Math Olympiad | How to solve for "a" and "b" in this problem ?
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- Опубліковано 1 жов 2024
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Umm, 2 unkowns, one equation. That means specific solutions are impossible, you need another equation! Is there an additional specification, like a and b have to be integers?
There are 6 solutions
a =25, b = 24
a = -25, b = -24
a = 25, b = -24
a = -25, b = 24
a = 7, b = 0
a = -7, b= 0
The squares cancel out all the negatives.
Есть еще решение : а=-7, в=0
@@ТатьянаАгнивцеваThis is solution #6.
This is pretty obviously a Pythagorean triple of the form 7-24-25. I noted that 49 is the square of 7, and the two variables are also squares, so they had to be 25^2 and 24^2.
My question is what did the Math Olympiad have as the correct answer?
Even more than 6: example (a,b) = (+/-9, sqrt(32)),+/8, sqrt(15),..., without definition D= for (a,b) there are many more, the presenter should limit D to integer only.
In this case of the equation a² - b² = (a + b) (a - b) = 49, in which the condition (a + b) ≥ (a - b)
must be satisfied, in addition to the factors 49 x 1 and 7 x 7, there are also the factors -1 x -49
and -7 x -7 that satisfy the condition, so that all possible values for (a, b) are: (25, 24), (-25, 24),
(7, 0) and (-7, 0).
This is not correct math. You should have shared your assumptions at the beginning. You are assumung both a and b to be >= 0 and are integers. Otherwise, in this case it is one equation with two variables. So there are multiple values for both a and b. If you assume a to be dependent and b is independent then for all values of a (both positive and negative), there is a value for b. Please each correct math.
He is assuming that b >= 0 and a > b (or else a^2 - b^2 =< 0), as well as that both are integers. And right, he should have said he was assuming these premises before starting solving the problem
Why take so long to derive such a simple result, after the above assumptions are given?
கரேக்ட்
This is not the correct solution for the math.
@@pas6295What language is this?
One equation n two variables it means infinite solutions !
Debió acotar al principio que las soluciones solo podían pertenecer a Z y por lo que parece, en su caso solo a N
Two unknowns, one equation. Cannot be done. The answer will be speculative. Clearly, a or b is 7, at the same time the other must be 0. The since we have two unknowns in one equation, we have no way of binding the solutions to the variables. Everything else done or said here is nothing but meaningless mathematical gymnastics
a=±7 b=0
a=±25 b=±24
Good, our youtuber friend made the tipical mistake of the beginner.
@@jaume65bcnon va pas faire de sport on est à l’école de mon père il va me donner une petite cuillère 😅😅😅😅😅😅😅😅😅😅😅😅😅😅😅😅😅😅😅😅😅
On va pas faire de sport on est à
it will give the same answer so that he skip this a step(Square cancels the negative)
Wrong! Wrong!
(a-b).(a+b)=49. The number 49 =49*1. So (a+b)=49 and (a-b)=1. İf we add (a+b) to (a-b) and so add 49 to 1 we can get 2a =50. We can get from this equilibrium that a=25 and b=24
2а=50. Из этого СРАЗУ получается а=25. Зачем сначала все делить на 2, потом сокращать 2, и получать 25. Для кого вы читаете математику? Тот кто смотрит это, тот уже кое что смыслит в математике. Эти подробности раздражают. Смотреть не хочется. Это какой-то незыблемый метод и без него сам лектор не может?
Works only if it is Natural numbers
There are 6 solutions as @purnamishra put it above.so please check again
El planteamiento está incompleto no se establece correctamente las condiciones o se están asumiendo condiciones no planteadas
Note; {a,b} are elements of Natural numbers which are counting numbers or positive integers,
Otherwise we will have many solutions
But the question does not say so. And you do not say so in the video. How does one know from looking at question?
Ok thanks for clarifying🙂
And maybe you can pin your comment so that more viewers know this.
Los número naturales no incluyen el cero. Son los números enteros.
@@gerar2158El cero es un número natural porque para mucha gente una cuenta bancaria de cero desafortunadamente es natural
shouldn t you add that a and b are positive or at most eqal to 0?
