Better - and MUCH easier (no quadratics) to write out x=a+bi and y=c+di and work out values of a,b,c,d. Realize immediately 1. b+d=0 2. ad+bc=0 3. ac-bd=36 4. a+c=6 1 and 2 give b=-d a=c Then 4 gives a = c = 3 Substituting into the 3 we get b=-d=sqrt27 No quadratics. CLEAN solution.
The solution is a pair of complex conjugates, 3 ± 3i✓(3), so it doesn't matter which variable has which exact complex number as long as the other variable holds its conjugate.
No solution in non-i domain. The biggest product of xy would be if both would be equal, so both would be 3. Ergo: 9 is the greatest possible product xy.
I have a very easy method to do this.. 1. Create a relation between a-b and a+b [(a+b)^2 -4ab]^-2 = |a-b| 2. Here we can see we'll get 2 values of a-b 3. Use those values to solve the linear equation in two variables two times Takes only a minute to do this
Did you know, 4xy= (x+y)² - (x-y)² Now, if xy = 36, x+y = 6....(i) then, (x-y)² = -108 or, x-y = 6i√3..... (ii) add (i) and (ii) x = 3 +3i√3 y = 3 -3i√3
Next video, where you didn't use symmetry! x and y are interchangable! When you know the solutions for x, those for y are the same, just not in the same cases! When x_1 = s1, then y_1 = s2. When x_2 = s2, then y_2 = s1. Or { x, y } = { s1, s2 }. Maybe more intuitively: x = 3 +/- 3 sqrt(3) i y = 3 -/+ 3 sqrt(3) i
Instead of x and y... say we had x₁ and x₂ x₁+x₂=6 x₁x₂=36 Let's consider a polynomial P(x)=(x-x₁)(x-x₂) P(x₁)=P(x₂)=0, furthermore, there are no other solutions for P(x)=0 [in tamed mathematics ab=0 implies a=0 or b=0.... possibly both] So if we solve (x-x₁)(x-x₂)=0... we will know x₁ and x₂ (x-x₁)(x-x₂)=0 x(x-x₂)-x₁(x-x₂)=0 x²-xx₂-x₁x+x₁x₂=0 x²-(x₁+x₂)x+x₁x₂=0 Substituting for sum and product: x²-6x+36=0 Then it's just solving for x... Note that the initial equations are symmetric, as long as x₁ is one of the solutions, x₂ is the other.
Alternatively, you can do: (x₁+x₂)² - 4x₁x₂=6²-4(36) x₁²+2x₁x₂+x₂² - 4x₁x₂ = -3(36) x₁²-2x₁x₂+x₂² = -3(36) (x₁ - x₂)² = -3 (36) x₁ - x₂ = ± 6i √3 Remember, since x₁ and x₂ are symmetric, if you switch them out you get (x₂ - x₁)=-(x₁ - x₂)... which checks out with how there is ± for the difference between them. (then you add the x₁ - x₂=... equation with the x₁ + x₂=... equation to isolate x₁)
We all know that both functions will NOT intersect in a (x, y) coordinate graph. Therefore, there will be two non-real solutions. x+y=6 xy=36 -----> y=(36/x) x+(36/x)=6 x²+36=6x x²-6x+36=0 x²-6x+9=-27 (x-3)²=-27 |x-3|=3i√3 Case 1 x-3=3i√3 x=3+3i√3 3+3i√3+y=6 3i√3+y=3 y=3-3i√3 (3+3i√3, 3-3i√3) ❤ Case 2 x-3=-3i√3 x=3-3i√3 3-3i√3+y=6 -3i√3+y=3 y=3+3i√3 (3-3i√3, 3+3i√3) ❤
Why is it not necessary that x must equal y. To approach the problem initially you can choose either Y= (6-X) or X=(6-Y) and get the same result mirror image
x + y = 6 xy = 36 => x + xy + y + 1 = 43 => x(y + 1) + (y + 1) = 43 => (y + 1)(x + 1) = 43 Realizing 43 is a prime number and x,y have equal roles. We have: 1. x + 1 = 1 & y + 1 = 43 => x = 0, y = 42 2. x + 1 = - 1 & y + 1 = - 43 => x = -2, y = -44 So the solution of the equation is: (x;y) = (0;42) = (-2;-44) and their permutations
@@kienvantran5769solutions don't necessarily have to be integers, assuming they are leads to these mistakes edit: for this reason, it's not safe to draw any conclusions from equations containing more than one variable; you just don't have enough information, as there are infinite pairs of real numbers expressible in terms of x+1 and y+1 whose product makes up 43.
