@@Rx800.0 Sure, most of such equations needs to be calculated properly, but knowing how such functions progress helps you with a good prediction where the solution should be. It will be tricky if the result is a complex value.
it's easily proofed, that one obvious solution is x= - 2 alternative solution 1: divide x³ -- x² + 12 by x+2 with the so called long division (polynomial division) to find the polynomial of degree 2. solve this quadratic equation to receive two complex solution if needed. alternativ 2: x³ -- x² + 12 = ( x +2)(x² + ax + 6). find the parameter a by multplication, and comparing both sides.
How nostalgic for school! I still remember the amazement when I realized that I was coming to the solution and the joy when I found it. It was just an equation in the end, but for me it was like conquering the world
If the goal of the olympiad is to solve the problem quickly, I would apply numerical theory first. It’s obvious that x^3 is more than x^2 when x is positive. So the equation can only be solved with x
Графічно. Побудуємо графіки y=x^2 та y=x^3+12. Вони мають одну точку перетину, очевидно аргумент відємний. Значить рівняння має один корінь. Підбираємо, х=-2. Все !!
By finding that (-2) is a real root we can use long division X^3-X^2+12/X+2 = X^2-3X+6 and then get the imaginary roots. Thank you sir for what you do. 14:2
Much appreciated. There' s also a simpler way to come up with the answer. If factoring out x^2 , then we have: x^2 (1-x)= 12 There would be just one possible choice left out of three positive pair factors of 12 that includes a perfect square. i.e. 4 and 3. So: x^2= 4 and (1-x)= 3 The only common answer of the above two equations is then: x=-2
Honestly, knowing that x has to be negative is a dead giveaway, cos it can't be positive (cos subtracting a cubed number (bigger) from the same but squared (smaller) number yet still getting a positive result => x has to be negative), and we also know it's not "1". Hmm what number comes next?
No entiendo porque tantos comentarios negativos. Es cierto que el problema es sencillo para ser de una olimpiada de matemáticas pero nunca dijo de qué nivel es... Además la solución es correcta y hay varias formas de llegar a ella. "Intuir" que -2 es una solución y luego factorizar es fácil pero, para mi, tiene más logro llegar a esa conclusión por una vía matemática. Y en las olimpiadas eso se valora.
@lopezpablo88, I think you deserve a standing and a clapping ovation from everyone @OnlinemathsTV as far this math challenge comments are concerned, hahahaha... I love it when people criticize and correct other constructively with a deep understanding of the point/s in question. Here, you have shown a deep level of mathematical prowess. Thanks a million for your wonderful contribution to the growth of this platform/channel. We all here love and salute your choice of words and wisdom in handling issues sir. Maximum respect and deep love to you from all of @OnlinemathsTV sir....❤️❤️💖💖💖💕💕😍😍😍.
The last 2 lines on the first column had errors, the last, serendipitously 'correcting' the second to last. In the second to last you forgot to make the 2^3 positive inside the parenthesis. In the last you forgot to make (from the erroneous equation) the x^3 negative. So erroring twice the 'corrected' the equation on the last line. So just erase the second to last equation and your good! All is well that ends well. You passed the test, making 2 bad operations!!!!
That depends on how you look at it. To go from the second to last to the last is wrong but it corrects the second to last wrong by being wrong. So if you remove the second to last, only then is the last right.@@proislam-co6pg
Nice. I would use a formal method to solve this problem: 1) observe -2 is a real root. 2) divide x^3-x^2+12 by x+2, the quotient then is Q(x)=x^2-3x+6, now it's quite easy to find other two complex roots of Q(x)=0.
