Once you find '3', a fun way to find the resulting quadratic equation to solve for the other two roots is ... long division. Just set up a long division for /a^3+a^2-36\a-3. First round: a^2, gives a^3-3a^2, subtract it to leave 4a^2-36. That goes 4a times, leaving 12a-36, etc until leaving rest 0. The new equation is a^2+4a+12, hence a^3+a^2-36 can be written as (a-3)(a^2+4a+12) .

If we are using the "guess a root" method, I think it is much easier if we just do that with a substituion. * Easy to guess a = 3 is a root * Now we let b = a - 3, and then substitute a with b + 3 and combine the terms. It is guaranteed to get an equation bf(b) = 0 in which f(b) is quadratic.

Thank you for the question. Once you determine that a - 3 is one root, isn't it a little more straightforward to find the other roots by dividing a^3+a^2-36 by a-3? Polynomial division is not so difficult.

There is an easier way: Rewrite as a^2 (a+1) = 36. if a is an integer then you can factor both sides so that the right hand side can be expressed as two factors as 36 x 1; 18 x 2; 9 x 4, 3 x 12, 6 x 6. One of these factors must be a^2 so this must be 9 or 4. 9 works as 9(3+1) = 36. I think it is clear the other two roots are complex but you can find them by dividing by the know factor (a-3) to give: a^2 + 4a + 12 = 0. Complete the square to give (a + 2)^2 = -8 which has two complex roots.

a³ + a² = 36 a²(a + 1) = 36

a² + a³ = 6² a² * (1 + a) = 3² * 2² a² shows that a can be 3 or 2, while (1 + a) indicates that a can only be 8 or 3. Logically, the answer is 3.

My favorite way is to think of it is as literally adding a cube and a square. The area of the cube is a by a by a and the extra a by a square is simply given. With this, the only way to make a perfect square is to place enough a squares to be square. With one a by a square already given you must add (n^2)-1 a by a squares to make a square. This means that a must be three since we added three square to complete the square. This can be put out further to state that any number a squared and cubed together in the set (n^2)-1, where n is 2 or greater will be equal to (n((n^2)-1))^2. e.g. ((2^2)-1)^3 + ((2^2)-1)^2 = (2((2^2)-1))^2 ---> this is 3^2 + 3^3 = 6^2 = (2*3)^2 ((3^2)-1)^3 + ((3^2)-1)^2 = (3((3^2)-1))^2 ---> this is 8^3 + 8^2 = 24^2 = (3*8)^2

I took every multiple of 36 first (1;2;3;4;6;9;12;18;36), then cut all those which had a the third power greater than 36 and got (1;2;3). Then I tried finding which number would satisfy (a+1)½ = 6÷a, which straight foward had to be 3, for the other numbers didn't have a perfect square for their sucessors.

You could also obtain that equation by simply dividing a^3 + a^2 - 36 by a - 3 and get a^2 + 4a + 12 with no remainder since a-3 is a factor you know in advance that the remainder is 0. It goes like this: First divide a^3 by a to get a^2 and you write a^2 as part of the solution and multiply it by a-3 to obtain a^3 - 3a^2 and you subtract this from the original equation a^3 + a^2 - 36 to get a^3 + a^2 - 36 - a^3 + 3a^2 = 4a^2 - 36 and you divide this by a to get 4a and write 4a as part of your solution and multiply it by a-3 to get 4a^2 - 12a and subtract this to get 4a^2 - 36 - 4a^2 + 12a = 12a - 36 and finally you divide this by a to get 12 and subtract 12a - 36 to get a remainder of 0. So the result is a^2 + 4a + 12. Dividing polynomials is essentially just like you divide numbers.

beside using factorization, we can do this: a^2(a+1)=36=3^2(3+1), a=3. the other 2 roots can then be easily followed

Use polynomial long division to get the quadratic equation (as mentioned before by other viewers).

