It is easier as follows: let A=✓x and B=✓(x-9). Then you get A+B=9 and A^2-B^2=9. Dividing gives A-B=1. From here, it follows that 2A=10, hence A=5. So x=25.
One can find a solution a different way that doesn't require writing anything down. Search first for whole number solutions for both terms on the left because that would make things easy. Thus we want the roots of both x and x-9 to be integers. Every odd number is a difference of consecutive squares and the odd number 9 happens when we take the difference of 4 squared and 4+1 squared. Thus we would want x so that x=5*5 and x-9=4*4. Thus x=25 is a POSSIBLE candidate. We then just check that it fits the original equation and see that it does. This doesn't guarantee there isn't another solution, but it can be done quickly and in your head.
Brother I just substituted every square number starting from 9 and found answer in 4 seconds and also this is solvable by hyperbolic function that's incredible
Before viewing: I see two square roots adding to form an integer. This implies (but does not by any means require!) that each square root is an integer. In such a case, both X and X-9 must be perfect squares, so I look for perfect squares that are 9 apart. 1 and 4 are 3 apart, 4 and 9 5 apart, 9 and 16 7 apart, and 16 and 25 9 apart. Therefore, the answer is X=25. We can plug it back in to check, and get sqrt25 + sqrt(25-9) = sqrt25 + sqrt16 = 5+4 =9.
I have easy method Let √x =a and √x-9 =b We have a^2-b^2=9 (1) And a+b=9 (1)=>(a+b)(a-b)=9 9(a-b)=9 a-b=1 Combine with equation a+b=9 we have a system of equation a+b=9 and a-b=1 If we solve this system of equation we have a=b Substitute a to √x we have √x =5 x=25
Multiply the numerator and denominator on the left side of the equation by (X)^1/2 - (X-9)^1/2. The result will be: (X)^1/2 - (X-9)^1/2=1. Then add this equation with the first equation, the result is: 2x(X)^1/2=10 and as a result X^1/2=5 or X=25. so easily
Multiple sqrt(x)-sqrt(x-9) in both side, and the the equation can be reduced to that sqrt(x)-sqrt(x-9) =1. So we add the original equation, then we have 2sqrt(x)=10. Then x=25 can be obtained. This method is almost same as "sqrt(x)=3 cosh theta" ....
Para resolver la ecuación $$\sqrt{x} + \sqrt{x - 9} = 9$$, sigamos los siguientes pasos: 1. Aislemos uno de los términos con raíz cuadrada: $$\sqrt{x} = 9 - \sqrt{x - 9}$$ 2. Elevemos al cuadrado ambos lados de la ecuación para eliminar la raíz cuadrada: $$(\sqrt{x})^2 = (9 - \sqrt{x - 9})^2$$ $$x = 81 - 18\sqrt{x - 9} + x - 9$$ 3. Simplifiquemos la ecuación: $$18\sqrt{x - 9} = 72$$ 4. Dividamos ambos lados por 18 para aislar la raíz cuadrada: $$\sqrt{x - 9} = 4$$ 5. Elevemos al cuadrado ambos lados de la ecuación nuevamente: $$(\sqrt{x - 9})^2 = 4^2$$ $$x - 9 = 16$$ 6. Finalmente, sumemos 9 a ambos lados para despejar \( x \): $$x = 16 + 9$$ $$x = 25$$ Por lo tanto, la solución de la ecuación es \( x = 25 \). Espero que esto te haya ayudado. Si tienes alguna otra pregunta, ¡aquí estoy para ayudarte!
