This was question 1 in the 1961 IMO in Hungary. I have made an attempt at solving this algebra problem using basic high school reasoning. Hope it makes sense.
Good evening sir, I have always hated studying since the exams in my country had killed my passion to learn or even understand why i was studying the subjects for. I found your channel about 5 months ago, although i do not watch all your videos, I have subscribed because of only 1 reason, your passion to teach and how you always smile while teaching. It has been an honor to be able to find a teacher/lecturer like you. Thank You!!
0:08 how ever you are talking about thanks for giveing us your E-Mail, probably you will recive a question from me within the next days. 0:19 when it looks easy always look twice! 0:45 while i was watching your intro, i thought you would be a great teacher 1:42 first mision for me understanding the task (my english ...) 2:21 yes the space on a bord is always tiny i knowe that i have a board by my self! 2:29 and thanks for expandeing the binomic formulars 4:07 ok yes that is solvable 6:15 a i love this formular! 8:25 looking very interesting! and long but that is fine! 8:46 i allready knew it that you can not compare complex nubers! 9:48 ok here we go now it make sence 10:36 nice time laps that always looks funny during calculations! 12:06 ok and now what are you gonna do? 14:05 ok make sence! 16:47 3 conditions ok lets see 17:51 understood continue 20:14 wait that had to be greater than 0 haven´t it? 21:18 your nice end sentencas again! and sorry i had sth. to do and comment the video later and that will be also for the next videos maight be the case yours sincerly K.Furry
Sorry, but I don't see a problem with D=0, x would remain positive (a²+b²)/(4a²) . Why was this case excluded? Maybe the variables wouldn't not be dustinct, but it's not obvious so far
I believe you overlooked a difference of 2 squares in the discriminant, you would get factorisation with less effort. Also I'm curious when it happens to solving equations x+y=S, xy=P, people on English TH-cam start from non-linear equation xy, thus getting a rational function that is to be fought heroically later on , while it's much easier to start from linear equation x+y, not causing any denominator to appear in the course of the solution?😮
11:39 could you explain how you made factoring mentally? Did you use the cross method (I've heard of it, though we have not learnt that at school where I live)
17:15 you have only shown under which conditions the discriminant is positive. For a2 + b2 + D obviously x will be positive, but for a2 + b2 - D it hasn't been shown that x is positive under the given conditions.
@@PrimeNewtons if you claim x is positive, and put +/-, you will need to show for both cases that x is positive. Now your claim has only been shown for the +, not for the -.
@@PrimeNewtons well, that's obvious, but not stated. The video leaves both options open. And it hasn't been analysed under which conditions the minus might give a positive x. Maybe under no condition, maybe for a certain range of a and b. Until analysed, we don't know.
x = i, y = 1/i, z = 1 is a solution with a = 1 and b = i PS: don't know if it's the only solution, I got this by solving these equations simultaneously. I got the condition, a² - b² = 2
Good evening sir,
I have always hated studying since the exams in my country had killed my passion to learn or even understand why i was studying the subjects for.
I found your channel about 5 months ago, although i do not watch all your videos, I have subscribed because of only 1 reason, your passion to teach and how you always smile while teaching. It has been an honor to be able to find a teacher/lecturer like you.
Thank You!!
Thank you 😊
A whole blackboard, in which the solution of the problem is included: THESE ARE PURE MATHEMATICS!!🤪 Good day from Greece!!
Solve jee advanced( entrance exam in india) questions are really good , will surely boost ur interest in mathematics 😶
I really like the way you are explaining those problems , thnak you
Sir, you are really the Rambo of the mathematics ! 🤗🤗🤗
0:08 how ever you are talking about thanks for giveing us your E-Mail, probably you will recive a question from me within the next days.
0:19 when it looks easy always look twice!
0:45 while i was watching your intro, i thought you would be a great teacher
1:42 first mision for me understanding the task (my english ...)
2:21 yes the space on a bord is always tiny i knowe that i have a board by my self!
