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) .
I see a lot of these types of videos, and nobody uses long division of equations. I'm not sure if it's even taught anymore. It's so much faster than the method shown.
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.
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.
We could try the Rational Roots theorm, which says (in this case) that if the cubic has a rational root, then it must be one of a = ±1, ±2, ±3, ±12, ±18 or ±36. Now we know from Descartes Rule of sign that there is one real root, so just try the positive values of a above noting that it can't be bigger than 4, as a³ = 64 where a² is positive, so we just need to try a = 1, 2 or 3 to find the rational root (if it exists), and it does when a = 3.
@@vitalibahdanovich8072 Guess and check DOES not constitute a solution, even if u are able to get an answer. U are only allowed to solve it algebraically.
@@jayjayDrm Well as a Math student, we are taught that guess and check method are never accepted as solution. One CANNOT use guess and check to solve problems.
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.
@@helrazy9026 What are you talking about? if you mean to say that the cube root of 13 plus the square root of 13 is 36 you're wrong and besides that is not the equation OP gave.
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.
Actually, if you assume a is an integer (which it might not be) then it turns out that we can use a²(1 + a) = 36 to find the value of a by a theorem. The greatest common divisor of a & (1 + a) is 1. I.e. gcd(a, 1 + a) = 1. But this also means that the gcd(a², 1 + a) = 1. Now a²(1 + a) = 36 = 6² where gcd(a², 1 + a) = 1 ⇒ 1 + a is a square, and it must be smaller than a², which leads us to deducing that 1 + a = 4, so a = 3, as a = 1 or 2 can't be a solution, as then 1 + a would not be a square.
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?
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.
If you can see the answer by inspection, why continue? Just skip the algebra and show by substitution that if a=3, then 3^3+3^2=27+9=36. What other proof is necessary? The question never asks for all complex values of a. The real numbers are suficient and implied unless explicitly expressed by the context (the picture is 2-D).
@@wdentondouglas Because I felt like commenting. Often the approaches or answers not asked for can be insightful, but apparently, infuriating to some 😂🤣.
@@KAMRANKHAN-xf7iq 1*a² = (b - 3)a², and as there is no "a" term, so 0*a = (c - 3b). The reason we can write a² + b*a + c as a factor is because if we have a linear factor of a - 3, then we must have a quadratic factor, which when multiplied by the linear factor results in a cubic expression.
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.
2 роки тому+1
It could have negative roots as well. However, since the function is monotonically increasing even for negative numbers, you know that it has only one solution
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
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
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?
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.
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
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...
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.
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
Yeah. No kidding. I knew that using 5 and 4 would not work because either of those cubed would be too high. Tried 3 and lo and behold, it worked. Serious overkill here.
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. :)
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.
You can do that systematically, and probably did so unconsciously: a^3 + a^2 = 36 => a^2(a + 1) = 36, plus the prime factorization of 36 guarantees you will find the answer.
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 :)
Wow so much nicely explained! Great technique. SO much impressed by the clean writing with different colours!!! Well done. Such a mesmerizing channel. #SUPERMATHS
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 ;)
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 .
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.
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.
But this can only be useful if you know there is an integer solution, otherwise there is no guarantee that plugging different values will eventually give something useful, which means that in the end dividing by a^2 doesn't isn't necessarily faster than just plugging values in the original equation. Since a^3 is the dominant coefficient, you can realize that only positive integers have a chance of being solutions, because negative ones will always leave a negative result, which means you can just try them and see if you get a solution before your result becomes too big. If you do, like in this case with a = 3, you can divide. If you don't, then you'll need another technique, like for example the cubic formula, that is absolutely awful, or a trick that works specifically for this question.
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?
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...
Consider a³ + a² - 36 = 0 By Descartes Rule of Sign, there is one positive real root, as the sign of the coefficients changes once when a is positive. By observation, a = 3 is a solution, so we have found the only positive real root of the equation. Applying Descartes Rule of signs again, then their are 0 or 2 negative real roots. However, a³ + a² is an increasing function except at the point where a = 0, where it has a point of inflection, hence there are no other real roots. We then factorise and solve for the other two complex roots.
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
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?
Because that's higher math. Purposely complicated to seem sophisticated. More like sophistry if you ask me Why solve it just using base integers when you can go on a long speil of all the numbers and formulas possible to make a circle peg into a square peg to fit into a triangle hole then say the its a circle because x(x³)=samolian
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
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.