Good methodology, but a successful guess of the first root would be a necessity which makes this approach dependent on scenarios. Wouldnt it better to modify the question to: knowing 3 is a root, solve all available roots?

I have used the graph method. Let a³+a²=f(a). And 36 is constant function. 36 is positive integer and f(a) is a monotonically increasing function on the interval x≥0. Therefore f(a)=36 has one integer solution.

I used the Rational Root Theorem with synthetic division to find that (a-3) was factor of the trinomial. This generated a=3. Setting the other factor equal to 0 and using the Quadratic Formula Theorem, the 2 complex conjugate roots were generated. You found 3 to be a root by guessing?

a^2(1 + a) = 36 which means all factors of 36 are equal to either (1 + a) or a^2. This gives the 3 desired roots of the cubic :)

Worked out in my head because its easy.

Another way in this specific simple case is to factor a² (a+1) = 36, and then we know that a number multiplied by itself and a number greater than one will give us 36. There are only so many ways to get 36 by multiplying 3 integers. 2 2 9, 3 3 4, 6 2 3. Only one of them satisfies our conditions so a must be 3 as one of the roots. And infact the only real root

Fascinating got the real root on my own, really needed help to get the complex roots.

Since by inspection a = 3 is a root, then a³ + a² - 36 can be factorised as (a - 3)(a² + ba + c) ⇒ a³ + (b - 3)a² + (c - 3b)a - 3c = a³ + a² -36 ⇒ (b - 3 = 1) & (c - 3b = 0) & (-3c = -36) ⇒ b = 4 & c = 12 Hence, (a - 3)(a² + 4a + 12) = a³ + a² - 36 = 0, and we can easily solve this for a. ⇒ a = 3, -2 - 2i√2, or -2 + 2l√2.

I just plugged "3" into the initial equation (math in my head) and realized it was correct. I struggled a bit with the long method.

I've a feeling that the imaginary roots can be found earlier in the procedure, but I'm not sure. I'll try again. Thank you.

The easiest way is to factor out the a^2 and then take the square root of both sides of the equation. This leaves a*(a+1)^1/2 = 6. The square root of six is excluded because the terms a and a+1 are different numbers. That leaves 2*3=6 as a potential solution. Trying a=2 does not work because it leaves a radical. But, comutatively, a=3 yields 3*2=6 as the correct solution.

if we simply approach a^3+a^2=36 a^3=36-a^2 a^3=(6+a)(6-a) since,a^3=a*a^2 if we put 6-a as a and 6+a as a^2 we get 3 as the only real satisfying solution.

a^3 + a^2 = 36 a + 1 = (6/a)^2 x = 6/a x^2 - (6/x) - 1 = 0 x^3 - x - 6 = 0 (x - 2)(x^2 + 2x + 3) = 0 x = 2 and complex roots. If x = 6/a then a = 3.

This was a pretty cool way to do it, and I will consider using it if I for some reason come across an equation like this again. Will have to say tho, I solved it in only about 1.5 minutes using simple guess n check logic: a^3 + a^2 = 36, and 36 relatively could not be solved with bases greater than 4 or something, so I know the numbers have to be relatively small. I started with a small number, 2, and got 12 as a result. I then decided to double that and try 4. I got 80, which was to big. If 2 was too small, and 4 was too big, 3 was the correct number. Now of course I don't suggest this method but if it's relatively small I suggest something like that if you dont wanna do something like in the video

In my opinion, once you have a root, doing long division is simpler than going through all that factoring.

This could be proven much simpler by applying graphics. Finding out 3 is a solution is easy enough, then it's only a matter of proving there are no others. 1. Since a^3 and a^2 are both exponential functions (in positives), the sum of them would also give us an exponential function, meaning that there is only one solution to this equation. 2. a couldn't be a rational number

Wow so much nicely explained! Great technique. SO much impressed by the clean writing with different colours!!! Well done. Such a mesmerizing channel. #SUPERMATHS

a = 3 by inspection. Long division. Quadratic formula.