Yes, my self, I'd always do that way because, in f(x)=0 where f(x) of power of 3 (let's denote it with f3(x)) must be in f1(x)*f2(x) to be solved. If the second f2(x) is factorized to f1(x)*f1(x), then 3 solutions all real right away. If not, f2(x) gives 2 solutions from the formula, either real or imaginary. That way we don't need x**3 +/- y**3 = (x +/- y)(x**2...) formula. The question is how fast you find the number for the f1(x) i.e. (x - ?). Typically the number is small, 1, 2, 3 or -1, -2, -3. Seldom odd case such as -7. I want to think of a systematic way to come up with this number, rather than applying try-and-hit way...
no, you dont need inspection. Descartes' Rule of Signs tells you that there IS one negative root. They can only be -1, -12, -2, -6, -4, -3. Try the edge cases first, that is -1 and -12. You see that it must be closer to -1. Try -2. That works. @@uthoshantm
@@uthoshantm In this case its pretty easy. x is clearly negative, and x^2 is a factor of 12, must be 4. If it had been, say 10 rather than 12, whole different kettle of fish.
@@dj_multiple_oneДа никак. Просто словами проговорил, что типа так неправильно. И в следующей строке написал, как надо. Но неверное представление не убрал.
When some terms were regrouped and put inside brackets the second line from the bottom left isn't correct but the line below is correct. It wasn't properly explained as to what was going on at that point. Someone who may be a bit weak in their math skills might get confused about that part and why the sign was changed from - to +.
... Good day sir, We could also solve the Complex part as follows: X^2 - 3X + 6 = 0 [ Applying Completing the Square ] ... (X - 3/2)^2 - 9/4 + 24/4 = 0 ... (X - 3/2)^2 = - 15/4 ... X - 3/2 = +/- SQRT(- 15/4) ... X - 3/2 = +/- SQRT((- 1)* 15/4) [ Applying i^2 = - 1 ] ... X - 3/2 = +/- SQRT(15/4 * i^2) ... X2.3 = 3/2 +/- SQRT(15) * i / 2 ... X2 = (3 + SQRT(15) * i) /2 v X3 = (3 - SQRT(15) * i) / 2 ... X2 and X3 are COMPLEX CONJUGATE SOLUTIONS, but are certainly NOT IMAGINARY SOLUTIONS, because in general Z = A + B * i is always COMPLEX, and when A = 0, then Z = B * i is both COMPLEX as IMAGINARY! In short : The set of Imaginary numbers (Z = B * i) is a SUBSET of the set of Complex numbers (Z = A + B * i ) ... great presentation by the way sir, and thanking you for your instructive math efforts ... best regards, Jan-W
По теореме о рациональных корнях уравнения можно сразу найти корень х=-2, затем поделить исходный многочлен на х+2, а дальше решить оставшееся квадратное уравнение. Стандартная школьная задача, что тут олимпиадного? :)
@@mnnkaz0 никак... Они все топят за "метод подбора", который на самом деле "метод пальца в небо". "Я угадал, потому что подошло" не равнозначно "я решил".
I'm a senior, ill in bed with flu and did this in less than a minute in my head - it's obvious from the equation that either x must be negative then you just need to fins a number that when it's square is added to it's cube yields 12 ( the fact it is negative means that minus the cube of it will become positive) is it not obvious that the number is 2? since 4 + 8 = 12. This is the third 'Olympiad' math question I've looked at this afternoon, being too unwell to do much else and all were easily do-able in my head rather than the long-winded solutions. I recently took a senior cognitive test (annual requirement at my age) and am amazed if this is how students are taught to solve these kind of puzlzles. Maybe I should start a "Granny shows you how" you tube series? 😆 Well I have learned something, I know I am not as quick-witted as I was 50 years ago, but I still have some of my marbles 😊
х^3-х^2-12=0 Челочисленные корни являются делителями 12. х=-2 является Схема Горнера (или деление многочлена на многочлен) и получаем квадраратное уравнение с D
Перенести все влево и исследовать функцию с помощью производной. Построить график и увидеть одну точку пересечения . Это -2. С помощью проверки убеждаемся , что -2 корень уравнения.
Some comments suggests, that x=-2 could be found within 10 seconds. And indeed, x=-2 is an obvious answer. But how to prove, that it is the only answer? We all know, that "x^2=4" has not only one answer. He broke the equation down to two: x+2=0 --> first solution 3x-6=x^2 --> potential alternative solution(s) Therefore the remaining question had been, if the second equation can be solved. Or not. And he listed correctly both complex conjugates --- even x2 and x3 are awkward. Well done.