Ppl don’t realize he used simple numbers to teach the basics. Of course you can solve this much easier but that was t the point . The point was to teach the basic that you could use for any numbers. 🙄🤦♂️
Попутно доказана теорема Виета. Для тех кто только знакомиться с комплексными числами, а эта задача именно для них, было бы очень полезно провести проверку подставив решения в оба уравнения.
wow. you have got my attention with this one. It's a quadratic equation but my brain is not nimble enough so i used open ai for the result, it is often correct even with the very hardest university exam problems. It says the solutions to the equation are: 3 ( 1 + sqrt(3i)) and 3( 1 - sqrt(3i))
Did anyone notice this thing where the video area just shows a black area and no matter how long you wait the video doesnt load? Refreshing once fixes it but sill🤔
What class/grade is this? I know everything, but the i part confuses me. Do you write i when solving the square root, 4th root, 6th root etc. of a negative number?
Obviously if x,y are real, they have the same sign, since xy is positive. Since x+y is positive, x, y > 0. Since xy = 36, at least one of x, y is >=6. , But then x+y>6, contradiction. Hence x,y are not real. If x, y are complex, it is easy to see that they must be conjugates. So the real part of x, y is 3. Then the imaginary part follows as pm sqrt(27).
Instead of solving these math, you should develop your relationships :) it is better.. My boss only know plus, substract, divide and multiply, but he owes a huge company now.. 😅
Okay, I'll be the idiot that asks the question: what the hell is the answer? I mean, using g real numbers, what is X, and what real number is Y? Using the imaginary number "i" produces an imaginary result. What value does that serve us in real life? This is why I was much better at Geometry than Algebra. At least Geometry was real.
Ой, вот это ты зря. Мнимые числа - это вещь. Не существовало бы их, не было бы прогресса в области физики, мы бы не знали, что такое переменный ток, и я бы сейчас не писал этот коммент, так как человечество вообще бы не знало, что такое телефон и современная техника. Те, кто считают, что у таких чисел нет смысла - это большой плевок в сторону Дель Ферро, Кардано, Гаусса, Эйлера и многих других. Просто запомни на всю жизнь: "И" в квадрате равняется минус единице! Это тождество открыло новое множество могучих чисел, а дальше - больше, как насчет кватернионов? Благодаря кватернионам и уже тремя мнимым числам ты можешь играть в 3D игры, а также строить 3D модели. Кардано вот такой мужик был, честь ему и хвала
People use the value of "i" for math equations, even though they don't always make sense in real life. Just because it doesn't translate well, doesn't mean it doesn't help with statistics and keeping data. Use the imaginary numbers in a way that will get you to the real result you want to eventually reach
"I" was invented because we fell for the mythical negative numbers, and then realized you couldn't find the square root of a negative number. If negative numbers were real, you could have -3 elephants in your yard, add 3 more and then have 0 elephants. I drove my boys nuts with that reasoning but they survived it.