All I remember is elementary algebra... so, I ended up going this way: a^3+a^2 = 36 a^3+a^2-36 = 0 At about that time I saw that quadratic form sitting in there, so... a^3+(a+6)(a-6) = 0 At that point, since in real integer terms 3 is the only number that is divisible by all those terms... 27+(3+6)(3-6) = 0 27+(9)(-3) = 0 27-27 = 0 0 = 0 So, a = 3. My brain still refuses to recall intermediate algebra with complex numbers, but, I watch these videos and work the problems in advance... slowly getting back to algebra, in memory of the best math teacher in the world...

Highly impressed with such a technical solution.. Well done.

yo you explained a cube minus b cube formulae, square, delta. but you didnt explain a2 a3 formulae that uses delta. you just typed directly the answer...

Great solution and i also really like the problem!!

I worked this one out via trial and error. 3 to the power of 3 is 27. 3 to the power of 2 is 9. 27+9=36. Therefore, a=3. It is usually quite simple to solve these problems by a process of trial and error if the unknown number is a whole number. If the unknown number is not a whole number, then it becomes much more complicated.

Well, using the terminology of Vieta's formula, one could simply write a² as a sum of -3a²+4a² and then can proceed factorization.

a=3. I did by solving it like a Diophantine Equation.

Ur trick is perfect,thank u so much

I mean, that can basically be solved by inspection. Haven't watched the video, but it boils down to: a^2 * (a+1) = 36. The factors of 36 are 2*2*3*3 and just fooling around for 20 seconds reveals that a=3. Obviously, there are two other solutions, complex probably, but meh. :)

While you're playing with your crayon I figured it out in my head. Larger numbers are a little harder but they still follow the same rules.

Once you find that a = 3 is root, you can see that the polynomial a ^ 3 + a ^ 2-36 will be equal to (a-3) g (a) where g (a) is a generic second degree polynomial, that is g (a) = ra ^ 2 + sa + t with r, s, t real numbers. Then, a ^ 3 + a ^ 2-36 = (a-3) g (a). By developing the product we obtain a third degree polynomial in function of the r,s,t parameters , using now the principle of identity of the polynomials, we can equalize the coefficients and obtain a system that allows us to calculate r, s, t. Thus, we find the same polynomial g (a) = a ^ 2 + 4a + 12 which admits the two complex roots found in the video :)

Simple way is.. a²(a+1)=36 And then comparing both sides.. easily you get the answer a=3

It can be solved by a little observation , by seeing the right hand side (36) we get that 'a' must be less than 4 as it's cube is 64 and greater than 2 and as 64 is a integer it would be 3 .

a with an exponent of 3 is 27 and a with an exponent of 2 is 9. The two a’s are 3!

Great olympiad. The olympiad of the Law.

Knowing that 3 is one of the roota. Use the briot ruffini to determine the equation as two terms multipling and then you'll have the roots in the simple way

Once you got your quadratic you could tell the roots are imaginary because in a quadratic equation if a=1 and c>(b/2)² then your variable is imaginary. 12>(4/2)².

This is the most simplest problem in Indian IIT exam. This is not an olympiad level question.

A different Approach, when I saw the equation I notice that maybe I will be able to write it as a difference of squares plus or minus a difference of cubes, then I notice that 36 can be expressed as. 27+9 that is a perfect square and a perfect cube. So a^3+a^2=36 => a^3-27. +. a^2-9 =0 =>. (a-3)(a^2+3a+9)+(a-3)(a+3)=0. =>. (a-3)(a^2+4a+12)=0 From that we get two complex solutions. -2+-2i*root(2) and one real solution. a=3

very well done, thanks for sharing

How would you apply this math problem to tangible relativity.

