Euler's formula: e^iθ = cosθ + i sinθ a^4=(a-1)^4 (a/(a-1))^4 = 1 = e^(2niπ) for n = 0, 1, 2, 3, ... take fourth root a/(a-1)= e^(iπn/2) = 1 , i, -1, -i giving no solution for 1 (it is an eqn of order 3, after all) but the others are a= i/(1-i) = (1+i)/2 a= 1/(1+1) = 1/2 a= -i/(1+i) = (1-i)/2
@@zyklos229 I have no idea why you would say that. Using Euler's formula is well established maths (and the easiest way to solve equations of the form (f(x)^n=g(x)^n for some order n).
Just from looking the equation, imediately I knew that a = 1/2 should be one solution. That's because the only way a number to the fourth power to be equal to this number minus one to the fourth power is if the number minus one is the same number with switched sign, that is, x^4 = (-x)^4, therefore a-1 = - a, and a = 1/2. Also, I noticed that the fourth power terms would cancel out on both sides of the equation, and the remaining equation would be a cubic. The other two solutions should be irrational, because there is no other possible real solution to a^4 = (a-1)^4, as I said above. So the only calculation needed is to divide the polynomial "a^4 - (a-1)^4 " by (a-1/2) and solve the resulting quadratic equation. Dividing - 4a³ +6a² - 4a + 1 by a - 1/2 gives a² - a + 1/2, and finding the roots gives a = (1+- i)/2. Very simple problem.
I had a similar thought. I saw 1/2 the same way you did. I carried out (a-1)^4, subtracted a^4 and got a cubic. I used synthetic division to factor out x - 1/2 and found the complex zeros of the resulting quadratic.
You wrote "I saw" - while this may be obvious, I believe it's better to deduce this solution. For example, you could take the fourth root, leading to an equation like |a| = |a-1|, which has a single real solution, a = 1/2. Then, you could divide the polynomials (a-1/2 is also a polinomial) and proceed as you have.
@@Kirkemuswe dont have to take actually take forth root. as soon as we “saw” 1/2 is an answer, we can substitute a=1/2 and confirm that both sides equal. and that’s it
Simple? Well, so please explain why the approach of applying the 4th root on both sides does not work. If you do that, I will agree you truly understood the problem.
@@lucianofinardi7222 what you mean "does not work"? It works, it gets one of the solutions, a = 1/2, which is the only real solution. The other ones are obviously complex numbers, then you have to divide the solution by (a-1/2) and solve the remaining polynomial.
You can take fourth roots on both sides, but the catch is that the results are only equal to each other when multiplied with one of the fourth roots of unity. So suppose that a^4 = (a-1)^4, then a = (a-1)*z, for some z such that z^4 = 1. Then, a(1-z) = -z, so a = z/(z-1). The fourth roots of unity are 1, -1, i and -i, so just plugging in these values for z you obtain the answers. Note that z = 1 does not give a solution because you are dividing by 0.
First expand the right hand side and cancel the a^4 term: a^4 = (a-1)^4 a^4 = a^4 - 4*x^3 + 6*x^2 - 4*x + 1 x^3 - 1.5*x^2 + x - 0.25 = 0 Now recognize that x = 0.5 is a root of the original equation, by inspection. So we can divide out (x - 0.5), which leaves a^2 - a + 2 = 0 The roots of that are 0.5 +/- j*0.5.
@@aristofer The Electrical Engineer has entered the chat. Just replace all the j with i and you're good to go. Electrical Engineers get a lot of practice with complex numbers so they will get the calculations correct except for that j thing.
You can do that very quickly ! a^4 = (a-1)^4 means : a = a - 1 or a = -(a-1) or a = i(a - 1) or a = -i(a-1) 1st equation : no solution and the 3 others are very easy to resolve
Much simpler solution. Make a change b = a - 1/2. Then you have (b-1/2)^4 = (b+1/2)^4. Open the binomial, and only the odd powers won't cancel out. You end up with 4b^3 + 4b = 0, which solutions are b=0, b=i/2, b=-i/2. Substitute back to a and voila, you have your answer.
By far the best and shortest solution. Use the quantity lying halfway between "a" and "a-1" to state the problem, i.e. " b = a - 1/2 ". Then the binomial expansions on the LHS and the RHS will be identical in form, except every second term will have a negative sign, causing half of the terms to cancel out. Then, as you say, ... voila !
A short way to the three solutions is as follows : you have the possible cases : i*x = x-1 ,-x= x-1 ,-i*x = x-1 ,which leads to the three solutions (1+i)/2 , (1 - i)/2, 1/2 .
My method: Since a number and its negative raised to an even power give the same solution, a = -(a-1) ==> a = 1/2 Expand the RHS Cancel a^4 Factor out (a-1/2) using long division Solve the other factor, a second order polynomial Get answers a = (1 +/- i) / 2
Note: this is not a quartic equation. It is a cubic, which is why it has three roots, by the FTOA. It's fastest just to expand the binomial, cancel the a^4 term, find the real root a = 1/2 (you can pretty much do this by inspection or guess and check) and then do simple polynomial division by the factor (2a - 1) to get the quadratic 2a^2 - 2a + 1, which gives you the two complex roots of (1 +/- i)/2..