@KaiUga-ni3hk you just pointed out a very salient point which I found difficult trying to make everyone to understand as far this math challenge is concerned. On behalf of OnlinemathsTV, I really want to say a very big thank you for your deep understanding of what actually prompt Onlinemaths TV to apply this approach that is being questioned by almost everyone here with a better understanding of things. You are good at what you do, maximum respect from everyone here for . Above all, we love you dearly and deeply sir....❤️❤️💖💖💕💕😍😍😍
you made a mistake. to jump from x^2-2^2-x^3-2^3=0 to (x^2-2^2)-(x^3-2^3)=0 is very wrong. it should have been (x^2-2^2)+(-x^3-2^3)=0 and then only you could have proceeded to (x^2-2^2)-(x^3+2^3)=0
yes, a mistake... the intention seems to be "we will group" but the parens are wrong. The way, say, SyberMath shows the intention to group is by underlining first.
With due respect Sir, Just by seeing the equation we can find the answer.. or hit and trial method can also be used.. Of course we will not be able to find complex answers mentally, but they are as it is rejected for the solution unless asked for in the exam. Thank you for sharing this video. Regards 🙏🙏
You must be Indian. Trial and error method is not real Mathematics. But you guys don't study or care about real Mathematics, nor appreciate it's beauty. To you, finding the answer faster than the other guy in an exam by any means possible is Maths. Higher score is victory. It is not methodical or exhaustive, it's just trick play. Like the difference between a cheap thrill seasonal action film and a timeless classic.
@@daakudaddy5453The method discussed in the video is also nit methodical, in the sense that it would only work for this cases. I guess the most reliable is the cubic formula here or numerical methods, everything else is tricks play. What's the point on insulting every student here when most of them are not even responsible for it? Besides, your criticism of students not being exposed to the beauty of math isn't only common in india but everywhere.
Согласен с @gorbachevaol. Имплементация алгоритма Горнера, подчеркнула бы структурную инвариантность, обеспечивает внедрение дифференциальных операторов в рамках алгебраического контекста. Так получаемые корни уравнения имеют больший математический смысл, по моему
J'ai eu de la chance de trouver plus rapidement ! On remarque que 12 = 3 x 4. Écrivons l'expression x**2 - x**3 = x**2 * (1 - x). Supposons xx*2 = 4 et (1 - x ) = 1 - (-2) , du coup on a bien x = -2 comme solution. (NB '**' signifie ici puissance d'un nombre et * multiplication)
Very beatiful explained and correctly solved problem. Because of the third degree term,there must be at least three roots for the equation. İf you draw the diagram of this equation in the analytical plane you will see the roots. I personally was not aware of the equation for root calculation and now learned it. Thank you very much for this informative video
@@AhmedsNjie-hg5du dude, it's an easy question, 4+8 us an intuitive solution so you immediately know that one root is -2, then you divide by (x+2) and you solve a quadratic. I don't know why the weird steps..
x² - x³ = 12 Let's see, a square and a cube that add/subtract to 12? Well 2 would work if it were a sum, not a difference: 4 + 8 = 12. But wait, we can make that happen just by changing the sign on the 2. x = -2 x² - x³ = (-2)² - (-2)³ = 4 + 8 = 12 Having found a root, the next step is to factor it out and solve the resulting quadratic. x³ - x² + 12 = (x + 2)(x² - 3x + 6) = 0 x = -2 or x² - 3x + 6 = 0 x = 3/2 ± ½i√15 So if you're looking for real solutions, x = -2 is the only one. Fred
Felicitaciones por el formidable desarrollo analitico. Sin embargo la solución (x=-2) adviene simplemente por tanteo una vez que te percatas de que DEBE ser un número negativo. Y el proceso mental para hallarla no lleva ni un minuto, sin necesidad de ser una especie de Ramanujan, ni tener un infinitesimo de su talento, por asi hablar
I think this way : 1. X must be negative, because if X positive, X^2 < X^3 => X^2 - X^3 never equals 12. 2. Negative but range ? X should be from 0 to -3 (I only consider integer), because -X^3 must < 12 itself to carry X^2. 3. Make a try with integer and easily find X = -2. Base on this to form the other equation to find another 2 X.