It depends on how we're asked to solve the problem. One way is to realize that the greatest common factor on the left hand side of the problem is a^2. a^2(a+1)=36 -> a = 36/a^2 - 1. For 1, 1 = 35. This is false. For 2, 2 = 8. This is also false. Lastly, 3 = 3. This is true. Knowing this we can then realize that while factoring a^3 + a^2 = 36 we can start with (a-3). Divide a^3+a^2-36 by (a-3). This leaves us with a^2 + 4a + 12. This is easier to factor. Using quadratic formula you can now find the other solutions. x=(-b+/-(b^2-4ac)^.5)/2a. Hence x=(-4+/-(16-4(1)(12))^.5)/(2(1)). This gives us -2+/-2i(2^.5) for our other 2 possible solutions. Let's prove it. Plus these solutions back into the original equation. When dealing with -2+2i(2^.5) a^2 = -8i(2^.5)-4. a^3 = 16i(2^.5)+8+16(2)-8i(2^.5). Add a^2 and a^3 to get 36. Do the same with the other solution. I'll leave this for you to prove.

Thanks for your exlanation. We can use factor theoreme n Horner's theoreme

L'elemento base interessato alla decrescenza numerica esprime il valore è la prevenzione?

Below is another way to solve this equation. Longer development (TARTAGLIA method) but universal. Imagine your equation was a^3 + a^2 = 37. Thank you very much for your video. This is for the fun of math. Question: a³ + a² = 36 a = ? (in R) Answer: a³ + a² = 36 a³ + a² - 36 = 0 third degree equation (template: a'x³ + b'x² + c'x + d' = 0) a' = +1 b' = +1 c' = 0 d' = -36 cancellation of second degree term (a²): a = x - b'/3a' a = x - 1/3 because a = x - 1/3 then a³ + a² - 36 = 0 becomes: (x - 1/3)³ + (x - 1/3)² - 36 = 0 ---===oo===--- (x - 1/3)³ = x³ - 3*x²*(1/3) + 3*x*(1/3)² - (1/3)³ (x - 1/3)³ = x³ - x² + x/3 - 1/27 ---===oo===--- (x - 1/3)² = x² - 2*x*(1/3) + (1/3)² (x - 1/3)² = x² - 2x/3 + 1/9 ---===oo===--- (x³ - x² + x/3 - 1/27) + (x² - 2x/3 + 1/9) - 36 = 0 x³ - x² + x/3 - 1/27 + x² - 2x/3 + 1/9 - 36 = 0 x³ - x² + x² + x/3 - 2x/3 - 1/27 + 1/9 - 36 = 0 x³ + x/3 - 2x/3 - 1/27 + 3/27 - 972/27 = 0 x³ - x/3 - 970/27 = 0 ---===oo===--- x³ - x/3 - 970/27 = 0 is based on the template x³ + xp + q = 0 p = -1/3 q = -970/27 TARTAGLIA formula: note: cube root of n = n^(1/3) x = [-q/2 + √(q²/4 + p³/27)]^(1/3) + [-q/2 - √(q²/4 + p³/27)]^(1/3) x = [-(-970/27)/2 + √((-970/27)²/4 + (-1/3)³/27)]^(1/3) + [-(-970/27)/2 - √((-970/27)²/4 + (-1/3)³/27)]^(1/3) x = 3.3333 a = x - 1/3 a = 3.3333 - 1/3 a = 3 Final result: a = 3 🙂

This is like going for a root canal (no pun intended) and having all the teeth pulled "for practice." Once you guessed 3 at the beginning, what was the point of doing all the other manipulations when the initial substitution solved the equation?

I worked it out in my head before they got to the quadratic equations!

I went brute force, but my solution only considered positive integers. a^3 = 36 - a^2, therefore a^3 < 36. There are only three positive cubes < 36, and those are 1^3, 2^3 and 3^3. Very quickly, 1 and 2 are eliminated from consideration, leaving 3, which works.