Incluso \sqrt{59} y \sqrt{10} son soluciones . Por eso se debe plantear bien el problema.
Te explico , en realidad hay infinitas soluciones . Las soluciones son todas todos los pares ordenados(x;y) que pertenecen a la grafica de la Hiperbola.
Почему не все ответы?
Если а= - 7, b=0 это одно из решений, которого нет в ответе, также а=-25, b=-24, тоже будут решение
Всего существуют 6 решений которые не указаны как решение
This is an Olympiad problem. So why weren’t negative and imaginary solutions considered?
625 - 576 = 49. 25^2 = 625; 24^2 = 576. Die Frage ist nur, welche 2 aufeinanderfolgenden Zahlen 49 ergeben. Die Berechnung funktioniert geht so auf, weil die Exponenten dieselben sind.
(a+b)(a-b)=49 olduqda heç də (a+b)>=(a-b) şərti həmişə doğru deyil.məs: a=25, b=-24 olarsa a+b
x²+y²=R²
x=b, y=7, R=a
therefore, the equation represents the circle b²+7²=a², and has infinitely many solutions in ℝ
Esta mal su comentario. Si hay infinitas soluciones y estas pertenecen a la grafica de la Hipérbola.
@@profemarcoresuelvebefore you say that someone is wrong, make sure that you are right.
X^2 + y^2 = r^2 will define a circle with radius r that is centered at the origin
A hyperbole has more variables to define the focal point
Totally makes sense. However, I view the problem not as one trying to find out how many solutions there are, but one that is trying to find specific solutions based on the relationship between a and b.
@@markdaniel8740 Primero hay que identificar quienes son las variables en el problema. Las variables son a y b, y 49 es la contante ,por consiguiente la igualdad a²-b²=7² que se esta planteando pertenece a la ecuación de la hiperbola que esta centrada en el origen.
Lo que ud ha planteado b²+7²=a², no es la ecuación del circulo amigo
That's only half correct. The equation x²+y²=r² only represents a circle if x and y are variable and r is the fixed radius.
Here you are fixing y=7 (to be more precise, you're fixing y²=7², i.e. |y|=7), and allowing x=b to be one variable, and the radius a to be variable too.
For any specific |a|>7, the equation actually represents the four points at which the circle x²+y²=a² intersects the two lines y=7 and y=-7.
For |a|
It ist not correct.
Pythagoras: a²+b2=c2
Here: c²+b²=a²
7²+b²=a²
When you change the value from b so you change automatically the value from a
So you have endless possibilities
Only person could solve is Modi.
Can you use the same method to solve this polynomial equation :
(X^2 + 4 ) ( X^2 +2X ) = -16
This is a quartic equation it’s very difficult.
Did you say that a and b are natural numbers?
Yes
@@mtbbk If so, then at what time?
@@is7728 Oh he did not say but from the answer we can say yes
@@mtbbk I understand but there are indeed infinitely many solutions if the question isn't well-written
What are some other solutions?
Я решил это задачу за 20 секунд. Мне 56 лет. Работаю сапожником 😅
We know that any odd integer = difference between the square values of two consecutive integers
This is very wrong. Also too many steps. Here is the right answer A=4 B= 3
4×4 =16 3×3= 9 . 16- 9=7.
There are an infinite number of solutions. I'm not satisfied with this approach.
-7 -25 -24 are all forgotten in your answers THUMBS DOWN
😅 да ты Энштейн !!!
Всю тетрадь исписал, чтобы решить одно маленькое уравнение 😅
Заграница всегда так решает, для дебилов.
a, b 모두 자연수?
When a and b are both natural numbers?