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.
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
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
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
Fun fact: I (think I) solved this problem without equations. I just thought of a number that to the power of 2 and 3 of 36. There could be countless attempts without equations, but you can see that such a number does not exceed 4, because 4²= 16 4³= 64, so it would not give the desired result. So I tried it with three, and it worked... 3³ = 27 3² = 9 27 + 9 = 36 Therefore, a = 3 I hope it's correct, because the result satisfied the given equality. Edit: Unfortunately, since it's an Olympics, you must have a solution to the answer. I don't know if the way I did it is ridiculous but, even so, I managed to solve it.
The only thing missing from your answer is showing that the equation has a unique solution (the solution is obviously unique if you draw the graph of f(a)=a^3+a^2. However, when mathematical rigour is demanded that might not be enough.). That is to say, a=3 and there exists no other solution. These theorems are called uniquness theorems in math. If you prove a uniquness theorem and then solve the equation by plugging in trial numbers, no mathematician would argue with you.
@@maazali1595 Out of courtesy, inert, leave here, this is a math class, not English. Nobody called you into the conversation. In my conversation you do not enter.
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
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
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|
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.
I use the following method so I don't hurt my brain: Using Matlab Symbolic Toolbox: syms a eqn = a^3 + a^2 == 36 solve(eqn,a) ans =
3 - 2^(1/2)*2i - 2 - 2 + 2^(1/2)*2i This method has the advantage that any system of equations can be solved, the solver never makes a mistake, and you can find the symbolic or numerical answer(s). Moreover, Matlab will then produce mcode,C/C++ code if you need to integrate it into a program. Nonlinear equations are also not a problem as well as nonlinear optimization, constrained optimization are all handled.
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.
OK ... Your method is a GOOD one, but maybe for a more complicated problem WHY ? .. Because we all know that 3 squared (3 x 3 X 3) = 27 and 3 squared (3 x 3) = 9 and 27 + 9 = 36 So A = 3
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.
a^3 + a^2 = 36 By inspection, we can see a = 3, so we factor out (a - 3) to get a quadratic equation: (a - 3)*(a^2 + 4*a + 12) = 0 Thus, a = 3, -2±i*2*√2 .
3 is a root. So if you divide a^3+a^2-36 by a-3 have a II degree equation. But this is a trick. Real solution is binomial approximation for general form.
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...
Some trial and errors matching a^2(a+1) with decomposition into prime factors 2^2 * 3^2 will do it too. 2^2 * (3^2) does not match. However 3^2 * (3+1) does.
I solved this problem in LESS THAN 1 MINUTE! It is sufficient to divide all the terms by a^2. and you get 36/a^2 - a = 1. Then you can try with a= 2 and you get 36/4 - 2 = 7, so it cannot be 2. But if you try with a= 3 you get: 36/9 - 3 = 4- 3 = 1, which is the correct solution, in less than 1 minute, and with the least calculation!
By synthetic division the other factor comes out to be a^2+4a-24???? But by long division it comes out to be a^2+4a-12. Why the error in synthetic division???
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) .
That is a 'fun' way? Never liked long division i guess! Clever though
That is how I would do it. It is the simplest.
Agree. The method shown in the video is far too complicated.
I see a lot of these types of videos, and nobody uses long division of equations. I'm not sure if it's even taught anymore. It's so much faster than the method shown.
@@solomonwilliams9708 It was in my math book in high school 5 years ago, but it wasn't taught, so we skipped over the chapter.
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.
You can even use synthetic division which is even faster
Yes. That is what came to my mind too
Exactly. Just do a long division, even your father can do it :)
@@BertGafas bro why are you guys doing long division instead of synthetic? It’s so much faster
@@Firefly256 Because everyone already knows how to do long division, it's intuitive
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.
Río
www.youtube.com/@mathsmadeeasy7794
I did it like that
Well done! That's what I suggested exactly.
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.
We could try the Rational Roots theorm, which says (in this case) that if the cubic has a rational root, then it must be one of a = ±1, ±2, ±3, ±12, ±18 or ±36.
Now we know from Descartes Rule of sign that there is one real root, so just try the positive values of a above noting that it can't be bigger than 4, as a³ = 64 where a² is positive, so we just need to try a = 1, 2 or 3 to find the rational root (if it exists), and it does when a = 3.