9*4=a^2(a+1)=36 -> a=3 After that Bhaskara to check for real solutions. In absence of that, complex ones. Just remember how to divide a polynomial a³+a²-36/a-3 = a²+4a+12 => no real roots, so Bhaskara will give the complex ones. This could also be solved using the cubic solution, but, since that method is slow, I won't bother.

a^2(a+1)=36=a*a*(a+1)=3*3*(3+1), so a=3 is obvious, then divide the equation by a-3 to get (a^2+4a+12) and solve with quadratic. Or even as it was laid out by streamer a^3-3^3 + a^2-3^2=0 so with C(onstant)=3, once against obvious a solution is a=C=3, and again divide by a-3 to get quadratic.

Ohhh this help me so much tysm...👍

That is like touching the ear around the neck. resolve to a^2(a+1) =36 a^2(a+1) =9x4 As a>1 a^2 has to be > (a+1) a^2=9 ; a+1 =4 a^2=9 => a=+3 or -3 a+1=3 => a=+3 So a = 3

Thanks 🙏

Nice video! I did it about the same way

Heres an easier way to factor a 3rd degree equation after finding one solution: You just plug the solution and the components of the 3rd degree equation onto a long division like table then Add components with results. And multiply results by solution. If you get a zero at the end. It means the solution you found is correct, so its also a check method Okay here it is: 3 | 1(a³) 1(a²) 0(a) -36(c) ------------------------------------------------------------------ | 0 3{1×3} 12{4×3} 36{12×3} | 1{1+0} 4{1+3} 12{0+12} 0{-36+36} Now without explainations it looks like: 3 | 1 1 0 -36 --------------------------------- | 0 3 12 36 | 1 4 12 0 And we get in the bottom line 1 4 12 0 Which stands for a²+4a+12=0 Here 2 more cool tricks for polynomials If components of polynomial will be called: a(x³), b(x²), c(x), d(constant). Then if: a + b + c + d = 0 then x=1 2nd trick: If a+c = b+d then x=(-1) Ex. 2x³-4x²+6x-4=0 Then ---> 2-4+6-4=0 ---> x=1 Also ---> 2+6 = -4-4 ---> false, x is not -1 Afterwards you can subtitute 1 into the table ive shown before to factor the polynomial 1 | 2 -4 6 -4 -------------------------- | 0 2 -2 4 | 2 -2 4 0 We get x²-2x+4=0 Good luck ;)

m²(m+1)=3²(3+1), so m=3 is a solution. Use that to factorize: (m-3)(m²+4m+12)=0. 16-48 is negative, so the other roots are complex.

very good👌

a3 = positive = only positive roots try a =1, then 2, then 3. solved by induction, because higher numbers = bigger result.

You can use a numerical method to find the zeros

If we get one factor a-3 second factor can be found ot by synthetic division

Thanks ma'am

In which board you are writing?

Thnku for question A=3

this is so cool

As stated by other viewers, this method is scenario based, and falls apart as soon as you can't find a good guess for a (e.g., a^3 +a^2 = π). Then again, I suppose it demonstrates how you can't always find the analytical solution of all equations...

Wow what a technique used. Highly impressed with such a technical solution.. Well done. Love yours content. #SUPERMATHS

Write the LHS as the product of 3 factors, aa(a+1), and note that the RHS, 36, is divisible by the prime numbers, 2 & 3. Then use the known result that if a product of integers is divisible by a prime number, then at least one of the products must itself be divisible by that prime. Considering just the prime 3, then either a is divisible by 3 or a+1 is. The numbers divisible by 3 are 3, 6, 9,. . . (We can ignore the negative numbers since that would give a negative number on the LHS.) a cannot be 6, 9, . . . since that gives a number on the LHS that is larger that 36. a+1 = 3, 6, 9, . . . also doesn't work, giving 12, 150, 576 , . . . on the left. But a=3 works: 3^3 +3^2 = 27 + 9 = 36. Once we have one solution, we can divide a^3+a^2 -36 by (a-3) to get a quadratic equation, which can be solved by the standard formula.