Lo resolví de dos formas: * Primero (algebraicamente): Raiz(x-9) = 9 - raiz (x) => x-9 = 81 + (x) - 18raiz(x) => 18raiz(x) = 90 ==> raiz(x) = 25 ==> x = 5 * Otro método (simple razonamiento aritmético, demora más en escribir que hacerlo en tu mente): 1) Pensar en 2 números X y (X-9) que tengan raíz exacta y sumen 9 2) Como las raíces son positivas, los pares de raíces no deben ser números tan grandes (1 y 8; 2 y 7; 3 y 6, 4 y 5). O sea, X podría ser 81, 64, 49, 36, 25, 16, 9, 4, 1. Descartados los 2 últimos porque (X-9) no puede ser negativo. 3) Tanteo: Raiz 81 + raíz 72?? Imposible Raíz 9 + raíz 0 tampoco Raíz 25 + raíz 16 SÍ (RPTA) X = 25
Теоре́ма Пифаго́ра - одна из основополагающих теорем евклидовой геометрии, устанавливающая соотношение между сторонами прямоугольного треугольника: сумма квадратов длин катетов равна квадрату длины гипотенузы.
Btw if we move √x to the right part of it we could say that the left part monotonously increasing and the right part monotonously decreasing so we should guess the single value
A simple trial and error with some logic does the trick. Both are square root that means both terms under the square root should be a perfect square. With the logic it could be easily seen that the answer is 25.
The only problem with this problem is that the equation equals to an integer. And you know by heart that the numbers have to be integers on the left. So you try 9, 16, then 25. That basically gets you the answer the fastest.
When I reached the x + sqrt(x^2 -9x) part I was like “Wait if I take sqrt(x) common then I can get the same thing I got on the beginning which is equal to 9” And then sqrt(x) + sqrt(x-9) = 45/sqrt(x) Then as I said earlier, LHS=9 so 45/sqrt(x) = 9 Since we would cross multiply in a fractional equality, I interchanged sqrt(x) and 9 to get sqrt(x) = 45/9 So sqrt(x) = 5 Finally, squaring both sides, x=25
Instead find two perfect squares with difference of 9. So we get 16 and 25. Put x=25 then x-9=16. And root 25 + root 16=5+4=9. Also verified...can be done in 30 seconds...waste of time solving like this if such questions come in competitive exams
Obvious answer is x=25. Which we see from looking at the equation. The question is if there are other values of x that satisfy the equation. By inspection we see no other integer number is possible as a solution. So we got other numbers. However, since sqrt of a non perfect square gives us a irrational number. And irrational + rational = irrational (and thus not 9). We can safely say we got no other real solutions. Since either sqrt(x) or sqrt(x-9) will produce an irrational number.
If the left there is an increasin function, in the right side is const. => there is only 1 solution for x and we have right to find a solution for x by selection method. X = 25
Please don't spread your weakness to the public . It might be possible that some of the students will follow your complicated process and they unnecessarily complicate their lives. Please keep it to yourself. When you have mastery of solution then please make videos and spread..
Siempre resolviendo ejercicios que funcionan solamente con los números indicados . Cualquier número del enunciado que sea primo arruina la forma de resolverlo.
Why insist on doing things the hard way? Move root of x to the opposite si9de of the equation. Now square both sides and reduce. You have root x equal to 5, giving x =- 25.
i was wrong. make it like this. (square root of x) = 9 - (square root of (x - 9)) x = 9^2 - x- 9 x = 81 - x - 9 x = 81 - 9 - x x = 72 - x 2x = 72 x = 36
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It is easier as follows: let A=✓x and B=✓(x-9). Then you get A+B=9 and A^2-B^2=9. Dividing gives A-B=1. From here, it follows that 2A=10, hence A=5. So x=25.
You are genius
Where does the A^2 - B^2 = 9 even come from? Is that an identity from somewhere?
@@jonathansweet2230 No identity, just observation that A^2 = x and B^2 = x - 9
So A^2 - B^2 = 9
No, that would make -9
@@cameroncurtis7261 I didn't read his comment carefully enough. Thanks
Unless you make it harder.
√x+√x-9 = 9
√x-9 = 9-√x
Squaring both the sides
x-9 = 81 + x - 18 √x
18√x = 90
√x = 90/18
√x=5
x=25
That's much easier...a cinch!