Let's solve the equation X2−X3=12X^2 - X^3 = 12X2−X3=12. Rearrange the equation to standard polynomial form: −X3+X2−12=0-X^3 + X^2 - 12 = 0−X3+X2−12=0 Multiply through by −1-1−1 to simplify: X3−X2+12=0X^3 - X^2 + 12 = 0X3−X2+12=0 This is a cubic equation, and we need to find the roots. Let's try to find the roots using the Rational Root Theorem, which suggests that any rational solution, in the form of pq\frac{p}{q}qp, is a factor of the constant term divided by a factor of the leading coefficient. Here, the constant term is 12 and the leading coefficient is 1, so the possible rational roots are the factors of 12: ±1,±2,±3,±4,±6,±12\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12±1,±2,±3,±4,±6,±12 We can test these possible roots by substitution to see if they satisfy the equation. Testing X=1X = 1X=1: 13−12+12=1−1+12=12≠01^3 - 1^2 + 12 = 1 - 1 + 12 = 12 eq 013−12+12=1−1+12=12=0 Testing X=−1X = -1X=−1: (−1)3−(−1)2+12=−1−1+12=10≠0(-1)^3 - (-1)^2 + 12 = -1 - 1 + 12 = 10 eq 0(−1)3−(−1)2+12=−1−1+12=10=0 Testing X=2X = 2X=2: 23−22+12=8−4+12=16≠02^3 - 2^2 + 12 = 8 - 4 + 12 = 16 eq 023−22+12=8−4+12=16=0 Testing X=−2X = -2X=−2: (−2)3−(−2)2+12=−8−4+12=0(-2)^3 - (-2)^2 + 12 = -8 - 4 + 12 = 0(−2)3−(−2)2+12=−8−4+12=0 So, X=−2X = -2X=−2 is a root. Now, we can factor (X+2)(X + 2)(X+2) out of the cubic polynomial: X3−X2+12=(X+2)(X2+aX+b)X^3 - X^2 + 12 = (X + 2)(X^2 + aX + b)X3−X2+12=(X+2)(X2+aX+b) To find aaa and bbb, we can perform polynomial division or use synthetic division. After factoring, we get: (X+2)(X2−3X+6)=0(X + 2)(X^2 - 3X + 6) = 0(X+2)(X2−3X+6)=0 Now we solve the quadratic equation X2−3X+6=0X^2 - 3X + 6 = 0X2−3X+6=0 using the quadratic formula: X=−b±b2−4ac2aX = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}X=2a−b±b2−4ac Here, a=1a = 1a=1, b=−3b = -3b=−3, and c=6c = 6c=6: X=3±(−3)2−4⋅1⋅62⋅1X = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}X=2⋅13±(−3)2−4⋅1⋅6 X=3±9−242X = \frac{3 \pm \sqrt{9 - 24}}{2}X=23±9−24 X=3±−152X = \frac{3 \pm \sqrt{-15}}{2}X=23±−15 X=3±i152X = \frac{3 \pm i\sqrt{15}}{2}X=23±i15 So, the solutions are: X=−2,X=3+i152,X=3−i152X = -2, \quad X = \frac{3 + i\sqrt{15}}{2}, \quad X = \frac{3 - i\sqrt{15}}{2}X=−2,X=23+i15,X=23−i15 4o
X = -2 calculated in my head! It is pretty obvious that the x³ can create an addition to the first term by raising a negative number to 3, thus getting an effective addition. Then it isn’t rocket science to figure out that (-2)² - (-2)³ = 4 - (-8) = 12 My two x:s
I solved this in 5 seconds in my mind with this simple logic: "If x is positive, the RHS should be negative, so x must negative so that x^3 is also negative, and the ( - ) in front of x^3 makes it positive again, so it adds to x^2." Then I just thought of simple numbers, and 2, seemed to be a good |x| value, and so, the answer with the logic is: -2.