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.
This is much better ☺️
@@vitalibahdanovich8072 Guess and check DOES not constitute a solution, even if u are able to get an answer. U are only allowed to solve it algebraically.
this solution is more than guess and check.
@@jayjayDrm Well as a Math student, we are taught that guess and check method are never accepted as solution. One CANNOT use guess and check to solve problems.
@@huiyinghong3073 ok, but i see nothing guessed here. solve a^2 = 3^2 or 2^2 and 1+a = 3^2 or 2^2. and then filter with logic.
a³ + a² = 36
a²(a + 1) = 36
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.
√³13 + √13 = 36. Easy peasy
@@helrazy9026 What are you talking about? if you mean to say that the cube root of 13 plus the square root of 13 is 36 you're wrong and besides that is not the equation OP gave.
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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 did? I did the same thing! Found the answer quickly.
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 :)
you are considering integer equations. But this is not an integer equation.
@@howareyou4400 yeah!
Actually, if you assume a is an integer (which it might not be) then it turns out that we can use a²(1 + a) = 36 to find the value of a by a theorem.
The greatest common divisor of a & (1 + a) is 1. I.e. gcd(a, 1 + a) = 1. But this also means that the gcd(a², 1 + a) = 1.
Now a²(1 + a) = 36 = 6² where gcd(a², 1 + a) = 1
⇒ 1 + a is a square, and it must be smaller than a², which leads us to deducing that 1 + a = 4, so a = 3, as a = 1 or 2 can't be a solution, as then 1 + a would not be a square.
That is how I solved it too. Assume that a is an integer a^2(a+1) = 4 x 9 => a = 2 or 3 substituting a=3
@@WagesOfDestruction except that a=2 is not a solution. There are two complex number solutions
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?
even if you didnt know. The 36 is 3^5 so you will make it 3^5= 3^3+3^2 and you continue
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.
If you can see the answer by inspection, why continue? Just skip the algebra and show by substitution that if a=3, then 3^3+3^2=27+9=36. What other proof is necessary? The question never asks for all complex values of a. The real numbers are suficient and implied unless explicitly expressed by the context (the picture is 2-D).
@@wdentondouglas
Because I felt like commenting. Often the approaches or answers not asked for can be insightful, but apparently, infuriating to some 😂🤣.
a^2 +ab+ c??? From where this factor comes out? I'm really confused
@@KAMRANKHAN-xf7iq
1*a² = (b - 3)a², and as there is no "a" term,
so 0*a = (c - 3b).
The reason we can write a² + b*a + c as a factor is because if we have a linear factor of a - 3, then we must have a quadratic factor, which when multiplied by the linear factor results in a cubic expression.
www.youtube.com/@mathsmadeeasy7794
Use polynomial long division to get the quadratic equation (as mentioned before by other viewers).
www.youtube.com/@mathsmadeeasy7794
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.
It could have negative roots as well. However, since the function is monotonically increasing even for negative numbers, you know that it has only one solution
@ there are no positive values of function that greater then f(a)=1 on interval x
Great olympiad. The olympiad of the Law.
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
how and yes.
Very good
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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.
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
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?
In cubic polynomials, the first root is taken out by Hit and Trial method. Which in a way is guesswork.
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
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.
In my opinion, once you have a root, doing long division is simpler than going through all that factoring.
Agree.
Can we use synthetic division method because it is simpler than long division method?
Yes divide
Alapból indulok a Diophanthoszi módszerrel.
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 by inspection. Long division. Quadratic formula.
Worked out in my head because its easy.
Thank you!
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
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...
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.
Highly impressed with such a technical solution.. Well done.
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
Yeah. No kidding. I knew that using 5 and 4 would not work because either of those cubed would be too high. Tried 3 and lo and behold, it worked. Serious overkill here.
Not reliable, the answer could be a fraction
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Well, using the terminology of Vieta's formula, one could simply write a² as a sum of -3a²+4a² and then can proceed factorization.
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. :)
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.
True, I figured that out as well.
You can do that systematically, and probably did so unconsciously: a^3 + a^2 = 36 => a^2(a + 1) = 36, plus the prime factorization of 36 guarantees you will find the answer.
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 :)
Wow so much nicely explained! Great technique. SO much impressed by the clean writing with different colours!!! Well done. Such a mesmerizing channel. #SUPERMATHS
Fascinating got the real root on my own, really needed help to get the complex roots.
