A Nice Algebra Equation | Math Olympiad Problem
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- เผยแพร่เมื่อ 24 ก.ย. 2024
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It is very easy to solve if you recognize that x=√(20+√(20+√(20+√(20+.........))), => √(20+√(20+x))=√(20+x)=x
So we have √(20+x)=x
You don't solve olympiad problems by brute force. Took me about 15 seconds to solve the problem.
I believe this is the standard solution, not a tricky solution.
Yep, recognize the repetition and look for solutions where sqrt(20+x)=x. x=5 or -4. Next factor the quartic equation to get to x^2+x=19 for the last two solutions.
√(20+√(20+x))=xl
Let y = √(20+x)
Then
√(y+20) = x >0
√(x+20) = y >0
This system is symmetric over x and y, so if x = a and y = b is the solution then x=b and y=a is the solution as well.
Square both sides
x²=y+20
y²=x+20
and subtract the first equation from the second
x²-y²=y-x
(x+y)(x-y)+(x-y)=0
(x+y+1)(x-y)=0
The solution may be splitted in two branches
• x+y+1=0
• x-y = 0
The first branch has no solutions because x > 0, y > 0, so
x+y+1 is always > 0
So x = y
x²=x+20
D= 1²+4•20 = 81 = 9²
x=(1±9)/2= {5;-4}
The only positive root may be used due to the restriction x > 0
So x = 5 is the only solution
sqrt(20 + sqrt(20 + x)) = x
(sqrt(20 + sqrt(20 + x)))^2 = x^2
20 + sqrt(20 + x) = x^2
sqrt(20 + x) = x^2 - 20
(sqrt(20 + x))^2 = (x^2 - 20)^2
20 + x = x^4 - 40 x^2 + 400
x^4 - 40x^2 - x + 380 = 0
by factorizing
(x - 5)(x + 4)(x ^2 + x - 19) = 0
therefore
x = - 4, 5 or (- 1 +- sqrt(77))/2
among them the only solution is
x = 5
Hi I really enjoy watching your videos. Could you please before you start to solve an equation describe the strategy you will use? That would be most helpful to me. Thank you and keep up the great work.
As for this particular problem, let f(x)=(20+x)^(1/2), then, then, the equation is f(f(x)=x. Since function f(x) is monopolously increase with x, the necessary condidion should be f=(20+x)^(1/2)=x. Hence, 20+x=x^2, namely x^2-x-20=(x+4)(x-5)=0. x=5 satisfy the sufficient condition as well. Hence, the answer is 5. By the way, the original equation can be fffffffff.... fff(x)=x. The answer to this case as well is x=5.
This channel brings me joy. Thank you 🙇
√(20 + √(20 + x)) = x
Condition 1: x ≥ -20 , x ≥ √20 , and x ≥ 0 is x ≥ 2*√5
√(20 + x) = x² - 20
Condition 2: x² ≥ 20 is x ≤ -2*√5 or x ≥ 2*√5
Combined conditions: x ≥ 2*√5
20 + x = x⁴ - 40*x² + 400
x⁴ - 40*x² - x + 380 = 0 , 380 = 2²*5*19
(x - 5)*(x³ + 5*x³ - 15*x² - 76) = 0 , 76 = 2²*19
(x - 5)*(x + 4)*(x² + x - 19) = 0
x = 5, -4, (-1 ± √77)/2 { remember x ≥ 2*√5 }
x = 5 { discard -4, (-1 ± √77)/2 }
I'm Japanese but the way you spoke was easy to understand.
Oh, Grim_Reaper_from_Hell responded same way, already ... ...... In case, there were no repetittion, we need to solve like presented here in your lecture.
Too complicated. It's better to solve the corresponding quartic equation.
At 1:20 and later you make the mistake that V(p^2)=p. Wrong, it is +p or -p.
Brilliant 😊
let Y=Sqrt (20+X), Y^2=20+X 1️⃣, Sqrt(20+Y)=X, 20+Y=X^2 2️⃣ , 1️⃣➕2️⃣ Y^2+Y=X^2+X, (X-Y)(X+Y+1)=0, X+Y+1!=0, X=Y, X^2=20+X, X=5 or X=-4
Excelente profesor
Very good problem. Thank you.
Great explanation
Thank you very much!!
Im glad that i didn't watch the video and guessed it that the answer is 5. And I skipped till the end, and it was 5.
Если x = f(f(х)), то х = f(x). Или я не прав? Кроме того, в данном случае x > 0. Упрощаем, возводим обе части в квадрат, получаем простое квадратное уравнение, выбираем положительные корни. Задача решается в уме за 30 секунд.
Но чего не сделаешь ради хайпа, да?
What if f(x) = k - x ? We see:
f(f(x)) = k - (k - x) = x .
@@danielleong1865 Все лошади - животные, но далеко не все животные - лошади. 😁
Əvəzləməni çox gözəl etdiniz.Əla həll etdiniz.
sqrt[x(x-1)+sqrt(x(x-1)+x)] = x
is always valid.
You can find many examples
(0) [20+(20+x)^0,5]^0,5=x>=0 ; (1) (20+x)^0,5=y>=0 therefore : (3) y^2=20+x ; (4) {x}^2={ [ 20+y ]^0,5 }^2=20+y ; subtract from from (3) - (4) . (5) (y-x)*(y+x)=x-y . Therefore : (6)x=y , or (7) x+y=-1 . (7) - contradicts (1) and (0). We are substitute (4) in (6) - we get (8) : x=x^2-20 , x^2-x-20=0 ………
With respect to, Lidiy
Quand les solutions ne conviennent pas trouvez un autre signe que X (pour éviter les confusions avec x )
Awesome!
