A beautiful math problem for advanced students | Equation Solving
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- เผยแพร่เมื่อ 3 ก.ค. 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
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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.
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
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 🤢🤮
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.
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
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 ,
It took me all of 3 seconds to know a = 3.
Your channel is very helpful
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!
a^3 + a^2 = (a^2)×(a + 1) = (3^2)×(3 + 1) = 9×4 = 36
V interesting and knowledgeable
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.
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.
This is not a mathematical way. If the answer was 1.23142765, then this method would not help.
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.
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...
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)
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
Thanks ❤
3, as a real root.
I just looked at it, didn't even watch the video, and immediately saw that a = 3. It's so obvious.
Muy interesante video sobre ecuaciones con exponentes cuadrados y cubos, muchas gracias por compartir.😊❤😊.
a^3+a^2 =36 ; 3^3+3^2=36 ; 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 equalitu 3
3 cubed plus 3 squared = 36
answer is a=3
a = 3
Value of of 'a' will be :- 3
I used substitution and found this answer in seconds
Why not just that solution?
a³ + a² = 36
a³ + a² = 27 + 9
a³ + a² = 3³ + 3²
a = 3
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 #
27+9=36
I would like to see a rigourous démonstration. This is not the case here
a=3 trivial solution
The value of a=3
3
Pô!! Essa eu fiz de cabeça 😂
You already solved it at 0:46
a=3......I solve these in my brain!
The equation has still 2 roots.
Can you do it again with the rhs =35
Then we use CARDANO's formula.
Для чего дальнейшие рассуждения когда D
Извлекаем и получаем мнимые корни.
Can't we say a = 3 , is that wrong?
Так это будет 3
Are these "Olympiad" problems intended for 7 year olds?
Please reform your writing method
You have solved this by long way, I could not understand. Bogus, go through simple way.
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.
For sure not beautiful and by no means for advanced students…
?
🤢🤮
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
"3" did it my head in like 10 seconds