@@johnpaullogan1365 A quartic equation in x with a x^4 = 0 _is_ a cubic equation. You wouldn't say the equation x - 1 = 0 is an 87th-degree polynomial in x except all the coefficients of x^2, x^3 etc. a re zero. No, you'd say it was a linear equation.
@@johnpaullogan1365 Firstly, there is no "formula" for finding the roots of an 87th degree polynomial. Even if there were, trying to apply it to a polynomial of lower degree will yield indeterminate answers (try setting a = 0 in the quadratic formula). But this is moot. x - 1 = 0 _is not an 87th degree polynomial_ , not even some weird degenerate version of such. An 87th degree polynomial with complex coefficients has 87 roots, up to multiplicity. A linear equation has one root. In fact that is a _definition_ of an nth degree polynomial with complex coefficients: it has n roots.
in the complex plane, a and a-1 have to lie on the same circle centered at the origin therefore the real parts of all solutions is 1/2. The angle these 2 points make with the origin has to be multiple of pi/2. When r = 0, angle is pi which gives 1/2+0i. As r gets larger the angle crosses p/2 and gives 1/2+1/2i and 1/2-1/2i. As r gets even larger, angle decreases and never crosses another multiple.
8:10 |a|=|a-1| means the distance between a and 0 is equal to the distance between a and 1, so a is the midpoint of 0 and 1,so you get a is equal to 1/2. I would not cross that out.
Take the quartic root of each side and you have +- a = +- (a - 1) ==> (1) a = a - 1 (2) a = 1 - a (3) -a = a - 1 (4) -a = 1 - a Equations 2 and 3 resolve to a=1/2
We can easily find a solution over R putting 2 parabolas on the same plane. They are essentially the same, (a-1)^4 being shifted by 1 to the right. So they have only 1 intersection between 0 and 1. And as it is one and only intersection, we just plug 1/2 in looking on the graph, and that's it.
if x^2=y^2 then x= plus or minus y. a^2= plus or minus (a-1)^2 a^2=plus or minus (a^2-2a+1) a^2=a^2-2a+1 or a^2=-a^2+2a-1. first you subtract a^2-2a from both sides to get 2a=1 which simplifies to a=1/2. the second becomes 0=-2a^2+2a-1. quadratic formula gives (-2(plus/minus)sqrt(4-8))/-4. goes to (-2(plus/minus)2i)/4 which becomes (1+i)/2 or (1-i)/2
Perhaps it is because I'm an engineer but I started in my head with trial and error - 2^4 cannot be equal to 1^4 - no point in getting bigger, similarly for a=-1 and below. However 0.5^4 would be the same as (-0.5)^4. That took about 20 seconds of thinking. Do mathematicians always have to take the long route? Runge-Kutta says "guess and answer and then guess a better one"
funny thing, if you try to expand the (a-1)^4 NORMALLY, (i.e. *WITHOUT* using the (x+y)*(x-y) identity ), you get a different equation on the right side that has no real solutions... Really weird... :O .
a = 1/2 is a trivial real solution expand the fourth power and note that a^4 cancels out divide for (a-1/2) because of Ruffini theorem solve for a with the quadratic formula I don't know why this looks overcomplicated. Edit: I don't know the english name of the theorem I used, the theorem is that given a polinomial P(x) with a complex solution ε, then P(x) is divisible by (x - ε)
That's not entirely right, because taking the fourth root is a good first step. It's helpful. This "naive" step gives you a first solution, a = 1/2. But then you must test the result and see that a - 1 is negative, so the fourth root of that must be a complex number. Now you can get going. Taking the fourth root is not the mistake, but not drawing the right conclusions is one.
"to convince myself it's actually just 3 solutions". Il faut faire attention aux apparences. En fait, ce n'est pas un polynôme de degré 4 mais de degré 3. a^4 = a^4 -4a^3+6a^2-4a+1 => 4a^3 -6a^2+4a-1=0
It is an OLYMPIAD Maths question but if you know how to apply from Pure Maths 3 concept, you will get all solutions, do not reject any value. Some of them shouldn't said that a=1/2, but we have 2 complex solution.
You can save yourself a few lines by making it squares in the first place: (a^2)^2=((a-1)^2)^2 a^2=(a-1)^2 v a^2=-(a-1)^2 a^2=a^2-2a+1 v a^2=-a^2+2a-1 -2a+1=0 v 2a^2-2a+1=0 I think this is far easier for possible students to comprehend, then making use of x^2-y^2=(x+y)(x-y) but thats just my opinion as a math teacher ;)
Sorry, you are not correct here. The equation is actually only of 3rd degree, because the a^4 terms cancel out if you expand the right-hand side. So only 3 roots.
If 2 powers with the same exponent are equal they should have the same base and in this particular case the exponent is even it is possible that the bases are opposite numbers. a is never equal to a-1 but -a = a-1 when a=1/2
Not true in general, not even in this particular case if you're solving over the field of complex numbers (which wasn't stated in the problem, to be fair, but follows from his explanations). Here you have to consider all 4 possible values of the (complex) 4th root (considered as a multi-valued function).
It’s not an Olympiad task. Students before university don’t know complex numbers. And if it’s only about R, then it’s obvious: one root, easy to find, easy to prove no more exist.
Obviously a != 0, so 1 = (1-1/a)^4, 1-1/a={1, -1, i, -i}, obviously 1 is dropped which is okay because this is a cubic equation. So 1/a={2,1+i,1-i}, a = {1/2, (1+i)/2, (1-i)/2}.
