is a mistake, please substitute any number for x and see for yourself what the results of this equation will be.

[@mercyfulsin]

Y=1/x being substituted back does not give x+1/x=1. You forgot the square. It's actually x^2+1/x=1 which is not a wrong statement

[@KrytenKoro]

It's not a mistake. It has a complex solution, not a non-complex solution

[@KrytenKoro]

Your answer is incorrect because you're excluding complex answers.

[@PopeVancis]

x+y=1 xy=1 x=1-y (1-y)y=1 y-y^2=1 -y^2+y-1=0 y=(-b +/- sqrt(b^2-4ac))/2a y=(-1 +/- sqrt(1^2-4(-1)(-1)))/2(-1) y=(-1 +/- sqrt(1-4))/-2 y=(-1 +/- sqrt(-3))/-2 y=(-1 +/- isqrt(3))/-2 y=1/2-(sqrt(3)/2)i, 1/2+(sqrt(3)/2)i y=0.5-0.866i, 0.5+0.866i For y=0.5-866i, x=1-(0.5-0.866i) x=0.5+0.866i For y=0.5+0.866i, x=1-(0.5+0.866i) x=0.5-0.866i So (x,y)=(0.5-0.866i, 0.5+0.866i), (0.5+0.866i, 0.5-0.866i)

[@oida10000]

xy=1 | x+y=1 --> y=1-x x(1-x)=1 | a(b±c)=ab±ac x-x^2=1 | -1 -x^2+x-1=0 | *(-1) x^2-x+1=0 | x^2±2ax=(x±a)^2-a^2 (x-1/2)^2-1/4+1=0 | -3/4 (x-1/2)^2=-3/4 | sqrt() x-1/2=±i*sqrt(3)/2 | +1/2 x=(1±i*sqrt(3))/2 --> y=(1∓i*sqrt(3))/2 Check x+y=(1±i*sqrt(3))/2+(1∓i*sqrt(3))/2=2/2=1 ((1∓i*sqrt(3))/2)*(1±i*sqrt(3))/2)=((1^2-(i*sqrt(3))^2)/(2*2)=(1+3)/4=4/4=1.

[@nikakhundadze]

can we do it using linear function and aka fracitonal funciton? x and y will be points of intersection, tell me if i am incorrect i just had a glance on problem and this was first thing i came up with

[@tharunsankar4926]

Let x and y be the roots of the following quadratic equation X^2 + X - 1 = 0 X = 1/2 +/- sqrt(5)/2 => phi, 1- phi where phi is the golden ratio x = phi, y = phi - 1

[@g2h5j3]

What about the real solution? (X,Y)1 = { x e R \ {1}, y = x/(x-1)} (X,Y)2 = { y e R \ {1}, x = y/(y-1)}

[@pi_xi]

This is not a solution, because you have a circular dependency between x and y.

[@g2h5j3]

@@pi_xi yeah it solves x+y=xy but none of them will be equal to 1...

[@pi_xi]

@@g2h5j3 Also none of them will be a real number.

[@g2h5j3]

@@pi_xi its set of real numbers?

[@g2h5j3]

@@pi_xi or could you say those sets are derived from the complex number solution?

[@cybersolo]

Starting with x = a+b*i y = c+d*i You get a+c = 1 b+d = 0 a*c-b*d = 1 b*c+a*d = 0 The solution is immediate.

[@IgiveStress]

WHAT HAPPENED TO HIS RIGHT HAND’S LAST 2 FINGERS???, THEY ARE PITCH BLACK

[@Aquafoxxo]

Probably a drawing glove

[@stevea.b.9282]

If this is a high school question, then I went to something else when I was 11

[@Denissium]

At first I was like "Oh. Well that is just square root 1" then I thought about it and OH fuck me it is not that

[@ruhi435]

Yeh I had solve it correctly

[@soundsoflife9549]

I get imaginary when life gets negative also.

[@supremeakahandsomelegit8187]

Ayo bro why's bro hand half black half white?