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L'elemento base interessato alla decrescenza numerica esprime il valore è la prevenzione?
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 ;)
Yes, you described the Horner method.
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=3.
I did by solving it like a Diophantine Equation.
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.
Thanks for your exlanation.
We can use factor theoreme n Horner's theoreme
a is 3... i just looked at it. So 3x3x3=27 + 3x3=9 means 27+9=36.
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
🙂
👍🏻👍🏻👍🏻
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.
But this can only be useful if you know there is an integer solution, otherwise there is no guarantee that plugging different values will eventually give something useful, which means that in the end dividing by a^2 doesn't isn't necessarily faster than just plugging values in the original equation. Since a^3 is the dominant coefficient, you can realize that only positive integers have a chance of being solutions, because negative ones will always leave a negative result, which means you can just try them and see if you get a solution before your result becomes too big. If you do, like in this case with a = 3, you can divide. If you don't, then you'll need another technique, like for example the cubic formula, that is absolutely awful, or a trick that works specifically for this question.
@@coltith7356 100% agreed.
@@coltith7356 Kan ook opgelost worden met iteratie als je geen integer hebt.
En je hebt niet perse computers nodig voor iteraties.
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?
Great solution and i also really like the problem!!
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...
a with an exponent of 3 is 27 and a with an exponent of 2 is 9. The two a’s are 3!
Consider a³ + a² - 36 = 0
By Descartes Rule of Sign, there is one positive real root, as the sign of the coefficients changes once when a is positive.
By observation, a = 3 is a solution, so we have found the only positive real root of the equation.
Applying Descartes Rule of signs again, then their are 0 or 2 negative real roots.
However, a³ + a² is an increasing function except at the point where a = 0, where it has a point of inflection, hence there are no other real roots.
We then factorise and solve for the other two complex roots.
You can use a numerical method to find the zeros
Like newton-rapshon
27+9 imediato! In fact, there are 3 roots. Excellent approach -- quite clear. What Olympiad??
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.
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
Wow what a technique used. Highly impressed with such a technical solution.. Well done. Love yours content. #SUPERMATHS
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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?
Because that's higher math.
Purposely complicated to seem sophisticated.
More like sophistry if you ask me
Why solve it just using base integers when you can go on a long speil of all the numbers and formulas possible to make a circle peg into a square peg to fit into a triangle hole then say the its a circle because x(x³)=samolian
Thnku for question
A=3
A=3 is correct:)
@@SuperYoonHo 😊
@@pranavamali05 🙃
A=3 is correct but there are two more solutions. Watch the video.
@@BertGafas i did watch it and i did solve it before i watched it and i know lol;)
Can't you just substitute (a) as 3 in the question itself?
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
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.
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.
So, does that mean a=36 to the root of 5? Which is roughly 2.0476?
Simple way is..
a²(a+1)=36
And then comparing both sides.. easily you get the answer a=3
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.
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
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
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
Fun fact: I (think I) solved this problem without equations. I just thought of a number that to the power of 2 and 3 of 36.
There could be countless attempts without equations, but you can see that such a number does not exceed 4, because 4²= 16 4³= 64, so it would not give the desired result.
So I tried it with three, and it worked...
3³ = 27
3² = 9
27 + 9 = 36
Therefore, a = 3
I hope it's correct, because the result satisfied the given equality.
Edit: Unfortunately, since it's an Olympics, you must have a solution to the answer. I don't know if the way I did it is ridiculous but, even so, I managed to solve it.
The only thing missing from your answer is showing that the equation has a unique solution (the solution is obviously unique if you draw the graph of f(a)=a^3+a^2. However, when mathematical rigour is demanded that might not be enough.). That is to say, a=3 and there exists no other solution. These theorems are called uniquness theorems in math. If you prove a uniquness theorem and then solve the equation by plugging in trial numbers, no mathematician would argue with you.
Olympiad you mean?
@@ishan9665 Yes.
Olympics💀💀💀
@@maazali1595 Out of courtesy, inert, leave here, this is a math class, not English. Nobody called you into the conversation. In my conversation you do not enter.
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
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
If we get one factor a-3 second factor can be found ot by synthetic division
How would you apply this math problem to tangible relativity.
Why not use the solution for the general cubic equation contained in the standard math tables.