It is called Completing The Square. It is a standard solution for real numbers too, simply make d=b/2 then add e to c such that c+e=d^2 y=x^2+6x+3 = x^2 + 2.dx + c+(d^2-c) = d^2-c y=x^2+6x+3 = 0 becomes x^2+6x+9=6 hence (x+3)^2=6 and x+3=sqrt(6) giving solutions -3+-sqrt(6) y=x^2+6x+3=0 quadratic solution is (-6+-sqrt(6^2-4.1.3) ) /2.1 = 1/2(-6+-sqrt(4.6)) = -3+-sqrt(6)
There is no real solution to this problem. If you are expecting a complex solution then you need to spell it out that the domain of x is the set of complex numbers.
@@rainerzufall42 Yes really. I have never heard of a rule saying that the domain of all functions is the set of complex numbers. Should we also include all n-dimensional vectors in the domain?
@@psionl0 You'll probably know, that IC is closed with respect of solutions of nth grade polynomials. IR is not. You don't have to expand to IH. And also IC^n or IR^n are not useful as default domain. Polynomials of grad n have n solutions in the domain, seems quite fundamental, if I just knew, which theorem would give us that...
@@rainerzufall42 Your rules are getting more complicated. Unless complex numbers are part of the question or, in the context of the problem, complex numbers are useful (eg wave equations), we should not simply assume that the domain includes complex numbers.
if u r a math teacher u can earn money by teaching those things. joke is aside in electric and electronic engineering complex numbers are used to solve many problems. especially alternative current circuits ( L;C circuits) and electromagnetic waves.
X + Y = 6 XY = 36 Possibilities X = 0,1,2,3,4, and 5 Y= 6,5,4,3,2, and 1 Imagine X times Imagine Y times added together iXiY = 0+1+2+3+4+5+6+5+4+3+2+1=36 🥸
Ой, вот это ты зря. Мнимые числа - это вещь. Не существовало бы их, не было бы прогресса в области физики, мы бы не знали, что такое переменный ток, и я бы сейчас не писал этот коммент, так как человечество вообще бы не знало, что такое телефон и современная техника. Те, кто считают, что у таких чисел нет смысла - это большой плевок в сторону Дель Ферро, Кардано, Гаусса, Эйлера и многих других. Просто запомни на всю жизнь: i в квадрате равняется минус единице! Это тождество открыло новое множество могучих чисел, а дальше - больше, как насчет кватернионов? Благодаря кватернионам и уже тремя мнимым числам ты можешь играть в 3D игры, а также строить 3D модели. Кардано вот такой мужик был, честь ему и хвала
if x=3+3√3,y=3-3√3 xy≠36 this question is actually NO SOLUTION because when he do the solution you can notice that (-6)²-4(1x36)= -108 which is smaller than zero and in ax²+bx+c, when b²-4ac
Ой, вот это ты зря. Мнимые числа - это вещь. Не существовало бы их, не было бы прогресса в области физики, мы бы не знали, что такое переменный ток, и я бы сейчас не писал этот коммент, так как человечество вообще бы не знало, что такое телефон и современная техника. Те, кто считают, что у таких чисел нет смысла - это большой плевок в сторону Дель Ферро, Кардано, Гаусса, Эйлера и многих других. Просто запомни на всю жизнь: "И" в квадрате равняется минус единице! Это тождество открыло новое множество могучих чисел, а дальше - больше, как насчет кватернионов? Благодаря кватернионам и уже тремя мнимым числам ты можешь играть в 3D игры, а также строить 3D модели. Кардано вот такой мужик был, честь ему и хвала
I understood everything up until he started.
😂😂😂
Better - and MUCH easier (no quadratics) to write out x=a+bi and y=c+di and work out values of a,b,c,d.
Realize immediately
1. b+d=0
2. ad+bc=0
3. ac-bd=36
4. a+c=6
1 and 2 give
b=-d
a=c
Then 4 gives
a = c = 3
Substituting into the 3 we get b=-d=sqrt27
No quadratics. CLEAN solution.
CLEAN, if you do it correctly.
-bd = 27 AND b = -d b = sqrt(27) with d = -sqrt(27) OR b = -sqrt(27) with d = sqrt(27).