Why you haven't applied polynom division?

a³+a²=36 --> a is an integer a²(a+1)=3²(3+1) --> a=3 To see the existence of any other value of a, note that as a is an integer then a²=3² and a+1=3+1 --> a=±3 a²=3+1 and a+1=3² --> a=±2, a=8 As check noting that a³+a²=a²(a+1) then: For a=-2, a³+a²=-8+4=-4≠36 Fod x=2, a³+a²=8+4=12≠36 For x=-3, a³+a²=-27+9=-18≠36 For a=3, a³+a²=27+9=36 For a=8, a³-a²=512-64=448≠36 Thus the solution is a=3

To get 24 +12= 36....what they got in common is the 6...but it's powered so I need to go different route..(6×6 - 2×6 to get 24.....and the other side of + can be counted as well ...hmmm √ in front of (...this happens also with after =36 but I don't think it's needed to be count..I think

This basically just "oh you have a polynomial of degree higher than two? Try to guess one zero and you are good". I mean if someone didn't know that method already, it is pretty useful, but this is not an olympiad level math

If only natural numbers allowed, the answer is 3. (a-3)(a^2+4a+12)=0 This function has a critical point on 0 and 2/3.

27+9 imediato! In fact, there are 3 roots. Excellent approach -- quite clear. What Olympiad??

Your way of solving is counter intuitive. How would someone who doesn't know the answer already solve this question? If you dismantle 36, it goes 3² and 2². Could you use a more intuitive way of solving this quastion? How would you use Bhaskara to solve this?

it can be solved by synthesis method also.

Go with the synthetic division of polynomials.

Here's my approach to the question: First i factored out an a² => a²(a+1) = 36 Then i took the square root on both sides => a•sqrt(a+1) = 6 Then i wrote down the factors of 6 => a•sqrt(a+1) = 1×6 = 2×3 Then i plugged in each factor of 6 to get the correct answer, 3.

I just looked up all perfect squares

🙏 good morning, vanishing process is most suitable,a=3,sir/madam thanks

Математика соотношений и пропорций. Если в реале появляется потребность решения такой задачки, это решаемо на "раз-два". Но заморачиваться тем, что давным-давно решено, нет уж. Есть гораздо-гораздо более насущные вопросы.

this problem is much less than difficult, esp if you have taken the interm alg class offered by aops. i would hesitate to label this level of algebra as "olympiad mathematics" but it is quite more difficult than much more than what core math standards would put one up to, i suppose. as many have already beat me to it, you could simply perform synthetic division/polynomial division after figuring out a=3 either through pure guessing or the rational root theorem. either way, great video & explanation of the solution

So, does that mean a=36 to the root of 5? Which is roughly 2.0476?

a2*(a+1) then find the divisors of 36 which are perfect sqares, then *9 then a=3, that is so easy.

Is there a way to find a=3 without iterating? I don't like brute-force equations

3 was my instant instinct as I paused at 2 seconds. I don't know why.. but then I thought of it 6×6 is 36. The ³ and ² are 5 proportions of that 36. And a+a must be the last in which 3+3 works to make the 6th proportion of 6. Then I came to the comments. I still don't know if Im right yet. But I saw -3 and have a feeling that is right

a is 3... i just looked at it. So 3x3x3=27 + 3x3=9 means 27+9=36.

step 1: take a sharp look at the equation to see that a=3 is in fact a solution. step 2: note that the left hand side is "clearly" monotonous for a>=0 and thus injective, meaning there is no other positive a satisfying this equation. step 3: note that a^3+a^2 is negative for a|a^2|), so that wont lead to a solution either step 4: note that for |a|

Started trying math on it, then decided just to do trial and error and got a=3 in about a minute.

This won't be in the SAT right..?

Basically the "trick" is guess a root and work from there.

Sounds more like "Maths Widow" ...