[@adamm8280]

What's tricky about this one?

[@DivinityAureole]

It's tricky to the average person

[@beardedemperor]

@adamm8280 it's tricky on its face because: X and Y can't both be negative; they sum to a positive number. X and Y cannot be one negative and one positive value; they give a positive number when multiplied. Neither X nor Y can be zero; they have a non-zero product. Thus both X and Y are positive numbers. Now, X and Y aren't both 1; they sum to 1, not 2. For two positive, non-1 values to have a product of 1, one (let's say X) must be greater than 1 and one (Y) less than 1 (e.g X=2 and Y=½; X=1⅔ and Y=⅗ etc.). But the sum of two positive numbers is always greater than either of those numbers. In other words X+Y > X > 1. Thus X+Y > 1. But also X+Y = 1. Thus no solution is found.

[@ayeflippum]

The video really didn't need to be this long.

[@learningwithayush3384]

Bro wrote the equation incorrectly

[@Genuenly]

Went from substitution method, quadratic equation , to wtf did I just witnessed? From a Grade 9 student

[@Jo_aJoaninha]

From a nerd 9° grade student, let me try explaining the process. 9° graders are not always familiarized with complex numbers. There's a thing in math called " i ", such that i = √(-1). When resolving the quadratic, an √(-3) will appear. It can be rewritten as √(-1·3), then split in two roots, as √(-1) · √(3). Now, if √(-1) = i, it is clever to make this substitution. i · √(3). By this process, therefore, we realise √(-3) = i·√(3). So we can make an substitution in the original quadratic: x = [1 ± i·√(3)]/2 then, the teacher splits this sum on the denominator as the sum of ratios. x = ½ ± i·√(3)/2 This is done just to place our results in *complex number form*. (if you're interested in understanding complex numbers, there's plenty of good videos about it). Then, it's just expressing the two values of x separately, one with a minus, and one with a plus. Getting the two values of X, we'll need the two values of Y. This is done by substituting X in any equation. Now we've found the two possible values for x and y.

[@Genuenly]

@@Jo_aJoaninha I understood everything until the X = [1± I • √(3)]/2 part

[@krusko5644]

​@@Jo_aJoaninha Tysm I now somewhat understand this (im 10th Grade)

[@yesiamrussian]

i already knew this but thanks for helping others (im 10th grade)

[@Emerald1]

@@Genuenly You're familiar with the quadratic formula, right? the x = [-b ± √(b^2 - 4ac) / 2a equation? That part under the square root, "b^2 - 4ac" is called the discriminant, which is what was found when we got -3 (in the video), so that was then put under the square root, but since that's a complex number (because of the negative in the square root), we did the process done in Jo_aJoaninha's reply. We just then substituted it back into the quadratic formula. That is what gave us x = [1 ± i·√(3)] / 2. The -b was -(-1) The √(b^2 - 4ac) was √(-3) which then became i·√(3) The 2a was 2(1), which simplified to just 2. After that, since the numerator is being split by addition/subtraction (±) with the same denominator, it was split into 1/2 ± i·√(3) / 2 The ± is meant to denote two solutions, where one has addition and one has subtraction, so the solutions for x become 1/2 + i·√(3) / 2 and 1/2 - i·√(3) / 2 From there you just plug those solutions back into y = 1 - x to find your possible solutions for y, which are quite literally the same thing as your solutions for x, but flipped, since you want x and y to be conjugates for this problem. A conjugate is when you have two expressions that are only different in the sign between the two terms. For example: the conjugate to an expression (a + b) would be (a - b)

[@a.s.vanhoose1545]

I have questions and they’re not related to the problem

[@scientious]

An imaginary solution.