2 solutions, both clean.
That was slick! 👍👍
@@rainerzufall42 You have a point but ...
The problem is symmetric for x and y!
@@Alekosssvr Yes. I wrote that (not here). Symmetry is king.
The solution is a pair of complex conjugates, 3 ± 3i✓(3), so it doesn't matter which variable has which exact complex number as long as the other variable holds its conjugate.
Symmetry!
Thank You for watching! Have a great day! Much love and respect❤❤❤ What do you think about this solution?
(3+_3V3i, 3_+3V3i).
How did you know that complex answers were acceptable?
No solution in non-i domain.
The biggest product of xy would be if both would be equal, so both would be 3. Ergo: 9 is the greatest possible product xy.
I have a very easy method to do this..
1. Create a relation between a-b and a+b [(a+b)^2 -4ab]^-2 = |a-b|
2. Here we can see we'll get 2 values of a-b
3. Use those values to solve the linear equation in two variables two times
Takes only a minute to do this
And this is why I am a carpenter.
Natural born carpenter! Now you are the master of carpenter
I understood everything up until he started.
This just makes me sad. I used to speak and write math. Now, I can barely follow along and just get the jist of it.
So "backwards c" times "regular c" plus y equals 6. Ok I'm with you so far...
Yes, one step to get x and y values, consecutively ...
| x + y = 6
perfection!
@@GillesF31 great just one question 😅 Like Does this work for all types of these questions?
Like what if it doesn't have roots... Then?
@@Im.Struggler Yes with any data. Below is an example I imagined just now (to provide you with a proof):
| x + y = -12
|
| xy = 12
S (Sum) = -12
P (product) = 12
roots of k² - Sk + P = 0 are x and y
k² - (-12)k + 12 = 0
k² + 12k + 12 = 0
/// simple quadratic equation resolution:
root #1: k = x = (-12 + √(12² - 4·1·4))/(2·1) = -12/2 + 4√6/2 = -6 + 2√6 = -1.101
root #2: k = y = (-12 - √(12² - 4·1·4))/(2·1) = -12/2 - 4√6/2 = -6 - 2√6 = -10.899
/// final results:
■ x = -1.101
■ y = -10.899
/// check:
x + y = (-1.101) + (-10.899) = -12
xy = (-1.101).(-10.899) = 11.999 = 12
All my best ...
🙂
@@GillesF31 thank uuuuu
Rubbish.
When a=1 and b is even one can use the pq-formula x = -(p/2) +- sqrtr(p/2)² - q). One gets directly: x = 3² +- sqrt(3²-36).
Did you know, 4xy= (x+y)² - (x-y)²
Now, if xy = 36, x+y = 6....(i)
then, (x-y)² = -108
or, x-y = 6i√3..... (ii)
add (i) and (ii)
x = 3 +3i√3
y = 3 -3i√3
I've watched several of your videos and got this one right on my own! I'm leaning from you. Thanks.
Next video, where you didn't use symmetry! x and y are interchangable! When you know the solutions for x, those for y are the same, just not in the same cases! When x_1 = s1, then y_1 = s2. When x_2 = s2, then y_2 = s1. Or { x, y } = { s1, s2 }.
Maybe more intuitively:
x = 3 +/- 3 sqrt(3) i
y = 3 -/+ 3 sqrt(3) i
Instead of x and y... say we had x₁ and x₂
x₁+x₂=6
x₁x₂=36
Let's consider a polynomial P(x)=(x-x₁)(x-x₂)
P(x₁)=P(x₂)=0, furthermore, there are no other solutions for P(x)=0 [in tamed mathematics ab=0 implies a=0 or b=0.... possibly both]
So if we solve (x-x₁)(x-x₂)=0... we will know x₁ and x₂
(x-x₁)(x-x₂)=0
x(x-x₂)-x₁(x-x₂)=0
x²-xx₂-x₁x+x₁x₂=0
x²-(x₁+x₂)x+x₁x₂=0
Substituting for sum and product:
x²-6x+36=0
Then it's just solving for x... Note that the initial equations are symmetric, as long as x₁ is one of the solutions, x₂ is the other.