Very good
That's easier thanks for this it is so fun to solve a math question for me
What a fuckingggg easyyy solution ❤❤❤❤ Love your channel ❤❤❤
Hello, I wanna learn more from you
One can find a solution a different way that doesn't require writing anything down. Search first for whole number solutions for both terms on the left because that would make things easy. Thus we want the roots of both x and x-9 to be integers. Every odd number is a difference of consecutive squares and the odd number 9 happens when we take the difference of 4 squared and 4+1 squared. Thus we would want x so that x=5*5 and x-9=4*4. Thus x=25 is a POSSIBLE candidate. We then just check that it fits the original equation and see that it does. This doesn't guarantee there isn't another solution, but it can be done quickly and in your head.
You’re making it harder than it looks
Why?
@@Jwjsje there is a much faster method to solve this.
@@Jwjsje In the first step, x^0.5 = 9 - (x-9)^0.5 would be a good start to simplify the solution
Replace √X with A, so easy.
Yes
√x = 3coshθ とすると
√(x-9) = 3sinhθであり
与式は
coshθ + sinhθ = 3 つまり
e^θ =3 と変形できます
よって√x = 5
x = 25
Brother I just substituted every square number starting from 9 and found answer in 4 seconds and also this is solvable by hyperbolic function that's incredible
Before viewing:
I see two square roots adding to form an integer. This implies (but does not by any means require!) that each square root is an integer. In such a case, both X and X-9 must be perfect squares, so I look for perfect squares that are 9 apart. 1 and 4 are 3 apart, 4 and 9 5 apart, 9 and 16 7 apart, and 16 and 25 9 apart.
Therefore, the answer is X=25. We can plug it back in to check, and get sqrt25 + sqrt(25-9) = sqrt25 + sqrt16 = 5+4 =9.
I have easy method
Let √x =a and √x-9 =b
We have a^2-b^2=9 (1)
And a+b=9
(1)=>(a+b)(a-b)=9
9(a-b)=9
a-b=1
Combine with equation a+b=9 we have a system of equation
a+b=9 and a-b=1
If we solve this system of equation we have a=b
Substitute a to √x we have √x =5
x=25
Right solution!
How do you find (1) =>(a+b)(a-b)=9 ? I don’t get it.
We use third constant equality
Х 25
@@robroyvanrobsna^2-b^2 can be factorised into (a+b)(a-b)
Multiply the numerator and denominator on the left side of the equation by (X)^1/2 - (X-9)^1/2. The result will be: (X)^1/2 - (X-9)^1/2=1. Then add this equation with the first equation, the result is: 2x(X)^1/2=10 and as a result X^1/2=5 or X=25.
so easily
Good
Multiple sqrt(x)-sqrt(x-9) in both side, and the the equation can be reduced to that sqrt(x)-sqrt(x-9) =1. So we add the original equation, then we have 2sqrt(x)=10. Then x=25 can be obtained. This method is almost same as "sqrt(x)=3 cosh theta" ....
Math professor here. Yes! Use conjugates!
分かり安い展開でした
ありがとうございます😃
Para resolver la ecuación $$\sqrt{x} + \sqrt{x - 9} = 9$$, sigamos los siguientes pasos:
1. Aislemos uno de los términos con raíz cuadrada:
$$\sqrt{x} = 9 - \sqrt{x - 9}$$
2. Elevemos al cuadrado ambos lados de la ecuación para eliminar la raíz cuadrada:
$$(\sqrt{x})^2 = (9 - \sqrt{x - 9})^2$$
$$x = 81 - 18\sqrt{x - 9} + x - 9$$
3. Simplifiquemos la ecuación:
$$18\sqrt{x - 9} = 72$$
4. Dividamos ambos lados por 18 para aislar la raíz cuadrada:
$$\sqrt{x - 9} = 4$$
5. Elevemos al cuadrado ambos lados de la ecuación nuevamente:
$$(\sqrt{x - 9})^2 = 4^2$$
$$x - 9 = 16$$
6. Finalmente, sumemos 9 a ambos lados para despejar \( x \):
$$x = 16 + 9$$
$$x = 25$$
Por lo tanto, la solución de la ecuación es \( x = 25 \). Espero que esto te haya ayudado. Si tienes alguna otra pregunta, ¡aquí estoy para ayudarte!