2:29 and thanks for expandeing the binomic formulars
4:07 ok yes that is solvable
6:15 a i love this formular!
8:25 looking very interesting! and long but that is fine!
8:46 i allready knew it that you can not compare complex nubers!
9:48 ok here we go now it make sence
10:36 nice time laps that always looks funny during calculations!
12:06 ok and now what are you gonna do?
14:05 ok make sence!
16:47 3 conditions ok lets see
17:51 understood continue
20:14 wait that had to be greater than 0 haven´t it?
21:18 your nice end sentencas again!
and sorry i had sth. to do and comment the video later and that will be also for the next videos maight be the case
yours sincerly
K.Furry
Dafuq?
@@giorgioripani8469 sorry i Do not understand what you want
At 9:18, I suggest that you could have used the difference of two squares in the square root to do more work. 😊
Si lo note también, era mejor entender al 4 como 2² y por multiplicación de potencia común ya estaba la estructura
I like the way you tackle the problem
Please tell me from where have you studied. I also want to study maths for my higher study
Wow! How long did it take to prepare for this video, i.e., how much sit-down time working toward the solution before the presentation?
I couldn't catch why exactly 3b²-a² couldn't be 0. For x to be a real number, isn't it also possible for 3b²-a² to equal 0?
You would not get distinct roots
@@PrimeNewtons Ah, thanks for the explanation.
If x + y + z = a, and x, y and z are distinct, then a >= 6
Thanks
Sorry, but I don't see a problem with D=0, x would remain positive (a²+b²)/(4a²) . Why was this case excluded? Maybe the variables wouldn't not be dustinct, but it's not obvious so far
4:38 Here we have Vieta formulas so we can easily write quadratic with roots x and y
Since x,y, and z are positive, and b is always squared, couldn't you have assumed (wlog) that b is positive and do away with the absolute values?
b² is positive. No evidence that b is not negative.
💚 from india
It’s quite an interesting maths but at the end I got confused😂
x,y, and z greater than zero was stated as a condition. However, that is what was supposed to be proved.
I believe you overlooked a difference of 2 squares in the discriminant, you would get factorisation with less effort.
Also I'm curious when it happens to solving equations x+y=S, xy=P, people on English TH-cam start from non-linear equation xy, thus getting a rational function that is to be fought heroically later on , while it's much easier to start from linear equation x+y, not causing any denominator to appear in the course of the solution?😮
Who is he? Fantastic!
What?
Again, there is an obvious factorization. The discriminant is a difference of squares.
I love Math but school just discourages me from solving problems as we are , in a way , forced to cram shit for the exams.
11:39 could you explain how you made factoring mentally? Did you use the cross method (I've heard of it, though we have not learnt that at school where I live)
Yes, I used the cross method.
At the end you could only conclude x, y and z are not all the same, but what about x=y and x=z
From the condition xy = z^2, if any two of them are equal, it will imply all 3 are equal...
@@deriklytten thanks for the explaination
@@deriklyttenvery thankful for pointing that out.🎉
I only would have gotten to the stage where a>absolute b😢
17:15 you have only shown under which conditions the discriminant is positive. For a2 + b2 + D obviously x will be positive, but for a2 + b2 - D it hasn't been shown that x is positive under the given conditions.
I don't need to
@@PrimeNewtons if you claim x is positive, and put +/-, you will need to show for both cases that x is positive. Now your claim has only been shown for the +, not for the -.
What condition will make x positive? If I choose the plus instead of the minus. That's all.
@@PrimeNewtons well, that's obvious, but not stated. The video leaves both options open. And it hasn't been analysed under which conditions the minus might give a positive x. Maybe under no condition, maybe for a certain range of a and b. Until analysed, we don't know.
x = i, y = 1/i, z = 1 is a solution with a = 1 and b = i
PS: don't know if it's the only solution, I got this by solving these equations simultaneously. I got the condition, a² - b² = 2
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