France l Olympiad Math problem l a/b=? l easy solution#maths #matheolympiad #mathcompetition #mathematics #mathcontest #mathoperation #education #mathviral
This seems like an easier method: Divide both sides by b. You get a/b + 1 = 6*sqrt(a/b). Now define u=sqrt(a/b). The equation becomes u^2+1=6u. Or u^2-6u+1=0. The solutions to this quadratic are u= 3+- 2*sqrt(2). Now a/b = u^2. When you square the two solutions to the quadratic equation for u you get the same answer as the video solution, 17 +-12*sqrt(2).
I did something similar. I solved for a in the original eq and got a = 6√(ab) - b. Then i divided that (a) by b to get a/b. That resulted in a/b = [6√(ab) - b]/b ⇔ a/b = 6√(a/b) - 1. Then I did the same substitution you did.
At 02:00, you can take a shortcut to divide each side by b² to have: a²/b² + 1 = 34*a/b , b ≠ 0 (a/b)² - 34*(a/b) + 1 = 0 This is a quadratic equation in a/b , where we solve with another shortcut to avoid squaring 34 and taking the square root of a large number at 06:20 later: a/b = [ --34 ± √((-34)² - 4*1*1) ]/(2*1) = 34/2 ± √(34² - 4)/2 = 17 ± √((34² - 4)/2²) = 17 ± √((34/2)² - 4/4) = 17 ± √(17² - 1) = 17 ± √((17 - 1)*(17 + 1)) = 17 ± √(16*18) = 17 ± √(4²*2*3²) = 17 ± 4*3*√2 = 17 ± 12*√2 We include the restriction that a, b > 0 because the original equation uses √(a*b) , which is the principal square root.
a + b = 6*√(a*b) , find a/b . a, b interchangeable, nonzero. Divide by b and solve. a/b + 1 = 6*√(a/b) (√(a/b))² - 6*√(a/b) + 1 = 0 √(a/b) = (6 ± √32)/2 = 3 ± 2*√2 a/b = (3 ± 2*√2)² = 17 ± 12*√2 Note that (17 + 12*√2) * (17 - 12*√2) = 1 . That is, the solutions are reciprocals of each other.
We know a/b ≥ 0 because the solutions are positive. That means, a and b have the same sign, which makes a*b positive, too. In the original equation, a and b are interchangeable, and both can be zero. However, finding a/b excludes zero because we can't divide by zero.
Dude, what's obvious for you might not be as obvious for everyone else. I teach calculus for years now and very often something is as clear as day for some but not for others... I can adjust how i teach based on the students on each class so in one class I might need to explicitly state a property i'm using while in other class the students don't need that. When you teach through an youtube video however you have no way of knowing who's watching and what level they are at.
...ależ z ciebie nudziarz, tak rozwlekłego rozwiązania powinieneś się wstydzić, bo należy obie strony podzielić przez b i już masz równanie kwadratowe ze zmienną sqrt(a/b), a z niego szybko wychodzi a/b=17 + - 12sqrt(2)
I don't know which world or which age you are living in. If this was an Olympiad question then both of you and me had met after succeeding that exam....but it didn't happen.
This seems like an easier method: Divide both sides by b. You get a/b + 1 = 6*sqrt(a/b). Now define u=sqrt(a/b). The equation becomes u^2+1=6u. Or u^2-6u+1=0. The solutions to this quadratic are u= 3+- 2*sqrt(2). Now a/b = u^2. When you square the two solutions to the quadratic equation for u you get the same answer as the video solution, 17 +-12*sqrt(2).
I did something similar. I solved for a in the original eq and got a = 6√(ab) - b. Then i divided that (a) by b to get a/b. That resulted in a/b = [6√(ab) - b]/b ⇔ a/b = 6√(a/b) - 1. Then I did the same substitution you did.
@@vinicus508 I like it
I like it
@@manassekouakou5835 thanks
Разделить на b
1+a/b=6sqr (a/b)
Замена
Sqrt (a/b) =x
1+x^2=6x
Элементарное квадратное уравнение
👍 Оригинально, но "долбежу многовато". Уж лучше арифметика больших чисел.
