A beautiful math problem for advanced students | Equation Solving
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- Опубликовано: 26 авг 2024
- What do you think about this problem? If you're reading this ❤️.
Also, i want to say sorry for a bad light during 1 minute. Thank You for understanding!
Equation: a^3 + a^2 = 36
Hello My Friend ! Welcome to my channel. I really appreciate it!
@higher_mathematics
#maths #math
Writting 36 as 27+9 presupposes you know the answer.
This is a very general method of solving cubic equations. You take an answer that is obvious and reduce the equation to second order. I know it is not very elegant, however if you don’t know the cubic formula it is often the best you can do unless the equation offers you very very VERY special circumstances.
I really wish people would stop using Olympiad as clickbait.
Exactly
It doesn't
@@foopthethird5784 but arent we missing (‘36’)^(i^(i+dimension number))
Just thinking, i might be wrong all the way
but arent we missing ***
Rewriting..
{R}^(i^(i*dimension size {in algebric same R})
A trivially simple problem for beginners. a=3 by inspection
Yes and then Horner's method to simplify and then discriminant to find the complex solutions.
a³+a² =36
a²(a+1)=36
a²(a+1)=9×4
a²>(a+1)
By way of comparison,
a²=9 and a+1=4
a=√9 and a=4-1
a=3 and a=3
Finally, a=3.
It's just a different perspective.
Thank you so much
a=1? No... a=2? No... a=3? Yes!!! Let's divide a³+a²-36 ÷ a-3 = a²+4a+12. So the other two solutions: 2a=-4±√-32, two more complex roots a=-2+i√8 and -2-i√8.
So, what you’re saying is: find the real root first, which by inspection or synthetic division is 3. Dividing by (x-3) gives you a quadratic, with two complex roots.
@@1234larry1 Exactly! It often helps if you can reduce the degree of the equations. The same goal is behind using the rational roots theorem, often used in similar situations.
Trial and error 🤢🤮
To really solve this just write a^2(a + 1) and notice that if a were to be an integer, it’s square must divide 36, and the only squares that divide 36 are 4, 9, and 36, so a must be 2, 3, or 6, from there, check
It took me all of 3 seconds to know a = 3.
Just replace 'a' by '1/b' to cast the equation in standard form then use Cardano (Tartaglia) equation to find one root and then divide by (b-root) to find the other two!
This method of solving the equation begins by guessing the correct answer! Why would anyone even continue once you got 27 & 9? That's your answer right there.
a^3 + a^2 = (a^2)×(a + 1) = (3^2)×(3 + 1) = 9×4 = 36
V interesting and knowledgeable
by faktoring , a^3+a^2 |+/-| n*a-36=0 , / 3*12=36 , trial n=3 / , a^3-3a^2+4a^2-12a+12a-36=0 , (a-3)(a^2+4a+12)=0 , a-3=0 , a=3 ,
/// for complex , a^2+4a+12=0 , /// , test , 3^2+3^2=27+9 , 27+9=36 , OK ,
a^2+4a+12=0 , a=(-4 |+/-| sqrt(16--48)/2 , a=(-4 |+/-| sqrt(-32))/2 , a=(-4 |+/-| sqrt(32)i)/2 , a=(-4 |+/-| 2*sqrt(8)i)/2 ,
a= -2+sqrt(8)i , -2-sqrt(8)i , solu. , a= 3 , -2+sqrt(8)i , -2-sqrt(8)i ,
Like they said, your solution presupposes to know the answer. Let’s do in another way.
a^3 = 36 -a^2 = (6-a)*(6+a)
Now, a^3 equals to a product of two factors; to get a product of two factors for a^3, you can decompose it this way only: a^2*a. Therefore
a^2= 6+a and a = 6-a.
From the second you get that a=3.
If you are given that the solution is in integers, then a^2(a+1) = 36 means a is in {1,2,3,6} - and process of elimination gives an answer. If you do not know that a is an integer, then you can try to identify where there are roots to a^3 + a^2 - 36 = 0 and you quickly get (a - 3)(a^2 + 4a + 12) = 0, or a = 3, -2 +/- 4✓(-2)
Put a=3,
See that a-3, will be a factor, then proceed
Thanks sir
Your channel is very helpful
by expanding this become so difficult 😭😂
Awesome explanation. If you went with just the obvious answer you wouldn’t have found the imaginary numbers. What I want to know is why hasn’t anyone commented about the most obvious part of the video. Your crazy good hand writing. I’d love to be that legible.