The root 1/2 is repeated root. Just take example of (x-1)^2=0 it has only one "solution" ie 1 but two roots 1 and 1 as it is a quadratic equation(necessarily 2roots)
If only the real roots are required for the solution, we can just cancel out the 4s in both parts with two possible cases: a = a -1 (no real roots here, discard) and a = -a + 1 => a = 1/2 (edit: accidentally wrote -a + a instead of -a + 1 at first)
Taking the square root of a negative number seems weird (the first reason is that (-i)^2 = i^2 = -1, so sqrt(-1) could be i or -i...). I have never seen that this is allowed, but maybe I don't know this formalism.
6 месяцев назад+3
I wouldn't expect complex solutions is what we are looking for. So, I'd just got both sides to the power of 1/4 and get |a| = |a - 1|. This tells us, a must be in the interval of [0, 1], which gives us solution a = 1/2.
My brain saw this as "How can the absolute value of a be equal to a-1, i.e. abs(a) = abs(a-1)" as even powers don't care about signs. Might not qualify for Math Olympiads, but a strong contender in every day life for sure 🤗
Ah! Thought there was an error expanding between the bottom LH column underlined, equation and it’s top RH column result. But no. It’s correct. I had not noticed the ‘-‘ in front of the LH bracket.
И не говори, я сам прифигел: думаю, чего тут решать аж на 10 минут видео, а он внезапно комплексные корни считает. Да ну, блин, дети на олимпиаде знают ТФКП, вот те на! )
Just like real life. Something = something else. Write stuff that says something to the power of the root minus one equals something else. Any expression = that expressions root raised to a power. When it really is all equal to zero.
@@ciiccci I suppose that one could take the square root, and then take the square root of those roots to get somewhere. Hence, Sqrt(a^4) = +/- Sqrt((a-1)^4), or a^2 = +/- (a-1)^2. Then, a = +/-(a-1), +/- (a-1)i. This would reduce to: Undefined, 1/2, 1/2 - i/2, and 1/2 + i/2.
Sorry, inkorrekt. Es ist in Wahrheit nur eine Gleichung dritten Grades, da sich die Terme a^4 links und rechts gegenseitig aufheben. Somit nur 3 Nullstellen.
I started by dividing by a^4. Then we get 1 = [(a-1)^4 / a^4] = [(a-1)/a]^4. That means [(a-1)/a]^2 = +/- 1. Multiplying by a^2 gives (a-1)^2 = +/- a^2. So a^2 -2a + 1 = +/- a^2. So first situation: 0 - 2a + 1 = 0. That leaves a = 1/2. Second situation is 2a^2 -2a + 1 = 0. And that ends with the complex solutions. Update: I see now I am not the first to do this 😅
Слишком просто для Олимпиады. Я решил уравнение и даже проверил устно, без бумаги, доски, и т.п. Решил чисто в уме. A1=0.5, A2,3=(1 +/- i)/2 Исходное представление намекает, что м.б. чётная степень съест знак. Было бы всё в левой части я мог бы не догадаться 😊 Итак abs(a)=abs(a-1), очевидно одного знака под модулем быть не может, имеем a=1-a, A1=0.5. Подставляем - первый корень есть. Один множитель есть, а 4я степень сократится в итоге будет квадратный многочлен. -4а^3+6a^2-4a+1=0. Делим на -4, затем на (a-0.5). a^2-a+1/4=0. Отсюда ещё 2 иррациональных корня. Проверить тоже можно устно - там вовсе легко, т.к. 1*1 и i*i сокращаются. Легкотня.
Olympiad? Pfft. Olympiads are supposed to test creativity and logical thinking skills. With enough practice this problem is very doable without the requirement of creativity or logical thinking. Take a look at the UK Olympiads, they actually test that instead of just testing how much someone has practiced.
@@ciiccci OK take the square root of both sides and you get a ± ambiguity. If you take the square root again you get ±1 and ±i. You have to look at the 4 possibilities.
Trivial as fuck. Just "see" that a=1/2 is one (rational) solution, then take out the linear factor (a - 1/2) and you get a quadratic equation with rational coefficients which you can solve the standard way to get the two complex conjugate roots which are 1/2*(1 +/- i). Done. The approach with substitution + binomial formula is nice if you absolutely want to avoid polynomial division, but is not really necessary. Hard to believe this is a problem from math olympiad.
I want you to cut three boards with the length of your three solutions and find if they have the same length. This is how shuttles fall down from the sky when top mathematicians work for NASA...
Euler's formula: e^iθ = cosθ + i sinθ
a^4=(a-1)^4
(a/(a-1))^4 = 1 = e^(2niπ) for n = 0, 1, 2, 3, ...
take fourth root
a/(a-1)= e^(iπn/2) = 1 , i, -1, -i
giving no solution for 1 (it is an eqn of order 3, after all) but the others are
a= i/(1-i) = (1+i)/2
a= 1/(1+1) = 1/2
a= -i/(1+i) = (1-i)/2
The e-trick is more of a guess, about all the results of forth root of 1
@@zyklos229 I have no idea why you would say that. Using Euler's formula is well established maths (and the easiest way to solve equations of the form (f(x)^n=g(x)^n for some order n).