Math is universal language!
a+b=6√ab, a^2+b^2+2ab=36ab, a^2+b^2-2ab= 32ab, a-b=± 4√2ab, (a+b)/(a-b)=6/(±4√2)=±3/(2√2),.
Using Componendo-Dividendo
a/b=(3±2√2)/(3∓ 2√2)= 17±12√2
At 02:00, you can take a shortcut to divide each side by b² to have:
a²/b² + 1 = 34*a/b , b ≠ 0
(a/b)² - 34*(a/b) + 1 = 0
This is a quadratic equation in a/b , where we solve with another shortcut to avoid squaring 34 and taking the square root of a large number at 06:20 later:
a/b = [ --34 ± √((-34)² - 4*1*1) ]/(2*1)
= 34/2 ± √(34² - 4)/2
= 17 ± √((34² - 4)/2²)
= 17 ± √((34/2)² - 4/4)
= 17 ± √(17² - 1)
= 17 ± √((17 - 1)*(17 + 1))
= 17 ± √(16*18)
= 17 ± √(4²*2*3²)
= 17 ± 4*3*√2
= 17 ± 12*√2
We include the restriction that a, b > 0 because the original equation uses √(a*b) , which is the principal square root.
a + b = 6*√(a*b) , find a/b .
a, b interchangeable, nonzero.
Divide by b and solve.
a/b + 1 = 6*√(a/b)
(√(a/b))² - 6*√(a/b) + 1 = 0
√(a/b) = (6 ± √32)/2
= 3 ± 2*√2
a/b = (3 ± 2*√2)²
= 17 ± 12*√2
Note that (17 + 12*√2) * (17 - 12*√2) = 1 .
That is, the solutions are reciprocals of each other.
It is a simple question below Olympiad level.
Все решается гораздо быстрей и проще
a/b = ((12(3a-((ab)^0,5))+b)/a
Yes there are easier methods we are given a+b. From there we can find a_b. Then by addition and subtraction we can find a and b.
Good 😊
Good, but you forgot to check ab>=0 a/b should be >=0 as well
We know a/b ≥ 0 because the solutions are positive. That means, a and b have the same sign, which makes a*b positive, too.
In the original equation, a and b are interchangeable, and both can be zero. However, finding a/b excludes zero because we can't divide by zero.
sorry but when you wrote out 6^2 = 6*6 = 36, I realized I couldn't watch this anymore. Really? You had to explain 6^2 = 6*6?
Dude, what's obvious for you might not be as obvious for everyone else. I teach calculus for years now and very often something is as clear as day for some but not for others... I can adjust how i teach based on the students on each class so in one class I might need to explicitly state a property i'm using while in other class the students don't need that. When you teach through an youtube video however you have no way of knowing who's watching and what level they are at.
Good!
Good video and the music create a good vibe. I like it
Thank you.
As many others mentioned, the first step is to divide the eq by b. This can be done because b!=0, otherwise a=b=0 and a/b is undetermined.
Use the '≠' symbol because what you wrote can be interpreted as b! = 0 .
Also, a = b = 0 is a solution to the original equation, but computing a/b eliminates that solution.
(a/b)^1/2 + (b/a)^1/2=6
let y = (a/b)^1/2
y + (1/y)= 6
y^2 - 6y + 1 = 0
y = 1/4 *( 6 +/- (36-4)^1/2)
y = 1/2* (3 +/- 2 * 2^1/2)
You found √(a/b) , but not a/b .
...ależ z ciebie nudziarz, tak rozwlekłego rozwiązania powinieneś się wstydzić, bo należy obie strony podzielić przez b i już masz równanie kwadratowe ze zmienną sqrt(a/b), a z niego szybko wychodzi a/b=17 + - 12sqrt(2)
I don't know which world or which age you are living in. If this was an Olympiad question then both of you and me had met after succeeding that exam....but it didn't happen.
U may be divided by b square.
Слишком просто для олимпиад. На олимпиадах постановка задач как из диссертаций, а это лажа.
Superb
Не похоже на олимпиадное
Divisioni e elevamenti al quadrato senza condizioni. Molta superficialità ed errori gravi....tutto ciò provoca DISINFORMAZIONE
Musica chata de fundo
lol