Re-write the problem as a^3=(6-a)(6+a) right hand side should be factorable by a therefore 6-a = 3 => a = 3.
Sir ,but at first when u substituted 3^3 and 3^2 at the place of 36 so we can also conclude this simply that a=3 on comparison of LHS and RHS .. for real roots and after all the answer is 3 ....
a=3 is the answer
3³+3² = 27+9=36
Are these "Olympiad" problems intended for 7 year olds?
Muy interesante video sobre ecuaciones con exponentes cuadrados y cubos, muchas gracias por compartir.😊❤😊.
3z3x3 =27
3×3 = 9
27 + 9 = 36
Yanı basit.
Başka,çözüm yolları sadece kafa karıştırır.
a = 3 is one answer . I will use remainder and factor theorem to find the other 2. Your method is also an excellent method. Thanks
a=3
a-3
Ok. La idea es pensar. Usar los productos notables. Como tarea debe ser direccionada y contestar con explicaciones en cada proceso . Así un estudiante mejora su capacidad cognitiva.
A = 3
Answer is a=3
a^3--27+a^2--9=0
(a--3)[a^2+3a+9+a+3] =0
a =3//
This is not a mathematical way. If the answer was 1.23142765, then this method would not help.
I just looked at it, didn't even watch the video, and immediately saw that a = 3. It's so obvious.
answer is a=3
a^3+a^2 =36 ; 3^3+3^2=36 ; a=3
Why not just that solution?
a³ + a² = 36
a³ + a² = 27 + 9
a³ + a² = 3³ + 3²
a = 3
Всё было ясно со второй строчки, но мы не ищем лёгких путей 🤣
3
Value of of 'a' will be :- 3
The moment you decide "out of the blue" to split the 36 into 27 and 9... Hmmm why not 3 and 33 ? . Just solve it by trying integer values for a starting at 2 . Job done and going home...
a= 3.
You already solved it at 0:46
Thanks ❤
I used substitution and found this answer in seconds
Haec a= 3 est.
3, as a real root.
Can't we say a = 3 , is that wrong?
You can just look at it, and know that it isnt 2, and isnt 4. So convoluted..
a=3 trivial solution
3 cubed plus 3 squared = 36
Can you do it again with the rhs =35
Then we use CARDANO's formula.
Which follows shuvon kar er fanki,,hater modhye chinte katle koto railo baki?
a=3......I solve these in my brain!
The equation has still 2 roots.
a³+a²=36
(x-3) (x²+4x+12)=0
(x-3) =0
x= 3 #
(x²+4x+12)=0
(x+2)²+8=0
(x+2)²=-8
(x+2)=±i√8
x= 2±i2√2 #
a equalitu 3
I would like to see a rigourous démonstration. This is not the case here
The value of a=3
Так это будет 3
Bagi yg pegusaha kecil dpt pinjaman tanpa bungga nabatas 10 juta.
When using equations over (^2) of itself, we are using (i and even a ^i) somehow ?
No need to laught at me, i am trying to understand how we can handles in 2d maths over ^2 equations without divisions of itself (singularities)
Pô!! Essa eu fiz de cabeça 😂
Для чего дальнейшие рассуждения когда D
Извлекаем и получаем мнимые корни.
Please reform your writing method
You have solved this by long way, I could not understand. Bogus, go through simple way.
27+9=36
Bla bla bla bla bla. Don’t need nine minutes.
What are you getting from being a dick? You aren’t impressed, move on with your life. Don’t rip the guy for being thorough. Videos like these are for people who don’t have a masters in mathematics.
Can u ans my q?
Sofism
lost me!
এতো পেচিয়ে অযথা দীর্ঘ করার প্রয়োজন আছে কি ?
For sure not beautiful and by no means for advanced students…
"3" did it my head in like 10 seconds
🤢🤮
a³+a²= 36
a²(a+1)= 9•4
a= ±3
a+1= 4 -> a= 3
(a³+a²-36)/a-3= a²+4a+12, a ∉ ℝ, a= 2(-1±𝒊√2)
😮
@@ningaiahkc3260 ridículo esse vídeo, muito mal explicado
?
a =3
3
a=3
a = 3
a=3
a=3
a = 3
3
3