Alternatively, you can do:
(x₁+x₂)² - 4x₁x₂=6²-4(36)
x₁²+2x₁x₂+x₂² - 4x₁x₂ = -3(36)
x₁²-2x₁x₂+x₂² = -3(36)
(x₁ - x₂)² = -3 (36)
x₁ - x₂ = ± 6i √3
Remember, since x₁ and x₂ are symmetric, if you switch them out you get (x₂ - x₁)=-(x₁ - x₂)... which checks out with how there is ± for the difference between them.
(then you add the x₁ - x₂=... equation with the x₁ + x₂=... equation to isolate x₁)
Also an example of how working with complex numbers can sometimes be EASIER than working with real numbers
True! If mathematicians hadn't already invented complex numbers, electrical engineers would have.
We all know that both functions will NOT intersect in a (x, y) coordinate graph. Therefore, there will be two non-real solutions.
x+y=6
xy=36 -----> y=(36/x)
x+(36/x)=6
x²+36=6x
x²-6x+36=0
x²-6x+9=-27
(x-3)²=-27
|x-3|=3i√3
Case 1
x-3=3i√3
x=3+3i√3
3+3i√3+y=6
3i√3+y=3
y=3-3i√3
(3+3i√3, 3-3i√3) ❤
Case 2
x-3=-3i√3
x=3-3i√3
3-3i√3+y=6
-3i√3+y=3
y=3+3i√3
(3-3i√3, 3+3i√3) ❤
🤕 Numbers salad
Glad I'm a Physicist
Great video . Q. Can the Solution be graphed ? thanks
I found this very fascinating and interesting. Thanks for breaking down this equation mabye it will be useful down the line who knows.
Why is it not necessary that x must equal y. To approach the problem initially you can choose either Y= (6-X) or X=(6-Y) and get the same result mirror image
The problem being symmetrical suggests the answer may be symmetrical; it does not suggest that X equals Y.
x + y = 6
xy = 36
=> x + xy + y + 1 = 43
=> x(y + 1) + (y + 1) = 43
=> (y + 1)(x + 1) = 43
Realizing 43 is a prime number and x,y have equal roles. We have:
1. x + 1 = 1 & y + 1 = 43
=> x = 0, y = 42
2. x + 1 = - 1 & y + 1 = - 43
=> x = -2, y = -44
So the solution of the equation is: (x;y) = (0;42) = (-2;-44) and their permutations
can you explain why my method has wrong solutions?
@@kienvantran5769solutions don't necessarily have to be integers, assuming they are leads to these mistakes
edit: for this reason, it's not safe to draw any conclusions from equations containing more than one variable; you just don't have enough information, as there are infinite pairs of real numbers expressible in terms of x+1 and y+1 whose product makes up 43.
@@funnysisyphusman Thanks for the explaination :)
@@kienvantran5769 glad I could help
カルダノの問題ですね
ガウス平面上に絶対値√36=6の円を描く
その円上で実数部が6/2=3になる点を見つける
その2点の座標
3+3√3i , 3-3√3i
が解になる
暗算で解く計算問題です
以上
This video reinforces the reason why I tested out of Algebra in college.
I like maths, but hate those kind of results
I love this kind of results because most results in the real world are ugly, although beautiful ones are better for learning
It's not as bad if he doesn't do it so slowly.
x=6-y
(6-y)y=36
6y-y^2=36
y^2-6y+36=0
y=[6+-rq(36-144)]/2
y=[6+-rq(-108)]/2
y=[6+-6•rq3i]/2
y1,2=3+-3rq3i
x=6-y
x=6-(3+-3rq3i)
x1,2=3+-3rq3i
Ppl don’t realize he used simple numbers to teach the basics. Of course you can solve this much easier but that was t the point . The point was to teach the basic that you could use for any numbers. 🙄🤦♂️
(3)+(3)=6 (y ➖ 3x+3). xy=36 6^6 3^2^3^2 1^1^3^2 3^2 (xy ➖ 3xy+2).
