A beautiful math problem for advanced students | Equation Solving 

A trivially simple problem for beginners. a=3 by inspection

Writting 36 as 27+9 presupposes you know the answer.

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.

3 to the power 3 + 3 to the power 2 is 36. Hence value of a=3

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.

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

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!

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.

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

Your channel is very helpful

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 ,

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.

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; -2-2V2*i; -2+2V2*i. Используя схему Горнера, получим первый корень: 3. И перейдем к квадратному уравнению a^2 + 4a + 12 =0.

a³+a²=36 =>a²(a+1)=36 Now 36=2²×3×3 Now put a=3 [By trail and error method] LHS,3²(3+1)=4×9=36=RHS

Thanks sir

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

Muy interesante video sobre ecuaciones con exponentes cuadrados y cubos, muchas gracias por compartir.😊❤😊.

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.

It took me all of 3 seconds to know a = 3.

Are these "Olympiad" problems intended for 7 year olds?

I just looked at it, didn't even watch the video, and immediately saw that a = 3. It's so obvious.

Put a=3, See that a-3, will be a factor, then proceed

I found all solutions in my head in a minute. First solution was ofc 3 just by checking the divisors of 36. Then all that was left to solve was a²+4a+12=0 (quadratic formula).

This is not a mathematical way. If the answer was 1.23142765, then this method would not help.

They love that letter i.

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^2+4a+12=a^2+2.2a+2^2+8=(a+2)^2+8>=8#0

Soln a³+a²=36 a³+a²=27+9 a³+a²=3³+3² a³=3³, a²=3² a=3 answer

by expanding this become so difficult 😭😂

Thanks ❤

Why not just that solution? a³ + a² = 36 a³ + a² = 27 + 9 a³ + a² = 3³ + 3² a = 3

Trial and true. 3³ + 3² = 36

3z3x3 =27 3×3 = 9 27 + 9 = 36 Yanı basit. Başka,çözüm yolları sadece kafa karıştırır.

a^3--27+a^2--9=0 (a--3)[a^2+3a+9+a+3] =0 a =3//

a^3+a^2 =36 ; 3^3+3^2=36 ; 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.

I would like to see a rigourous démonstration. This is not the case here

a=3 is the answer 3³+3² = 27+9=36

a=3 (3X3X3+3x3 )=27+9=36

Can you do it again with the rhs =35

Для чего дальнейшие рассуждения когда D

Can't we say a = 3 , is that wrong?

I used substitution and found this answer in seconds

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 #

Haec a= 3 est.

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)

3, as a real root.

Answer is a=3

a-3

Value of of 'a' will be :- 3

3

3.

A. a 3

3 cubed plus 3 squared = 36

a=3......I solve these in my brain!

Всё было ясно со второй строчки, но мы не ищем лёгких путей 🤣

a=3 trivial solution

The value of a=3

FFS, it took me 5 seconds to solve it.

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

Pô!! Essa eu fiz de cabeça 😂

a is equal to 3

a=3（3X3X3+3x3）= (27+9)=36

You can just look at it, and know that it isnt 2, and isnt 4. So convoluted..

Which follows shuvon kar er fanki,,hater modhye chinte katle koto railo baki?

Don,t say brainless problems

It’s simple give 3 to then do other things. No need to talk a lot

a = 3.

Higher Maths 😅

a equalitu 3

Bagi yg pegusaha kecil dpt pinjaman tanpa bungga nabatas 10 juta.

27+9=36

Swarati puja kintu bhasan na, some one else.

Please reform your writing method

a

Bla bla bla bla bla. Don’t need nine minutes.

Так это будет 3

You have solved this by long way, I could not understand. Bogus, go through simple way.

It is not proper solution.....stop it.

Can u ans my q?

Don't show that type of false mathmetics....

Sofism

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)

এতো পেচিয়ে অযথা দীর্ঘ করার প্রয়োজন আছে কি ?

lost me!

"3" did it my head in like 10 seconds

?

You are not a teacher....

answer is a=3

3

a = 3

a= 3.

a=3（3X3X3+3x3）=27+9=36

3

3