Someone loved complex analysis
Almost right, you just forgot, at the beginning, to say that a can't be equal to 1, so you can divide by a-1.
the good way of solving any problem
Just from looking the equation, imediately I knew that a = 1/2 should be one solution. That's because the only way a number to the fourth power to be equal to this number minus one to the fourth power is if the number minus one is the same number with switched sign, that is, x^4 = (-x)^4, therefore a-1 = - a, and a = 1/2. Also, I noticed that the fourth power terms would cancel out on both sides of the equation, and the remaining equation would be a cubic. The other two solutions should be irrational, because there is no other possible real solution to a^4 = (a-1)^4, as I said above. So the only calculation needed is to divide the polynomial "a^4 - (a-1)^4 " by (a-1/2) and solve the resulting quadratic equation. Dividing - 4a³ +6a² - 4a + 1 by a - 1/2 gives a² - a + 1/2, and finding the roots gives a = (1+- i)/2. Very simple problem.
I had a similar thought. I saw 1/2 the same way you did. I carried out (a-1)^4, subtracted a^4 and got a cubic. I used synthetic division to factor out x - 1/2 and found the complex zeros of the resulting quadratic.
You wrote "I saw" - while this may be obvious, I believe it's better to deduce this solution. For example, you could take the fourth root, leading to an equation like |a| = |a-1|, which has a single real solution, a = 1/2. Then, you could divide the polynomials (a-1/2 is also a polinomial) and proceed as you have.
@@Kirkemuswe dont have to take actually take forth root. as soon as we “saw” 1/2 is an answer, we can substitute a=1/2 and confirm that both sides equal. and that’s it
Simple? Well, so please explain why the approach of applying the 4th root on both sides does not work. If you do that, I will agree you truly understood the problem.
@@lucianofinardi7222 what you mean "does not work"? It works, it gets one of the solutions, a = 1/2, which is the only real solution. The other ones are obviously complex numbers, then you have to divide the solution by (a-1/2) and solve the remaining polynomial.
You can take fourth roots on both sides, but the catch is that the results are only equal to each other when multiplied with one of the fourth roots of unity. So suppose that a^4 = (a-1)^4, then a = (a-1)*z, for some z such that z^4 = 1. Then, a(1-z) = -z, so a = z/(z-1). The fourth roots of unity are 1, -1, i and -i, so just plugging in these values for z you obtain the answers. Note that z = 1 does not give a solution because you are dividing by 0.
Take the 4th root 9f both sides now a=a-1, yes
First expand the right hand side and cancel the a^4 term:
a^4 = (a-1)^4
a^4 = a^4 - 4*x^3 + 6*x^2 - 4*x + 1
x^3 - 1.5*x^2 + x - 0.25 = 0
Now recognize that x = 0.5 is a root of the original equation, by inspection. So we can divide out (x - 0.5), which leaves
a^2 - a + 2 = 0
The roots of that are 0.5 +/- j*0.5.
Not "j". Its "i".
j and k is quaternions
@@aristoferElectric Engineers denote √-1 as j (as i means current in eee) :p
@@SamiulIslam-z3d ok, but this is math, not electric enginery
@@aristofer The Electrical Engineer has entered the chat. Just replace all the j with i and you're good to go. Electrical Engineers get a lot of practice with complex numbers so they will get the calculations correct except for that j thing.
@@davidg4288😂
You can do that very quickly ! a^4 = (a-1)^4 means : a = a - 1 or a = -(a-1) or a = i(a - 1) or a = -i(a-1)
1st equation : no solution and the 3 others are very easy to resolve
Its like x²=y², x=-y or x=y, but for fourth [word]. Thanks, good to know.
What word should I write?
I thought the same thing. But perhaps it’s not rigorous? I could not believe the song and dance in the video!
Power@@aristofer
@@laertasgaming Oh thanks, I just forgot
Question: Solve for the value of a.
Answer: x = ...
Wait... what?!!! 😱🤔
lol
This means the problem is not solved.
Wo ist denn der mittelfinger wo ist denn der mittelfinger, hier bin ich , hier bin ich .....😂
@@Archik4let x=a
x is unknown, what's new 😅
Much simpler solution. Make a change b = a - 1/2. Then you have (b-1/2)^4 = (b+1/2)^4. Open the binomial, and only the odd powers won't cancel out. You end up with 4b^3 + 4b = 0, which solutions are b=0, b=i/2, b=-i/2. Substitute back to a and voila, you have your answer.
By far the best and shortest solution. Use the quantity lying halfway between "a" and "a-1" to state the problem, i.e. " b = a - 1/2 ". Then the binomial expansions on the LHS and the RHS will be identical in form, except every second term will have a negative sign, causing half of the terms to cancel out. Then, as you say, ... voila !
a⁴=(a-1)⁴
Obviously a=½ is a solution.
4a³-6a²+4a-1=0
By dividing it by (2a-1) we get
(2a-1)(2a²-2a+1)=0
D=4-8=-4
Theres no way this is an olympiad problem. This is like algebra ii at most.
finding the fourth root makes it a maths olympiad problem
@@pieterkok7486 How? This is high school algebra, even in the US that has bad math instruction.
@@pieterkok7486u gotta be joking 😂
this cant possibly be a math olympiad question, thers no way
Maybe its math olympiad for school?