Попутно доказана теорема Виета. Для тех кто только знакомиться с комплексными числами, а эта задача именно для них, было бы очень полезно провести проверку подставив решения в оба уравнения.
Spoilers for answer:
x=3+sqrt(27)i, y=3-sqrt(27)i
thats only high school level, not higher mathematics. quadratic polynom with a little twist.
Easy for you!
If the answer is limited to Real numbers; there is no solution. Once seeing sqrt of a negative, we can just stop and say there is no solution.
I didn't bother multiplying the 36 and 4. The b² and 4ac both had a 36 so I factored it out.
Really great 👍
Always the same way to solve this kind of problem : x and y are the solutions of x^2-sx+p=0
When I was 18, this would have been easy. Now, I was fast forwarding to the end.
z²-6z+36 = (z-3)²-(i √27)² = (z-3-i 3 √3)(z-3+i 3 √3),
∵ z = 3 (1±i √3), ∴ Say, x = 3 (1+i √3), y = 3 (1-i √3).
It looks complicated because it is.
One of them is 3(1+isqrt(3)) and the other is 3(1-isqrt(3))
wow. you have got my attention with this one. It's a quadratic equation but my brain is not nimble enough so i used open ai for the result, it is often correct even with the very hardest university exam problems. It says
the solutions to the equation are: 3 ( 1 + sqrt(3i)) and 3( 1 - sqrt(3i))
Did anyone notice this thing where the video area just shows a black area and no matter how long you wait the video doesnt load? Refreshing once fixes it but sill🤔
y ^2 - 6y + 36 = 0
y = 1/2*( 6 +/- (36- 144)^1*2)
y = 3 +/- 3 * 3^1/2 i
What class/grade is this? I know everything, but the i part confuses me. Do you write i when solving the square root, 4th root, 6th root etc. of a negative number?
High school math... useful lesson involving imaginary numbers (i) in something the students already know (quadratic formula).
Gigo , WHO Cares for complex solutions
it's much higher than i can handle lol
What part of any of that falls at all into the category of simply "looks" complicated.
Strictly speaking, it's ( 3+/- 3iSQRT 3).
There is no need to do the right section,since it holds no solution other then complex solutions.
Je suis français et même dans votre langage c'est plus intéressant que mes anciens cours de math !
Moi aussi (mais je suis American).
I understood everything until he started making up stuff.
Surely there has to be a simpler solution
Using x=6cos^2t and y=6Sin^2t, you can getting simple solution.
Can you explain it to me?
@@1987genisxy=(costsint)^2=1, and then sin2t=2,
@@1987genisand drawing triangle, and using formula of cos2t, you can getting value of x and y
Solved it in my head just based on logical considerations before watching the video. 😊 Who else?
Me
Got as far as I could looking for real solutions; my experience with imaginary numbers is a bit dated.
Me this is the most easiest maths question in my entire engineering life i had seen
So what.
Obviously if x,y are real, they have the same sign, since xy is positive. Since x+y is positive, x, y > 0. Since xy = 36, at least one of x, y is >=6. , But then x+y>6, contradiction. Hence x,y are not real. If x, y are complex, it is easy to see that they must be conjugates. So the real part of x, y is 3. Then the imaginary part follows as pm sqrt(27).
Instead of solving these math, you should develop your relationships :) it is better.. My boss only know plus, substract, divide and multiply, but he owes a huge company now.. 😅
Okay, I'll be the idiot that asks the question: what the hell is the answer? I mean, using g real numbers, what is X, and what real number is Y? Using the imaginary number "i" produces an imaginary result. What value does that serve us in real life? This is why I was much better at Geometry than Algebra. At least Geometry was real.