Interesting... 👍🏽👍🏽
But i think in order to solve the problem, the statement must be provided. Otherwise, it wont make much sense right??
a3 = positive = only positive roots
try a =1, then 2, then 3. solved by induction, because higher numbers = bigger result.
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|
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.
I use the following method so I don't hurt my brain:
Using Matlab Symbolic Toolbox:
syms a
eqn = a^3 + a^2 == 36
solve(eqn,a)
ans =
3
- 2^(1/2)*2i - 2
- 2 + 2^(1/2)*2i
This method has the advantage that any system of equations can be solved, the solver never makes a mistake, and you can find the symbolic or numerical answer(s). Moreover, Matlab will then produce mcode,C/C++ code if you need to integrate it into a program.
Nonlinear equations are also not a problem as well as nonlinear optimization, constrained optimization are all handled.
Gigachad.
Can we use logarithm?
Sure, but it is not necessary.
Ur trick is perfect,thank u so much
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'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.
OK ... Your method is a GOOD one, but maybe for a more complicated problem
WHY ? .. Because we all know that 3 squared (3 x 3 X 3) = 27
and 3 squared (3 x 3) = 9
and 27 + 9 = 36 So A = 3
very good👌
it can be solved by synthesis method also.
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.
Basically the "trick" is guess a root and work from there.
a^3 + a^2 = 36
By inspection, we can see a = 3, so we factor out (a - 3) to get a quadratic equation:
(a - 3)*(a^2 + 4*a + 12) = 0
Thus, a = 3, -2±i*2*√2 .
very simple:
let a = 1, 1^3 + 1^2 = 2 != 36.
let a = 2, 2^3 + 2^2 = 8 + 4 = 12 != 36.
let a = 3, 3^3 + 3^2 = 27 + 9 = 36 == 36 Bingo!
How much time? 2 minutes. I'm clever, yes? 😀
You'r missing 2 solutions. The idea is to use the one you know (3) to easily calculate the complex solutions
x^4+X^4 = 2X^4 = X256 if say X is 1
how come not same exponants ads up or used in process off rule
Correct solution: a^3 + a^2 = 36; Factor both sides: a^2(a + 1)= 9 x 4; Then, a^2 = 9 ; a = 9^1/2 ; Therefore, a = 3; Likewise, a + 1 = 4; a = 4 - 1 ; a = 3 . (Answer) . This is the correct solution.
3 is a root.
So if you divide a^3+a^2-36 by a-3 have a II degree equation.
But this is a trick. Real solution is binomial approximation for general form.
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...
I solved this in 10 seconds. Seriously. I thought I would try a whole number. I started with 3 and it worked.
Nice video! I did it about the same way
me too:)
but just used long division
it is faster than her's
@@SuperYoonHo Right the factoring is kind of slower
@@owlsmath u got it 👌
Is there a way to find a=3 without iterating? I don't like brute-force equations
Some trial and errors matching a^2(a+1) with decomposition into prime factors 2^2 * 3^2 will do it too. 2^2 * (3^2) does not match. However 3^2 * (3+1) does.
I solved this problem in LESS THAN 1 MINUTE! It is sufficient to divide all the terms by a^2. and you get 36/a^2 - a = 1. Then you can try with a= 2 and you get 36/4 - 2 = 7, so it cannot be 2. But if you try with a= 3 you get: 36/9 - 3 = 4- 3 = 1, which is the correct solution, in less than 1 minute, and with the least calculation!
You can solve it by iterative method:
1) Rearrange equation for one of the x's:
x = ( 36 - x² ) ^ (1/3)
2) Rearranged equation with indexes:
x[n+1] = (36-x[n]²)^(1/3)
3) Iterative method:
x[0] = 1 ⬅ start
x[1] = 3,27106631
x[2] = 2,93567218
x[3] = 3,01407567
x[4] = 2,99686145
x[5] = 3,00069693
x[6] = 2,99984510
x[7] = 3,00003442
x[8] = 2,99999235
x[9] = 3,00000170
x[10] = 2,99999962
x[11] = 3,00000008
x[12] = 2,99999998
x[13] = 3 😊
##############
Stephan Schlatow
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.
By synthetic division the other factor comes out to be a^2+4a-24???? But by long division it comes out to be a^2+4a-12. Why the error in synthetic division???
Why you haven't applied polynom division?