[@jonathanrivas3004]

just complete squares ._.xd

[@the_ant_guy]

The longest paper in the west

[@thevalarauka101]

just tried it in my head; is it 1/2 plus or minus sqrt(3)/2 i? just simple simultaneous equations

[@JSSTyger]

Correct...and whichever one gets the "minus", the other will get the "plus" and vice-versa

[@hiteshnachankar1084]

What a ez equation Btw was this supposed to be hard? I'm genuinely curious Wel if u didn't study maths in highschool then definitely this should be magical to u A great mind once said "don't study physics chemistry and maths in school and live in a world filled with magic and miracles"

[@vanoelizbarashvili6655]

araa 11 wuti sachiro

[@Ctrl_Alt_Sup]

xy=1 so x and y have the same sign x+y=1 so x>0 and y>0 x²−x−1=0 is the equation of the golden ratio so x=φ and y=1/φ with φ=(1+√5)/2 S={(1; φ) ; (φ;1)}

[@olteanumihai1245]

in boolean algebra that would work just fine if x=y=1

[@paolovolante]

Got this answer from a LLM AI, Claude 3.5 Sonnet: Let's solve the system: x + y = 1 x * y = 1 Here's the step-by-step process: From the first equation, we express y in terms of x: x + y = 1 y = 1 - x We substitute this expression for y into the second equation: x * y = 1 x * (1 - x) = 1 We expand the equation: x - x² = 1 We rearrange it into the standard form of a quadratic equation: x² - x + 1 = 0 This is a quadratic equation in the form ax² + bx + c = 0, where: a = 1 b = -1 c = 1 We use the quadratic formula to solve it: x = [-b ± √(b² - 4ac)] / (2a) x = [1 ± √(1 - 411)] / (2*1) x = [1 ± √(-3)] / 2 We simplify: x = [1 ± i√3] / 2 So, the two solutions for x are: x₁ = (1 + i√3) / 2 x₂ = (1 - i√3) / 2 To find the corresponding y values, we use y = 1 - x: y₁ = 1 - x₁ = 1 - (1 + i√3) / 2 = (1 - i√3) / 2 y₂ = 1 - x₂ = 1 - (1 - i√3) / 2 = (1 + i√3) / 2 Therefore, the complete solutions of the system are: (x₁, y₁) = ((1 + i√3) / 2, (1 - i√3) / 2) (x₂, y₂) = ((1 - i√3) / 2, (1 + i√3) / 2) These solutions satisfy both equations of the original system.

[@douglaswolfen7820]

@@paolovolante nicely done! Just to check though: you do realise that your "two" solutions are identical, except for the order of the two numbers? I think that most people wouldn't consider it important to determine which of the two numbers was x, and which of the two numbers was y, so I think this would usually be treated as one solution

[@mikko_mikko_mi]

NO.

[@gundummies]

When op wants to say first, he says "pers" When op wants to say pairs, he says "pers" Good job with your accent.

[@ioi0001]

And where is the tricky math question?

[@HQdefault64]

The way this guy draws x…

[@powertaco2867]

This is completely subjective. But i actually cannot stand the way you are writing X

[@TrackedFish]

How would you write it?

[@powertaco2867]

@TrackedFish with 2 straight lines. Like a normal person

[@willpina]

Pretty sure that's how murderers write an x 😂😂😂

[@andysharpei]

Proper x thanks not a junior school times sign like most math voss

[@haykor7165]

Not all people are braindead enough to write X as 2 lines...

[@blockman_games17]

what

[@josephang9927]

😭

[@erwinkharisma9657]

Easy. 0+1= 1 01 = 1 😂😂😂 Its true, but i know the question meant x . y = 1.

[@wolfgangtempl8623]

Nothing special about it. Just plain algebra..

[@lazynomad42]

So annoying

[@tomrhein-jc8bm]

Its annoying because only people that have a sense of maths see this stuff on youtube.

[@pengutiny6464]

You are :-/

[@danielnoriega6655]

The way your write the X 🤢🤮

[@Duck13e]

Unnecessary

[@pi_xi]

It's okay to make it more distant from × (multiplication operator / Cartesian product operator) and χ (Greek letter chi).