There is no purely real number solution. That is not to say that complex solutions are not useful, they show up "in real life" all the time too.
Ой, вот это ты зря. Мнимые числа - это вещь. Не существовало бы их, не было бы прогресса в области физики, мы бы не знали, что такое переменный ток, и я бы сейчас не писал этот коммент, так как человечество вообще бы не знало, что такое телефон и современная техника. Те, кто считают, что у таких чисел нет смысла - это большой плевок в сторону Дель Ферро, Кардано, Гаусса, Эйлера и многих других. Просто запомни на всю жизнь: "И" в квадрате равняется минус единице! Это тождество открыло новое множество могучих чисел, а дальше - больше, как насчет кватернионов? Благодаря кватернионам и уже тремя мнимым числам ты можешь играть в 3D игры, а также строить 3D модели. Кардано вот такой мужик был, честь ему и хвала
People use the value of "i" for math equations, even though they don't always make sense in real life. Just because it doesn't translate well, doesn't mean it doesn't help with statistics and keeping data. Use the imaginary numbers in a way that will get you to the real result you want to eventually reach
"I" was invented because we fell for the mythical negative numbers, and then realized you couldn't find the square root of a negative number. If negative numbers were real, you could have -3 elephants in your yard, add 3 more and then have 0 elephants. I drove my boys nuts with that reasoning but they survived it.
@@Slonyara1337 its all Greek to me 😕
I wasn’t at all familiar with the second solving method. Is there a name for it?
I mean, the only difference is that the second method used remarkable identities rather than the quadratic formula
It is called Completing The Square.
It is a standard solution for real numbers too, simply make d=b/2 then add e to c such that c+e=d^2
y=x^2+6x+3 = x^2 + 2.dx + c+(d^2-c) = d^2-c
y=x^2+6x+3 = 0 becomes x^2+6x+9=6 hence (x+3)^2=6 and x+3=sqrt(6) giving solutions -3+-sqrt(6)
y=x^2+6x+3=0 quadratic solution is (-6+-sqrt(6^2-4.1.3) ) /2.1 = 1/2(-6+-sqrt(4.6)) = -3+-sqrt(6)
@@ianhigh4354 Thank you for replying with this information.
An example of math that is only useful in physics, chemistry and mathematics. Cannot use it safely in programming or architecture.
"Only" mathematics, physics, chemistry, and electrical engineering is like saying "only" the stuff that makes the modern world possible.
@@kennethsizer6217 never said it was useless, but it is definitely niche. Still curious how this one would be used in electrical Engineering however.
3+3V3*i; 3-3V3*i.
@@КатяРыбакова-ш2д И наоборот.
Oh but what it you did y = 36/x instead
There is no real solution to this problem. If you are expecting a complex solution then you need to spell it out that the domain of x is the set of complex numbers.
Not really. If not specified, the domain is always IC. But you could ask "show me the real solutions of ..."!
@@rainerzufall42 Yes really. I have never heard of a rule saying that the domain of all functions is the set of complex numbers. Should we also include all n-dimensional vectors in the domain?
@@psionl0 You'll probably know, that IC is closed with respect of solutions of nth grade polynomials. IR is not. You don't have to expand to IH. And also IC^n or IR^n are not useful as default domain. Polynomials of grad n have n solutions in the domain, seems quite fundamental, if I just knew, which theorem would give us that...
Let's state, that we totally agree about these questrions, if the domain is explitly given. Can we agree on this?
@@rainerzufall42 Your rules are getting more complicated. Unless complex numbers are part of the question or, in the context of the problem, complex numbers are useful (eg wave equations), we should not simply assume that the domain includes complex numbers.
Дискременант
@@JisaMail Тогда используем мнимые числа.