Okay~
Đây là dạng hàm số
Y=(X*2+ C)^2
Cho biến số X mọi giá trị sao cho X*2 lớn hơn hoặc bằng C ( nếu C nhỏ hơn không), thì đạt được giá trị tương ứng của Y. Tồn tại vô số cặp nghiệm
Cặp nghiệm nguyên dễ nhận ra nhất của đẳng thức
a*2 - b*2=49 là
a =+ - 7
b =0 ( a phải khác 0)
a^2 - b^2 =49
50-1=49 => a^2 =50 , b^2=1 => a=V50 , b=V1
60-11=49 => a^2=60 , b^2=11 => a=V60 , b=V11
70-21=49 => a^2=70 , b^2=21 => a=V70 , b=V21
Cuidado... Estas llegando a unas conclusiones matemáticas que no tienen logica
Глупость несусветная.Господи,хорошо,что я училась в советское время.
Pythagoras. a^2 = b^2 + 7^2. Oneindig veel (reële) oplossingen
S,a=25;b=-24 is one of the answer.
1
It is not the method to solve the equation these are only permutation combination.to find the value of two variables there must be two equations
How about a 7, 24, 25 common right angle triangle
I spotted that in a fraction of a second, after noting that 49 is 7^2 and that both the variables are also squared.
It's recommended that students memorize the smallest Pythagorean triples for college entrance exams. These are 3-4-5, 5-12-13, 8-15-17, and 7-24-25.
Let a=n and b=n-c, where c is a constant.
Thus a^2 - b^2 = 49 can be written as n^2 - (n-c)^2 = 49.
Solving for c gives,
c = (n^2 - 49)^(1/2) + n.
Plug c back into b. Thus,
b = -(n^2 - 49)^(1/2).
So, if a = n, then b = -((n^2 - 49)^(1/2)).
So, a^2 - b^2 = 49 = n^2 - (-(n^2 - 49)^(1/2))^2 = 49, for n >= 7 and n
49 = 24.5 x 2? Who said a, b must be integer?
Đánh đố nhau chẳng giúp gì trong tư duy mở!
전제 조건이 자연수?
You forgot to specify a and b must be integer.
Thank you, very comprehensible explanation.
How to unsee that. And this channel. Omg.
I haven't heard that a and b should be integer numbers.
Very poor skill of teaching.
Do have maht adulation besayd school?
a=7^1/2 and b=i42^1/2 is also an answer?
a=7
b=0
How Simplier?
The difference of 2 consecutive squares is an odd number.
a and b differ 1 it's 49/2, so that leads to 24-25
Are there any other solutions?
a and b differ 2 is no good, the squares differ by two odd numbers, so that's even.
could they differ 3? That would be (x-2)+ x + (x+2), average x, 49 is not divisable by 3.
Neigther is it divisable by 5.
But it is divisable by 7, so (x-7)+(x-5) etc.
average x=7
1+3+5+7+9+11+13=49, so 0 and 7.
Written down like this it seems complicated, but take the first and fourth line and write that in a table
Only b>0, Plus a =0, b =-7
Lots and lots of possibilities. The first that comes to mind is ±7 and 0
The next is ±8 and ±sqr15
Next one ±9 and ±sqr32
But also ±sqr50 and ±1
Etc..........
±25 and ±24
Etc.........
I think the presenter wanted to consider only integer values of a and b
There are an infinity of solutions. One equation, 2 unknowns.
Much easier way to find a solution in integers: a^2 = b^2 + 7^2 and there's a well-known pythagorean triple, 7^2 + 24^2 = 25^2, so a = 25 and b = 24. Note this doesn't guarantee to have found all solutions.
It took me a few seconds to solve this problem in my head .
A =. 25 or -25, B = 24 or -24.
I doubt that.
a=25 b=24
(25+24) (25-24)= 49×1=49
Making more complication.
What if b is negative number?
Para la reflexión de los amigos. ¿Son esas las únicas soluciones?. Por ejemplo: a=14/3(3)^1/2; b=7/3(3)^1/2, ¿es solución?.Si es correcto, ergo, hay infinitas soluciones. salvo que se pidan solo soluciones en el campo de los Z (números enteros) exclusivamente, cosa que no se aclara.