@@onlineMathsTV Thanks 😀. My logic works, but your method is really useful if the answer was not as simple. Because it was just -2, I could guess, but if it was a more difficult value, your method is very useful.
*Actually it was more simple than you did* Sir, we can also solve it by factorising 12 to 4×3 Also factorising x²-x³ into x²(x-1) Now,x²(1-x)=4×3 =>x-1=3 =>-x=2 so, x=-2 And,(-2)²=4 tooo But, still there was soo much to learn in this video thanks for that 🙏🙏
Графически и ананалитически задача быстрее решается. - 2 сразу видно решение, простым подбором. Не в комплексных числах, конечно, там ещё могут быть решения
X^2 - X^3 = 12 X^2(1-x) = 2^2*(1-(-2)) , or X^2(1-x) =(-2)^2*(1-(-2)) If we compare left and right sides of equations, It is obvious, that true version of equation is X^2(1-x) = (-2)^2*(1-(-2)) That is x = -2
You prolonged the problem. You could have factored the problem out earlier without going through all those additional steps. But I understand you were trying to show the entire process of thought. Good job.
These type of problems in exams, usually have a simple solution that by simple looking can be obtained (for example (-2) is obviously a solution of the problem). Then you can divide the equation by x+2 and easily solve the obtained second order equation for other two roots.
This is brilliant. However, at 3:16, when we introduce the second bracket. the sign should have changed to +. We did not need to wait until the next step..
Уважаемый автор канала! Задание очень простое, решение методом группировки. Ответ минус два. Ранее было размещено уравнение- перевёртыш, тоже самое, но со знаком плюс. Это простейшая задача, она не олимпиадная, можно решить методом устного счёта.
X is -2 X^2(1-x) =12 Since rhs is +tive meaning lhs is also positive Hence 1-x is positve which means x is a negative number Now by put -1, then -2 we get our answer
Trial and error method much faster than any methods in this case since you can tell x is a small negative number intuitively. So plug in x = -1,-2,-3. Therefore, x = (-2) is the solution
Let P(x)= x^3-x^2+12. One can easily verify that x=-2 is a solution of the equation, so (x+2) is a factor of p(x). Use long division we see that p(x)=(x+2)(x^2-3x+6) ....
” Ｘ² ーＸ³ ＝１２”
１st left → Ｘ² ーＸ³ ＝ Ｘ²（１ーＸ）
２nd right → １２＝４×３＝ ４×（１＋２) ）＝ ( -2) ² ×（１ー ( -2) ）
３rd X＝ -2
The equation third degree have three roots (-2) not enough??
So he gets another two roots, Imaginary roots.😊
Well done 👏
＞@@dajo3032
Thank you so much.
I solved it the same as you
I have an another easy way to solve it. And i prove it easyly
It's pretty clear right from the start that x has to be negative, otherwise the right term would be
what about the complex solutions
@@jagzey Whatever complex number you put into x, it will never be equal to 12.
You can find it up to a certain number by trial and error. But what do you do when asked for a larger number?
@@Rx800.0 Sure, most of such equations needs to be calculated properly, but knowing how such functions progress helps you with a good prediction where the solution should be. It will be tricky if the result is a complex value.
Ми не домовлялися, а я вирішив таким же способом за 20с
Excellent math question and smart expanation. Muchas gracias
it's easily proofed, that one obvious solution is x= - 2
alternative solution 1:
divide x³ -- x² + 12 by x+2 with the so called long division (polynomial division) to find the polynomial of degree 2. solve this quadratic equation to receive two complex solution if needed.
alternativ 2:
x³ -- x² + 12 = ( x +2)(x² + ax + 6). find the parameter a by multplication, and comparing both sides.
How nostalgic for school! I still remember the amazement when I realized that I was coming to the solution and the joy when I found it. It was just an equation in the end, but for me it was like conquering the world
WOW! I LIKED THIS VIDEO...GOD BLESS YOU...