A short way to the three solutions is as follows : you have the possible cases : i*x = x-1 ,-x= x-1 ,-i*x = x-1 ,which leads to the three solutions (1+i)/2 , (1 - i)/2, 1/2 .
The problem statement must include specification of a field in which solution are to be sought. There is no reason to silently assume it is C.
Has no roots in GF2. 😢
My method:
Since a number and its negative raised to an even power give the same solution, a = -(a-1) ==> a = 1/2
Expand the RHS
Cancel a^4
Factor out (a-1/2) using long division
Solve the other factor, a second order polynomial
Get answers a = (1 +/- i) / 2
Note: this is not a quartic equation. It is a cubic, which is why it has three roots, by the FTOA. It's fastest just to expand the binomial, cancel the a^4 term, find the real root a = 1/2 (you can pretty much do this by inspection or guess and check) and then do simple polynomial division by the factor (2a - 1) to get the quadratic 2a^2 - 2a + 1, which gives you the two complex roots of (1 +/- i)/2..
isn't it technically a quartic equation of the form ax^4+bx^3+cx^2+d=0 where a=0.
@@johnpaullogan1365 A quartic equation in x with a x^4 = 0 _is_ a cubic equation. You wouldn't say the equation x - 1 = 0 is an 87th-degree polynomial in x except all the coefficients of x^2, x^3 etc. a re zero. No, you'd say it was a linear equation.
@@davidgillies620 i would expect a formula for finding x for an 87th degree polynomial to find the correct root x=1.
@@johnpaullogan1365 Firstly, there is no "formula" for finding the roots of an 87th degree polynomial. Even if there were, trying to apply it to a polynomial of lower degree will yield indeterminate answers (try setting a = 0 in the quadratic formula). But this is moot. x - 1 = 0 _is not an 87th degree polynomial_ , not even some weird degenerate version of such. An 87th degree polynomial with complex coefficients has 87 roots, up to multiplicity. A linear equation has one root. In fact that is a _definition_ of an nth degree polynomial with complex coefficients: it has n roots.
in the complex plane, a and a-1 have to lie on the same circle centered at the origin therefore the real parts of all solutions is 1/2. The angle these 2 points make with the origin has to be multiple of pi/2. When r = 0, angle is pi which gives 1/2+0i. As r gets larger the angle crosses p/2 and gives 1/2+1/2i and 1/2-1/2i. As r gets even larger, angle decreases and never crosses another multiple.
Be Careful not to change the "a" to an "x" in your answer!
Use Binomial Expansion rule, it is more safer
For 4! and 3!, they are 24 and 6, respectively. You do not need any single calculator for this and must be precise and accurate in this question
Substitute a = b + 1/2, and solve for b. Answer falls out much more easily.
How did you get this substitution? It is obvious why it has to work, but...? Perfect intuition?
Lol, symmetry.
The oddity of this solution is a 4th-order polynomial with only 3 roots.
Try expanding it out. The quartic terms cancel.
@@chaosredefined3834 I realize that. It's just interesting that the equation "appears" it would have 4 roots.
Technically, I think it does have four roots, two of them being coincident roots at a=½
@@trueriver1950 no because at the end you have a cubic
8:10 |a|=|a-1| means the distance between a and 0 is equal to the distance between a and 1, so a is the midpoint of 0 and 1,so you get a is equal to 1/2. I would not cross that out.
Take the quartic root of each side and you have
+- a = +- (a - 1)
==>
(1) a = a - 1
(2) a = 1 - a
(3) -a = a - 1
(4) -a = 1 - a
Equations 2 and 3 resolve to a=1/2
We can easily find a solution over R putting 2 parabolas on the same plane. They are essentially the same, (a-1)^4 being shifted by 1 to the right. So they have only 1 intersection between 0 and 1. And as it is one and only intersection, we just plug 1/2 in looking on the graph, and that's it.
You are overthinking this.
ABS(a) = ABS(a-1)
a a-1
Therefore
-a = a -1
a = 0.5
if x^2=y^2 then x= plus or minus y. a^2= plus or minus (a-1)^2 a^2=plus or minus (a^2-2a+1) a^2=a^2-2a+1 or a^2=-a^2+2a-1. first you subtract a^2-2a from both sides to get 2a=1 which simplifies to a=1/2. the second becomes 0=-2a^2+2a-1. quadratic formula gives (-2(plus/minus)sqrt(4-8))/-4. goes to (-2(plus/minus)2i)/4 which becomes (1+i)/2 or (1-i)/2
Perhaps it is because I'm an engineer but I started in my head with trial and error - 2^4 cannot be equal to 1^4 - no point in getting bigger, similarly for a=-1 and below. However 0.5^4 would be the same as (-0.5)^4. That took about 20 seconds of thinking. Do mathematicians always have to take the long route? Runge-Kutta says "guess and answer and then guess a better one"
And the non-real solutions? I guess you being an engineer these don't exist for you?
@@henkn2 Correct - they're not real.
funny thing, if you try to expand the (a-1)^4 NORMALLY, (i.e. *WITHOUT* using the (x+y)*(x-y) identity ), you get a different equation on the right side that has no real solutions...
Really weird... :O
.
a = 1/2 is a trivial real solution
expand the fourth power and note that a^4 cancels out
divide for (a-1/2) because of Ruffini theorem
solve for a with the quadratic formula
I don't know why this looks overcomplicated.