So "i" is included... thanks for the algebra exercise, but to no practical solution. Working in physics, this is not satisfying. But have a nice day
What is the real-life benefit of this??? Please bataye
If you want to crack any math college entrance exam its useful model
if u r a math teacher u can earn money by teaching those things. joke is aside in electric and electronic engineering complex numbers are used to solve many problems. especially alternative current circuits ( L;C circuits) and electromagnetic waves.
Fun!
X + Y = 6
XY = 36
Possibilities
X = 0,1,2,3,4, and 5
Y= 6,5,4,3,2, and 1
Imagine X times Imagine Y times added together
iXiY = 0+1+2+3+4+5+6+5+4+3+2+1=36
🥸
x + y = 6
xy = 36
x? y?
=======
x(6-x) = 36
6x - x² = 36
x² - 6x + 36 = 0
D = (-6)² - 4*1*36 = -108
D < 0 --> the roots are imaginer number 😅😅😅
I just want to see what x and y equal.
@@chromidius5339 x and y are complex numbers.
bro just use a calculator in order to simplify the calculation
Нет решения у этой системы! При отрицательном дискриминанте нет решений, тк из отрицательно числа нельзя извлечь корень 🤦
Euler enter the chat.
en.wikipedia.org/wiki/Complex_number#/media/File:Circle_cos_sin.gif
Ой, вот это ты зря. Мнимые числа - это вещь. Не существовало бы их, не было бы прогресса в области физики, мы бы не знали, что такое переменный ток, и я бы сейчас не писал этот коммент, так как человечество вообще бы не знало, что такое телефон и современная техника. Те, кто считают, что у таких чисел нет смысла - это большой плевок в сторону Дель Ферро, Кардано, Гаусса, Эйлера и многих других. Просто запомни на всю жизнь: i в квадрате равняется минус единице! Это тождество открыло новое множество могучих чисел, а дальше - больше, как насчет кватернионов? Благодаря кватернионам и уже тремя мнимым числам ты можешь играть в 3D игры, а также строить 3D модели. Кардано вот такой мужик был, честь ему и хвала
Ha, cute, I can see by the thumbnail that the solution is not complicated, just complex!
if x=3+3√3,y=3-3√3
xy≠36
this question is actually NO SOLUTION
because when he do the solution you can notice that (-6)²-4(1x36)= -108 which is smaller than zero
and in ax²+bx+c, when b²-4ac
@@star_bacon3677 x, y are other numbers.
What is I?
이정도는 학교서도 갈켜주는데 영상까지 찍을일임?
He’s wrong.
@@timwynne No .
Unreal! A somewhat fake question.
As indian this is very easy to understand for me
(I am also bad/weak in maths 😂)
What a moronic thing to state.
@@CaptianVeryObvious moron ?
Bro I'm 15
@@CaptianVeryObvious and keep your opinion in your ass
虚数の時点で解なしだろ？
Нет решения
@@ЕвгенийВовк-й7я Евгений, решения есть, но они мнимые.
Ой, вот это ты зря. Мнимые числа - это вещь. Не существовало бы их, не было бы прогресса в области физики, мы бы не знали, что такое переменный ток, и я бы сейчас не писал этот коммент, так как человечество вообще бы не знало, что такое телефон и современная техника. Те, кто считают, что у таких чисел нет смысла - это большой плевок в сторону Дель Ферро, Кардано, Гаусса, Эйлера и многих других. Просто запомни на всю жизнь: "И" в квадрате равняется минус единице! Это тождество открыло новое множество могучих чисел, а дальше - больше, как насчет кватернионов? Благодаря кватернионам и уже тремя мнимым числам ты можешь играть в 3D игры, а также строить 3D модели. Кардано вот такой мужик был, честь ему и хвала
Yyyy yyy
If you would specify the domain the problem woild be trivial. I am not going to watch the rest of the clip, thus saving 10 minutes of time.
英語わからねー（笑）
You skip too many steps
Бред сивой кобылы
In future videos, talk less and write less and say more.
What's this dude ? 3rd grade?