If the goal of the olympiad is to solve the problem quickly, I would apply numerical theory first. It’s obvious that x^3 is more than x^2 when x is positive. So the equation can only be solved with x
IF 0
X = -2
X² * ( 1 - X¹ ) = 12 ( para dar esse valor tem que ser 4 * 3 = 12
Então, (-2)² - (-2)³ = 12
+4 + 8 = 12
Bingo from Brazil!!!!
인도늠들이 수학을 잘한다고 들었는데 참말이네요..그런데 그 지경으로 살고 있는게 이상하네요
Wrong
exactly, thats what i wanted to write here - Noice
Графічно. Побудуємо графіки y=x^2 та y=x^3+12. Вони мають одну точку перетину, очевидно аргумент відємний. Значить рівняння має один корінь. Підбираємо, х=-2. Все !!
فقط القليل من التخمين بوضع 12=4×3
bommmbaaaa!!!! que fantástico, muy bien profesor. su didáctica es muy clara y enseñadora. muchas gracias
By finding that (-2) is a real root we can use long division X^3-X^2+12/X+2 = X^2-3X+6 and then get the imaginary roots. Thank you sir for what you do. 14:2
Much appreciated. There' s also a simpler way to come up with the answer. If factoring out x^2 , then we have:
x^2 (1-x)= 12
There would be just one possible choice left out of three positive pair factors of 12 that includes a perfect square. i.e. 4 and 3. So:
x^2= 4 and (1-x)= 3
The only common answer of the above two equations is then: x=-2
Bravo!!! you are good at it.
Thank you sir 🙏
This is just a guess and check method. The preferred way to solve any polynomial equation is to solve for 0 first.
Honestly, knowing that x has to be negative is a dead giveaway, cos it can't be positive (cos subtracting a cubed number (bigger) from the same but squared (smaller) number yet still getting a positive result => x has to be negative), and we also know it's not "1".
Hmm what number comes next?
@@randymills2660Neither is that a guess and check not is solving for 0 is the preferred way
No entiendo porque tantos comentarios negativos. Es cierto que el problema es sencillo para ser de una olimpiada de matemáticas pero nunca dijo de qué nivel es... Además la solución es correcta y hay varias formas de llegar a ella. "Intuir" que -2 es una solución y luego factorizar es fácil pero, para mi, tiene más logro llegar a esa conclusión por una vía matemática. Y en las olimpiadas eso se valora.
@lopezpablo88, I think you deserve a standing and a clapping ovation from everyone @OnlinemathsTV as far this math challenge comments are concerned, hahahaha...
I love it when people criticize and correct other constructively with a deep understanding of the point/s in question.
Here, you have shown a deep level of mathematical prowess.
Thanks a million for your wonderful contribution to the growth of this platform/channel.
We all here love and salute your choice of words and wisdom in handling issues sir.
Maximum respect and deep love to you from all of @OnlinemathsTV sir....❤️❤️💖💖💖💕💕😍😍😍.
Explained in very good way. Thank you so much!
The last 2 lines on the first column had errors, the last, serendipitously 'correcting' the second to last. In the second to last you forgot to make the 2^3 positive inside the parenthesis. In the last you forgot to make (from the erroneous equation) the x^3 negative. So erroring twice the 'corrected' the equation on the last line. So just erase the second to last equation and your good! All is well that ends well. You passed the test, making 2 bad operations!!!!
Yeah ! I was wondering how x^3-2^3=x^3+2^3 !
only the second last is wrong
That depends on how you look at it. To go from the second to last to the last is wrong but it corrects the second to last wrong by being wrong. So if you remove the second to last, only then is the last right.@@proislam-co6pg
luckly i decided to read the comments before judjing my retirred maths abilities
Cuando abris paréntesis por primer vez, 2 al cubo pasa a positivo.
Nice. I would use a formal method to solve this problem: 1) observe -2 is a real root. 2) divide x^3-x^2+12 by x+2, the quotient then is Q(x)=x^2-3x+6, now it's quite easy to find other two complex roots of Q(x)=0.
that quotient method is really usefull solving high power equations.
Yes, my self, I'd always do that way because, in f(x)=0 where f(x) of power of 3 (let's denote it with f3(x)) must be in f1(x)*f2(x) to be solved.