Why are you complicating the solution.. see the easier method:
✓x-9 = 9-✓x
Squaring both sides,.
x-9= 81+x-18✓x
18✓x = 90
✓x = 5
x= 25
9차이 나는 제곱수 생각하면 25가 바로 떠오르지 읺는가? 단순 계산은 널리 알려져있어서..
Factoring is very important, it helps us simplify problems.
Lo resolví de dos formas:
* Primero (algebraicamente):
Raiz(x-9) = 9 - raiz (x)
=> x-9 = 81 + (x) - 18raiz(x)
=> 18raiz(x) = 90 ==> raiz(x) = 25
==> x = 5
* Otro método (simple razonamiento aritmético, demora más en escribir que hacerlo en tu mente):
1) Pensar en 2 números X y (X-9) que tengan raíz exacta y sumen 9
2) Como las raíces son positivas, los pares de raíces no deben ser números tan grandes (1 y 8; 2 y 7; 3 y 6, 4 y 5). O sea, X podría ser 81, 64, 49, 36, 25, 16, 9, 4, 1. Descartados los 2 últimos porque (X-9) no puede ser negativo.
3) Tanteo:
Raiz 81 + raíz 72?? Imposible
Raíz 9 + raíz 0 tampoco
Raíz 25 + raíz 16 SÍ (RPTA) X = 25
Just by hit and trial it could be easily solved within seconds
1,朧げながらにx=25を悟る
2,√(x-9)は単調増加、9-√xは単調減少より解がただ一つであることが分かる
I'm glad it was that simple.
Теоре́ма Пифаго́ра - одна из основополагающих теорем евклидовой геометрии, устанавливающая соотношение между сторонами прямоугольного треугольника: сумма квадратов длин катетов равна квадрату длины гипотенузы.
X=25 very easy
Btw if we move √x to the right part of it we could say that the left part monotonously increasing and the right part monotonously decreasing so we should guess the single value
A simple trial and error with some logic does the trick. Both are square root that means both terms under the square root should be a perfect square. With the logic it could be easily seen that the answer is 25.
Wag mo nang hanapin si x sasaktan ka lang ulit nyan😂😂😂
25
The only problem with this problem is that the equation equals to an integer. And you know by heart that the numbers have to be integers on the left. So you try 9, 16, then 25. That basically gets you the answer the fastest.
Excelent
When I reached the x + sqrt(x^2 -9x) part I was like “Wait if I take sqrt(x) common then I can get the same thing I got on the beginning which is equal to 9”
And then sqrt(x) + sqrt(x-9) = 45/sqrt(x)
Then as I said earlier, LHS=9 so 45/sqrt(x) = 9
Since we would cross multiply in a fractional equality, I interchanged sqrt(x) and 9 to get sqrt(x) = 45/9
So sqrt(x) = 5
Finally, squaring both sides, x=25
Instead find two perfect squares with difference of 9. So we get 16 and 25. Put x=25 then x-9=16. And root 25 + root 16=5+4=9. Also verified...can be done in 30 seconds...waste of time solving like this if such questions come in competitive exams
Наслаждение для глаз
Oh please !
Just look for two squares 9 apart.
That'll be 25 and 16.
It works.
Don't overcomplicate a simple puzzle.
Not sure what your doing! But that’s a hallmark of math. Thanks for sharing:)
In 10 second i was like it's 25😂, you have to make a perfect square to get a integer as the result... BOOM 😂
Sidhe root x ko right main le jao sbs kare fr ek bar aur sbs hoga then hmko single value milegi
ง่ายๆ
Seems like a long way to do all that, but I liked your hand writing so thanks.
It is easier to move Sq rt x to the right side then squaring both sides
And quicker.
Nice😊
Можно рациональнее решить: перенести один корень вправо и после возвести в квадрат. Обе части уравнения
Obvious answer is x=25. Which we see from looking at the equation.