This is mediocre way of doing math. And the fellow thinks it is Olympiad level maths.😅
The easiest solution is a=7 and b=0
If you need fast answer. a = 7 b = 0
피타고라스.정리 7 24 25 역순 플마 25 플마 24
Very complicated 😂
Also 7 and 0 : (7+0)(7-0)=49
¿Cómo salen las soluciones?. Si se hace a=bk (con “k”constante). De ahí, operando algebraicamente se desprende que b=(7*((k+1)(k-1))^1/2)/(k+1)(k-1). En consecuencia, para cada valor de “k" existirá un par ordenado (a,b) que es solución. para A^2-b^2=49
a^2 - b^2 = 7^2 => a=7, b=0. Not perfect hindu logic
In common right angle triangle
7^2= 24+25, consecutive nos in the middle, as per vedic
7^2+24^2=25^2, rest is best known
Mukundsir
Mr devdas,
It is simple, 49=7^2=24+25
7^2+24^2=25^2, this is applicable 4 all odd no squares
Mukundsir
4:50 для нахождения b проще воспользоваться вторым уравнением системы:
▫a-b=1;
▫-b=1-a; /-1
▫b=a-1=25-1=24.
bro forgor plus or minus
(25;24)(7;0)(-25;-24)(-7;0)
7, 0
There are inf solutions along the circle of radio 7 ; a2 + b2 = 7^2
a+ b = 7 ; a- b = 7 ;a +b = a - b , 2b = 0 b = 0 ; a + b = 7 , a - b = 7 ответ : а = 7 , b = 0 .
Bad solution
a or b must be sero.
Yeah !
I did it ッ！
OK so. . .idiot question. Why only use the positive factors (1 × 49)(49 × 1)(7 × 7)?
(-1 × -49) = 49. (-49 × -1) = 49. (-7 × -7) = 49
For example, using -1 and -49 (-1 > -49)
a + b = -1
a - b = -49
2a = -50
a= -25
b = -24
Check:
a^2 - b^2 = 49
-25 can be squared (-25)^2 = 625
-24 can be squared (-24)^2 = 576
625-576 = 49
Try -7 and -7
a + b = -7
a - b = -7
2a = -14
a = -7
b = 0
Check
a^2 - b^2 = 49
(-7)^2 - 0^2 = 49
-7 can be squared (-7)^2 = 49
So a^2 is 49 if a = -7
49 - 0^2 = 0
0 squared is 0
So b squared = 0
49 - 0 = 49
So i got (a,b) = (7, 0) or (25, 24) or (-7, 0) or (-25, -24).
So. . .why are negative factors not factors?
Where is the error in my math?
I was going to ask the same question as well.
Uygun bir soru degil, hemde kolay
( a + b) (a- b)=7
a=7 b=0 , a is not -7 ,it is imposible, I solved the problem
Two more answers. a+b = -7. a-b = -7. In this case a = -7; b = 0.
A=25, B= 24.
a= 24, b=23.
This is very very correct❤ thank you
You are so welcome ❤️
So glad! 😊
Pay it forward
a=+/- 7,b=0
-7._7=49
Я в уме без решения нашел корни. И так ясно a=7, b=0.
Now solve it using calculus.
All those who think calculus can't be used to solve this, raise their hands.
कैसे संभव है यदि a+b बड़ा होगा a-b से तो a+b=7 and a - b=7 not possiable
a+b>=a-b means a+b>a-b or a+b=a-b
7, 24,25 pythagorean triplet
Thêm điều kiện của bái toán a and b phải là số nguyên
Apparently nonnegative, in fact.
a=56 b=7 a-b = 49 Ok, so I guessed ... it was sure as hell easier that the above. Ok, I am not a math person, but am I wrong???
Yes , you are right that you are wrong. It is a^2-b^2=49=7^2.
Don't ignore the squaring and don't only look for one solution if not specified.
Solution by insight
64-25=49
a=8, b=5
64-25=39.
It's not completely
a=± 7 & b= 0
a cannot be -7. a=7 b=0 I solved the problem
@@tanklar1 why?
@@entertainmentzone6838 because a+b=a-b=7
@@entertainmentzone6838 a-b=a+b=7
@@tanklar1 a+b=-7 & a-b=-7 also
Можно решить в уме. Спасибо образованию СССР. Мне 60,решила гораздо быстрее.
7 24 25 pysagor