If the second f2(x) is factorized to f1(x)*f1(x), then 3 solutions all real right away. If not, f2(x) gives 2 solutions from the formula, either real or imaginary.
That way we don't need x**3 +/- y**3 = (x +/- y)(x**2...) formula. The question is how fast you find the number for the f1(x) i.e. (x - ?). Typically the number
is small, 1, 2, 3 or -1, -2, -3. Seldom odd case such as -7. I want to think of a systematic way to come up with this number, rather than applying try-and-hit way...
Well, that's assuming you can pick up the -2 by inspection.
no, you dont need inspection. Descartes' Rule of Signs tells you that there IS one negative root.
They can only be -1, -12, -2, -6, -4, -3.
Try the edge cases first, that is -1 and -12.
You see that it must be closer to -1.
Try -2. That works.
@@uthoshantm
@@uthoshantm In this case its pretty easy. x is clearly negative, and x^2 is a factor of 12, must be 4.
If it had been, say 10 rather than 12, whole different kettle of fish.
Отлично так минус на плюс заменил, бро! 👍🏻
я тоже не понял как он так
@@dj_multiple_oneДа никак. Просто словами проговорил, что типа так неправильно. И в следующей строке написал, как надо. Но неверное представление не убрал.
@@broomska1 Вот то, что не убрал - это большущий косяк! Вероятно у них в колхозе так учат.
I love your method of teaching. Thanks
When some terms were regrouped and put inside brackets the second line from the bottom left isn't correct but the line below is correct. It wasn't properly explained as to what was going on at that point. Someone who may be a bit weak in their math skills might get confused about that part and why the sign was changed from - to +.
YES,my brother,he has bigggggggggggg fault
and his solution is completly fault
Muito bom, parabéns. Continue postando vídeos. 👏👏👏
... Good day sir, We could also solve the Complex part as follows: X^2 - 3X + 6 = 0 [ Applying Completing the Square ] ... (X - 3/2)^2 - 9/4 + 24/4 = 0 ... (X - 3/2)^2 = - 15/4 ... X - 3/2 = +/- SQRT(- 15/4) ... X - 3/2 = +/- SQRT((- 1)* 15/4) [ Applying i^2 = - 1 ] ... X - 3/2 = +/- SQRT(15/4 * i^2) ... X2.3 = 3/2 +/- SQRT(15) * i / 2 ... X2 = (3 + SQRT(15) * i) /2 v X3 = (3 - SQRT(15) * i) / 2 ... X2 and X3 are COMPLEX CONJUGATE SOLUTIONS, but are certainly NOT IMAGINARY SOLUTIONS, because in general Z = A + B * i is always COMPLEX, and when A = 0, then Z = B * i is both COMPLEX as IMAGINARY! In short : The set of Imaginary numbers (Z = B * i) is a SUBSET of the set of Complex numbers (Z = A + B * i ) ... great presentation by the way sir, and thanking you for your instructive math efforts ... best regards, Jan-W
many hate comments here. I wanna appreciate that you explain it and i solved it very well.
По теореме о рациональных корнях уравнения можно сразу найти корень х=-2, затем поделить исходный многочлен на х+2, а дальше решить оставшееся квадратное уравнение. Стандартная школьная задача, что тут олимпиадного? :)
But division by zero is undefined?😮
I love your teaching skills
Задача решается за 5 секунд в уме... Вот бы у меня в школьные времена были такие простые задачи))
здравствуйте, а как можно в уме быстро решить? объясните, пожалуйста!
@@mnnkaz0обратить внимание на 4 + 8 = 12 и на минус в выражении, и просто понять что х = -2.
C'est bien expliqué mais il fallait préciser l'ensemble dans lequel on travaille dans la question.
Да, за 5 секунд, если вы до этого решали схожие примеры....
@@mnnkaz0 никак... Они все топят за "метод подбора", который на самом деле "метод пальца в небо". "Я угадал, потому что подошло" не равнозначно "я решил".