The question is if there are other values of x that satisfy the equation. By inspection we see no other integer number is possible as a solution.
So we got other numbers. However, since sqrt of a non perfect square gives us a irrational number. And irrational + rational = irrational (and thus not 9). We can safely say we got no other real solutions. Since either sqrt(x) or sqrt(x-9) will produce an irrational number.
Cho tôi hỏi ở đất nước các bạn đây là bài toán cho người bao nhiêu tuổi
👍👍👍👏👏👏👏👏
Man, this was really simple, I didn't understand why you did so much. As soon as I saw the cover image of the video, I realized that X was 25.
Good details work. Alternative would be to move root x to other side and solve it much quicker.
I just solved it in only 5 sec. Simply I assumed x=25😅
Hay una forma mucho más sencilla
Maybe this is easier to do, but it looks pretty relaxing 😄
Интересно автор пишет
0:18 Put x=options(x=25) gives the ans. Use your experience and practice not formulae
Satisfying
(a+b) × (a - b) is more simple!
❤❤gracias ❤❤❤ me ayuda mucho que desarrolles paso a paso el ejercicio ❤❤❤
X is 9
Solution by insight
5+4=9
x=25
I have easy method step-squaring on both side.
O que você fez com os dois X. Corrige
You choose a very hard solution, you can use " Difference of Two Squares " in 2 lines
Понятно
Perfect
Replace x=y2 and find for y..then it's easily solvable
Almost ran out of ink
Its easy use approximation just 10 sec 😊
The first thing that went thru my head when i saw the problem was "Pythagorean theorem." Then, I immediately substituted 5 for x.
Ну это для нас
X=5
One mistake it's not major. X²-2A.(-X) also +. Negative sign not match for this sum
Thanks.
Слева возрастающая функция, икс находим подбором. Всё
3-4-5 triangle is smile 🙂. (it's very easy)
x=25，so simple.
by observation, x=25
平方根符號書寫錯誤
Зачем такие сложности?
Его можно решить гораздо быстрее, просто подставляя цифры😅
If the left there is an increasin function, in the right side is const. => there is only 1 solution for x and we have right to find a solution for x by selection method. X = 25
if 2 increasing functions are given, then the root exists only 1 root is 25
Please don't spread your weakness to the public . It might be possible that some of the students will follow your complicated process and they unnecessarily complicate their lives. Please keep it to yourself.
When you have mastery of solution then please make videos and spread..
Are yar bahut tripical bana diya
3 4 5
9 16 25
X = 25
Быстрее можно методом подбора)))))
Siempre resolviendo ejercicios que funcionan solamente con los números indicados . Cualquier número del enunciado que sea primo arruina la forma de resolverlo.
25 without using pen
minutes 2:00
why do you put 2x?
X》9 , sinon votre raisonement est completement faux
You can also solve this using geometry, i believe
x=25
I always hate too much details for the simple steps.
Why insist on doing things the hard way? Move root of x to the opposite si9de of the equation. Now square both sides and reduce. You have root x equal to 5, giving x =- 25.
Always
This is why I hate algebra, lol
All these extra numbers and letters out of nowhere
X=25. 3 seconds test
x = y^2 и решение в три строки. Не благодарите.
as I look at 345 triangles for years, I know x is 25 at first eye......
Why so long???? Move the root x on rhs n then square its so wuick then
x + x - 9 = 81
2x = 90
x = 45
easy
LoL
i was wrong. make it like this.
(square root of x) = 9 - (square root of (x - 9))
x = 9^2 - x- 9
x = 81 - x - 9
x = 81 - 9 - x
x = 72 - x
2x = 72
x = 36
Found 2 in seconds, next
Need to sutes to complete?😅
수학은 정말 미친짓이다. 😂
Only 3 sec, I have the solution, x=25
算這種題只是在浪費生命，只是技巧，學生只會背題
Or, and hear me out, you just learn your perfect squares. 1, 4, 9, 16, 25, 36...