Edit: I don't know the english name of the theorem I used, the theorem is that given a polinomial P(x) with a complex solution ε, then P(x) is divisible by (x - ε)
a^4=(a-1)^4 so a = w*(a-1) where w is any 4-th root of unity. Then, a*(w-1)=w so a = 1+1/(w-1). Valid solutions: w=+-i, w=-1
Bro I don't understand,this doesn't have 4th root .what is w. Why u find w
it is basically a-1= |a| (a has to be negative otherwise we’ll get -1 = 0), so it is a - 1 = -a, a = 1/2
You can make it simplier by looking for the for roots
That's not entirely right, because taking the fourth root is a good first step. It's helpful. This "naive" step gives you a first solution, a = 1/2. But then you must test the result and see that a - 1 is negative, so the fourth root of that must be a complex number. Now you can get going. Taking the fourth root is not the mistake, but not drawing the right conclusions is one.
Just divide both side a^4 and recognise that square root of 1 can be -1
Far too easy for an Olympiad. Main problem I had was to convince myself it's actually just 3 solutions but to get the solutions were very easy.
8:07 why ?
"to convince myself it's actually just 3 solutions". Il faut faire attention aux apparences. En fait, ce n'est pas un polynôme de degré 4 mais de degré 3. a^4 = a^4 -4a^3+6a^2-4a+1 => 4a^3 -6a^2+4a-1=0
0.5 (1/2) - the first solution, which can be intuitively found in a second :)
You change a to x , then we see a =2 (but this is other "a").
You also do not precise that a is complex numer (my first assumpion is a is real number)
my direct impulse:
1. a1 = 0.5
2. a^4 cancels out -> polynomial 3rd order
3. polynomial division by (x-0.5)
4. abc (or pq) formula
(I'm German)
I just watched up to minute 2 and now I'm sad
What do you mean with a^4 cancels out and how do you achieve 3rd order polynomial from that?
@@scull3208 Use binomial theorem to rewrite (a - 1)^4 in an expanded form. Just by the first glance you can notice the first term is going to be a^4
@@scripter3579 Ahh I see thanks
It's easier to replace a by x + ½. Then the powers to the 4 reduce immediately to 2x³ + ½x = 0
Die Argumentation mit den Graphen und Ihrer Symmetrie ist doch viel einfacher! Und es lässt sich zeigen, dass es genau einen Schnittpunkt gibt.
a=0 is the forth root of the equation. 0^4 = 1 and (-1)^4=1.
1)a= -(a-1) so that a = 1/2.
2)If we factorize the given formula, (2a-1)(2a^2 -2a +1)=0. We got the solutions (1+i)/2, (1-i)/2
The solution can be shorter. Let b = a - 1/2, so the equation will be (b+1/2)^4 = b-1/2)^4 , it can be transformed to b*(b^2 + 1/4) = 0
It is an OLYMPIAD Maths question but if you know how to apply from Pure Maths 3 concept, you will get all solutions, do not reject any value. Some of them shouldn't said that a=1/2, but we have 2 complex solution.
Feels like a problem from the International Olympiad in Algebra 2
I will use Binomial Expansion rule instead of couple rule, to eliminate one of the variables.
You can save yourself a few lines by making it squares in the first place:
(a^2)^2=((a-1)^2)^2
a^2=(a-1)^2 v a^2=-(a-1)^2
a^2=a^2-2a+1 v a^2=-a^2+2a-1
-2a+1=0 v 2a^2-2a+1=0
I think this is far easier for possible students to comprehend, then making use of x^2-y^2=(x+y)(x-y) but thats just my opinion as a math teacher ;)
also 4 roots of unity. if x^4=y^4 x= +y, -y, iy, or -iy. plugging in gives one n solution and the others give the solutions we are looking for
I would suggest that two coincident real roots exist at a=1/2. There have to be four roots!
Sorry, you are not correct here. The equation is actually only of 3rd degree, because the a^4 terms cancel out if you expand the right-hand side. So only 3 roots.
If 2 powers with the same exponent are equal they should have the same base and in this particular case the exponent is even it is possible that the bases are opposite numbers. a is never equal to a-1 but -a = a-1 when a=1/2
Not true in general, not even in this particular case if you're solving over the field of complex numbers (which wasn't stated in the problem, to be fair, but follows from his explanations). Here you have to consider all 4 possible values of the (complex) 4th root (considered as a multi-valued function).
It’s not an Olympiad task. Students before university don’t know complex numbers. And if it’s only about R, then it’s obvious: one root, easy to find, easy to prove no more exist.
Obviously a != 0, so 1 = (1-1/a)^4, 1-1/a={1, -1, i, -i}, obviously 1 is dropped which is okay because this is a cubic equation. So 1/a={2,1+i,1-i}, a = {1/2, (1+i)/2, (1-i)/2}.
don't overthink it. a = 1/2 is one answer.
Multiply both sides by 16 = 2^4.
So, (2a)^4 = (2(a - 1))^4.= (2a - 2)^4.
Let u be the arithmetic mean of 2a and 2a - 2; i.e., u = 2a - 1.
2a = u + 1.
2a - 2 = (2a - 1) - 1 = u - 1.
This, (u + 1)^4 = (u - 1)^4.
