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

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

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.

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.

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

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.

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!

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

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 ,

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

a^3 + a^2 = (a^2)×(a + 1) = (3^2)×(3 + 1) = 9×4 = 36

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.

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)

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

Your channel is very helpful

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.

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

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

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

by expanding this become so difficult 😭😂

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

V interesting and knowledgeable

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.

Haec a= 3 est.

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

a-3

Thanks sir

a =3

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

Answer is a=3

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

A = 3

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

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

Value of of 'a' will be :- 3

answer is a=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...

3

a= 3.

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

3, as a real root.

You already solved it at 0:46

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 #

I used substitution and found this answer in seconds

a=3 trivial solution

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

a equalitu 3

3 cubed plus 3 squared = 36

Thanks ❤

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

The value of a=3

Так это будет 3

Can you do it again with the rhs =35

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

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

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

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)

27+9=36

Pô!! Essa eu fiz de cabeça 😂

Для чего дальнейшие рассуждения когда D

Please reform your writing method

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

Sofism

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

Can u ans my q?

For sure not beautiful and by no means for advanced students…

এতো পেচিয়ে অযথা দীর্ঘ করার প্রয়োজন আছে কি ?

lost me!

🤢🤮 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)

?

"3" did it my head in like 10 seconds

a=3

3

a = 3

a=3

a=3

a = 3

3

3