Good explanation. Brilliant, especially, you prove to drop the solution x=(-1+sqrt(77))/2. Thank you.
x=(-1+sqrt(77))/2が除外される理由がわかりません。
それと20=uと置き換えても解答に遠回りとなるだけだと思います。
Thank you, although I can’t read Japanese :)
Thank you, although I can’t read Japanese :)
@@royalrocky1969: The principal square root accepts nonnegative values, and returns nonnegative results. Thus, √n accepts n ≥ 0 , and yields √n ≥ 0 .
The innermost term is √(20+x) . Thus, √(20+x) ≥ 0 and (20+x) ≥ 0 .
In both cases, x ≥ -20 .
The next level is √(20 + √(...)) = x .
We know √(20 + √(...)) ≥ 0 and √(...) ≥ 0 , but x = √(20 + √(...)) , so x ≥ √20 . That's x ≥ 2*√5 . Note that 2*√5 ≈ 4.4721359549995...
For x = (√77 - 1)/2 , we can estimate the bounds for √77 as:
√64 ≤ √77 ≤ √81
8 ≤ √77 ≤ 9
(8-1)/2 ≤ (√77-1)/2 < (9-1)/2
3.5 ≤ x ≤ 4
{ Calculator: x ≈ 3.8874821936960... }
But we know x ≥ 4.4721359549995... , so x = (√77 - 1)/2 is rejected.
@royalrocky1969 : The principal square root accepts nonnegative values, and returns nonnegative results. Thus, √n accepts n ≥ 0 , and yields √n ≥ 0 .
The innermost term is √(20+x) . Thus, √(20+x) ≥ 0 and (20+x) ≥ 0 .
In both cases, x ≥ -20 .
The next level is √(20 + √(...)) = x .
We know √(20 + √(...)) ≥ 0 and √(...) ≥ 0 , but x = √(20 + √(...)) , so x ≥ √20 . That's x ≥ 2*√5 . Note that 2*√5 ≈ 4.4721359549995...
For x = (√77 - 1)/2 , we can estimate the bounds for √77 as:
√64 ≤ √77 ≤ √81
8 ≤ √77 ≤ 9
We adjust it to convert it to x :
(8-1)/2 ≤ (√77-1)/2 < (9-1)/2
3.5 ≤ x ≤ 4
{ Calculator: x ≈ 3.8874821936960 ... }
But we know x ≥ 4.4721359549995... , so x = (√77 - 1)/2 is discarded.
I used 3 secs to solve this quatn. 5 is clear to be the result
Ich auch 😂
5
Checked!
Sir, please which app do you use for your thumbnail
I don't understand why x must be greater than or equal to the root of 20. Please respond.
Sqrt (20 + a positive number or zero) equals X so X must be greater than Sqrt(20)
y = sqrt(u+x), y > 0
x = sqrt(u+y)
y^2 = u + x
x^2 = u + y
(x-y)(x+y) = -(x-y)
(x-y)(x+y+1) = 0
How x,y > 0
x = y
x^2 - x - u = 0
If u = 20, x1 = 5 and x2 = -4
x must be non negative and therefore x2 = -4 is not a solution.
X=5
Как всегда, решение избыточно подробно обьясняется, а проверка скомкана.
Take (√20-x)=x and solve then we get X=5
5.
U
Whats the first square number bigger than 20? Could it be 25? Hmm 20 + 5 equals 25 and 5 is the sq root of 25.. I'm gonna take a wild guess that the answer is gonna be 5
There's no guarantee that x is an integer.
X=5.
Como te complicas, Ecotú.
X=5 solved in 3 seconds
The way you check your solutions is very confusing, and doesn't make sense.
Why don't you check by direct substitution both your solutions? Simply substitute with x = - 4.
There are 4 solutions to a quartic equation. He eliminated 2 because they were negative, which can't be a result of the principal square root of a real number. He then used approximate values to eliminate x = (√77-1)/2 . That left only x = 5 , which he checked.
This is stupid, if you can solve sqrt(20 + X) = X then you have a solution for sqrt(20 + sqrt(20 + X) = X. This gives the equation X*X - X - 20 = 0 and X >= 0 so the solution is X = 5.
Sorry but your last calculation is far too complex : much easier to keep 20= x(x-1), where an obvious solution is 5 and the other -4.
Не решая один корень. х=5
不难得出x=5
solution compliquée
X=18, (x-1)4=x=19 = 104.976
Х=5 🙉🙉🙉🙉🙉🙉🙉
Как можно догадаться, что нужно заменить 20 на u?
Это стандартный метод решать сложное уравнение если коэффициенты можно записать в более простой форме, в данном случае коэффициенты можно записать в форме квадратного уравнения.
You can leave the number 20 in the calculations, but it's just a convenience to use a symbol for it.
5 😂
X=-4
√(20 + √(20 + -4)) =?= -4
√(20 + √16) =?= -4
√(20 + 4) =?= -4
√24 =?= -4
2*√6 ≠ -4
🤮 How do do simple things most complicated.
X=5
What is so difficult?????
Браво, брависимо.
5
x=5
5
X=5
x=5