0 = (u + 1)^4 - (u - 1)^4
= ((u + 1)^2 + (u - 1)^2) * ((u + 1)^2 - (u - 1)^2)
= (u^2 + 2u + 1 + u^2 - 2u + 1) * ((u + 1) + (u - 1)) * ((u + 1) - (u - 1)
= (2u^2 + 2) * 2u * 2
= 4u * 2 * (u^2 + 1)
= 8u * (u + i)(u - i).
Divide both sides by 8.
u = 0 or u = ±i
2a - 1 = 0 or 2a - 1 = ±i.
2a = 1 or 2a = 1 ± i.
a = 1 / 2 or a = (1 ± i) / 2.
its just expanding (a-1)^4 , getting a trinomial. done, factor out solve, you will get 1/2 and (1+-i)/2
How should you factor it?
@@clashking8412 You get to 1 = a^2 ( 4a + 2 ), only real solution is 1/2 after guessing.
@@MrGerdbrecht while solving it I realized I could just use difference of squares but thanks :)
Simpler method, take square root of both sides,
a^2 = ± (a-1)^2
expand RHS and simplify
+ve root gives a = 1/2
-ve root gives (1 ± i)/2
What happened to the 4th answer?
It's a cubic in disguise because when you multiply it out and set it to zero you get a³ as the highest power
The root 1/2 is repeated root.
Just take example of (x-1)^2=0 it has only one "solution" ie 1 but two roots 1 and 1 as it is a quadratic equation(necessarily 2roots)
Back to School!@@GooogleGoglee
Not really @@KG1_007
The forth answer has no solution
If only the real roots are required for the solution, we can just cancel out the 4s in both parts with two possible cases: a = a -1 (no real roots here, discard) and a = -a + 1 => a = 1/2
(edit: accidentally wrote -a + a instead of -a + 1 at first)
Please use a for your answer at the end and not a. Also this can be trivialized by using complex 4th root of unity
Taking the square root of a negative number seems weird (the first reason is that (-i)^2 = i^2 = -1, so sqrt(-1) could be i or -i...). I have never seen that this is allowed, but maybe I don't know this formalism.
I wouldn't expect complex solutions is what we are looking for.
So, I'd just got both sides to the power of 1/4 and get |a| = |a - 1|. This tells us, a must be in the interval of [0, 1], which gives us solution a = 1/2.
1/2 is an obvious solution.
a⁴ = (a−1)⁴ ⇒ 4a³ - 6a² + 4a - 1 = 0
f(a) = 4a³ - 6a² + 4a - 1 is strictly increasing,
so a=1/2 is the only solution.
you can easily solve the math . Root on both side. and then it will be easy to you
My brain saw this as "How can the absolute value of a be equal to a-1, i.e. abs(a) = abs(a-1)" as even powers don't care about signs. Might not qualify for Math Olympiads, but a strong contender in every day life for sure 🤗
Could consider another interesting way for solvin it taking the sustitution u=a-1/2.
Dilly dalliy approach lengthening the presentation!
Ah! Thought there was an error expanding between the bottom LH column underlined, equation and it’s top RH column result. But no. It’s correct. I had not noticed the ‘-‘ in front of the LH bracket.
I don't believe this is a olympiad question.
No it could has olympaid is based on time management not only about hardness of question
This task is solved by 2 steps with geometric considerations
Great interprétation
I can't see anything wrong with: a = 1/2
a⁴ = 0.0625000
(a - 1)⁴ = 0.0625000
Is it an exercise of problem solving or problem making?
a basic school task, we clicked these very quickly
Lol, why not give the whole task?
I have already passed complex analysis, but how shall I know that some German kids also know it?
И не говори, я сам прифигел: думаю, чего тут решать аж на 10 минут видео, а он внезапно комплексные корни считает. Да ну, блин, дети на олимпиаде знают ТФКП, вот те на! )
Just like real life.
Something = something else.
Write stuff that says something to the power of the root minus one equals something else.
Any expression = that expressions root raised to a power.
When it really is all equal to zero.
The fsktotization from the firdt step eorks for the 2nd step too. In the second term insert -i^2 before a
The final answers are basically wrong because you are solving for a & not x-That's what my teacher would have said💔😭
First get both sides four-rooted, you end with a=a-1. The answer is nope.
🤣🤣🤣
What is the name of the method you used? I thought negative numbers don't have square roots. Is there a reason the "i" is i ?
This is complex numbers. Try to learn about it.
Almost correct explanation…
√-4 = √(-1×4) = √(i² × 4) = √i² × √4 = i × 2 = 2i.
Stating that √-1 = i is a sloppy shortcut ;)
Why does a fourth degree equation have only three solutions? The basic theorem of algebra is violated, however.
a^4-a^4=0 and the equation becomes 3 degree equation ok
@@Safiyakhatunkhatunbro can u plz explain me why we can't 4✓a^4= with other side as he tell in end of the vedio
@@JackPullen-Paradox7:50 . Can u plz too
@@ciiccci I suppose that one could take the square root, and then take the square root of those roots to get somewhere. Hence, Sqrt(a^4) = +/- Sqrt((a-1)^4), or a^2 = +/- (a-1)^2. Then, a = +/-(a-1), +/- (a-1)i. This would reduce to: Undefined, 1/2, 1/2 - i/2, and 1/2 + i/2.
This is due to the fact that i/(i-1) = i(i+1)/-2, etc. and i/(i+1) = i(i-1)/-2 = (1+i)/2.
In what works would this be an Olympiad problem? Takes literally a couple of seconds to find a solution and prove it’s the only one.
Gleichung vierten Grades,
also vier Nullstellen/Wurzeln ...
a = 1/2 ist als doppelte Nullstelle zu berücksichtigen.
Sorry, inkorrekt. Es ist in Wahrheit nur eine Gleichung dritten Grades, da sich die Terme a^4 links und rechts gegenseitig aufheben. Somit nur 3 Nullstellen.
Н-да, если бы такие задачи были на школьных олимпиадах в мое время, то на городские олимпиады попадали бы целиком все школы, включая двоечников )
I started by dividing by a^4. Then we get 1 = [(a-1)^4 / a^4] = [(a-1)/a]^4. That means [(a-1)/a]^2 = +/- 1. Multiplying by a^2 gives (a-1)^2 = +/- a^2. So a^2 -2a + 1 = +/- a^2. So first situation: 0 - 2a + 1 = 0. That leaves a = 1/2. Second situation is 2a^2 -2a + 1 = 0. And that ends with the complex solutions. Update: I see now I am not the first to do this 😅
Слишком просто для Олимпиады. Я решил уравнение и даже проверил устно, без бумаги, доски, и т.п. Решил чисто в уме. A1=0.5, A2,3=(1 +/- i)/2
Исходное представление намекает, что м.б. чётная степень съест знак. Было бы всё в левой части я мог бы не догадаться 😊
Итак abs(a)=abs(a-1), очевидно одного знака под модулем быть не может, имеем a=1-a, A1=0.5. Подставляем - первый корень есть.
Один множитель есть, а 4я степень сократится в итоге будет квадратный многочлен. -4а^3+6a^2-4a+1=0. Делим на -4, затем на (a-0.5). a^2-a+1/4=0. Отсюда ещё 2 иррациональных корня. Проверить тоже можно устно - там вовсе легко, т.к. 1*1 и i*i сокращаются.
Легкотня.
Зачем проверять устно, когда это можно сделать молча в уме?
@@eugnsp Не понял вопроса? Я делал именно молча, в уме. Термин "устно" использовал как противоположеность "письменному" решению или "у доски".
Bro there would be a 4th answer as x is raised to the power 4 and not only 3 amswerd
It's actually only of 3rd degree, because the two a^4 terms cancel each other out. So only 3 solutions expected.
Olympiad? Pfft. Olympiads are supposed to test creativity and logical thinking skills. With enough practice this problem is very doable without the requirement of creativity or logical thinking. Take a look at the UK Olympiads, they actually test that instead of just testing how much someone has practiced.
Try a=(a-1), -(a-1), i(a-1) , -i(a-1)
a=1/2, (1-i)/2, (1+i)/2
1/(1-i)
Where did u get that √i
@@ciiccci there’s no √ in mine. You and I agree
@@DavidMFChapmansorry i mean ✓-1 .
@@ciiccci OK take the square root of both sides and you get a ± ambiguity. If you take the square root again you get ±1 and ±i. You have to look at the 4 possibilities.
So we say in the hypothesis (a belongs to C - complex one).
interesting that a 4th equation has 3 roots, not 4?
The a^4 cancels. It’s a cubic in disguise
(a-1)^4=a^4+4... etc. Therefore we have a^4 on both sides, so they cancel out leaving the highest term as -4a^3
2:55 why?
Trivial as fuck. Just "see" that a=1/2 is one (rational) solution, then take out the linear factor (a - 1/2) and you get a quadratic equation with rational coefficients which you can solve the standard way to get the two complex conjugate roots which are 1/2*(1 +/- i). Done.
The approach with substitution + binomial formula is nice if you absolutely want to avoid polynomial division, but is not really necessary.
Hard to believe this is a problem from math olympiad.
|a| = |a-1| -> a=a-1; a=-a+1 -> a=1/2
What is the 4th solution?
Look at carefully this is 3 degree equation ok
@@Safiyakhatunkhatun thanx: a^4 on both sides cancel a^4 out!
Why were there only 3 roots?
4th order polynomial has 4 roots not 3. !
It's not 4th order - a^4 is cancelled from both sides.
At a quick glance I start by taking the square root of both sides. Then a^2 = a^2 -2a +1.Then 2a -1 = 0 and -2a + 1 = 0. Then a = + 1/2 and - 1/2
Meth olympaid*
a^4=(a-1)^4
Raising both sides to the power 1/4
a=a-1
1=a-a
1=0
🫡
I want you to cut three boards with the length of your three solutions and find if they have the same length. This is how shuttles fall down from the sky when top mathematicians work for NASA...
a = +/- (a-1)
a^4=(a-1)^4 imply a=|a-1| if a>1: -1=0 so there is no solution if a
i used just the rule: square root(x^2)=|x|
It's obviously 0.5 . Why would you want to spend 10 minutes solving it?
Also, you have to prove this is the only solution. That's why you can't just say the answer
|a| = |a-1| has the unique solution, a = ½
На какой уровень образования составлена эта олимпиадная задача?
Очевидно, на тот, в котором школьники знают ТФКП.
Good