I'm a senior, ill in bed with flu and did this in less than a minute in my head - it's obvious from the equation that either x must be negative then you just need to fins a number that when it's square is added to it's cube yields 12 ( the fact it is negative means that minus the cube of it will become positive) is it not obvious that the number is 2? since 4 + 8 = 12. This is the third 'Olympiad' math question I've looked at this afternoon, being too unwell to do much else and all were easily do-able in my head rather than the long-winded solutions. I recently took a senior cognitive test (annual requirement at my age) and am amazed if this is how students are taught to solve these kind of puzlzles. Maybe I should start a "Granny shows you how" you tube series? 😆 Well I have learned something, I know I am not as quick-witted as I was 50 years ago, but I still have some of my marbles 😊
х^3-х^2-12=0
Челочисленные корни являются делителями 12.
х=-2 является
Схема Горнера (или деление многочлена на многочлен) и получаем квадраратное уравнение с D
What!!! Write normal letters or shut up and stay in Soviet union!!!
@@hannukoistinen5329 retarded yenkee lmao
@@hannukoistinen5329 I want to stay in Soviet Union)
@@hannukoistinen5329А причем тут Советский союз?
@@hannukoistinen5329junge, soviet Union ist schon lange vorbei. In welchem Jahr wohnst du?
Great solution, Dear
wow. that was really cool! thank you for your videos
Glad you like it, you welcome
Great explanation
well done sir , I learnt a lot
फेक्टर मेथडं है थोड़ा लम्बा है 12=-4-8 कर घात बनाकर हल करने से नया फन्डा समझ मे आया। Thankyou.
Перенести все влево и исследовать функцию с помощью производной. Построить график и увидеть одну точку пересечения . Это -2. С помощью проверки убеждаемся , что -2 корень уравнения.
Ingenioso. Y si, hay varios caminos válidos 😃
То чувство, когда решил в уме😁
Так видно же что x отрицательный, там подбором решается за секунду
@@neokripte да. Но ещё нужно доказать, что других корней нет.
interesting wow it is correct
My solution take only 30 second 😂. I feel very intelligent 😂😂😂 thank you for primary school level olimpic questions 😅😅
😂😂😂
Is fthis really for Primary school level? I hope so as I was quite concerned at the fall in scholastic standards
@@chrissyday67 this question is easy to answer but you are right. Unfortunately, Every year student standarts are falling down more...
12 = 8 + 4 = 2^3 + 2^2
-x^3 + x^2 = 2^3 + 2^2
(-x)^3 + (-x)^2 = 2^3 + 2^2
-x = 2 => x = -2 - first solution
Further it's a simple quadratic equation, i.e. mathematical craftsmanship...
Very good exercise, thank you.
Some comments suggests, that x=-2 could be found within 10 seconds. And indeed, x=-2 is an obvious answer. But how to prove, that it is the only answer?
We all know, that "x^2=4" has not only one answer.
He broke the equation down to two:
x+2=0 --> first solution
3x-6=x^2 --> potential alternative solution(s)
Therefore the remaining question had been, if the second equation can be solved. Or not. And he listed correctly both complex conjugates --- even x2 and x3 are awkward.
Well done.
@KaiUga-ni3hk you just pointed out a very salient point which I found difficult trying to make everyone to understand as far this math challenge is concerned.
On behalf of OnlinemathsTV, I really want to say a very big thank you for your deep understanding of what actually prompt Onlinemaths TV to apply this approach that is being questioned by almost everyone here with a better understanding of things.
You are good at what you do, maximum respect from everyone here for .
Above all, we love you dearly and deeply sir....❤️❤️💖💖💕💕😍😍😍
2乗の数から３乗数を引いてプラスになる数はマイナスの数であると解る。-１では12ににならない。-２で即+４と-８とで１２と解る。
Thanks for your efforts
Thank you sir
x^2-x^3=13
x^3-x^2+12=0
factors of 12:±2 , ±3 , ±1
-2 satisfy the equation -> x+2 factor
(x+2)(x^2-3x+6)=0
x=-2 , (3/2)±sqrt(15)i/2
Excelente
Very good explanation. Thank you.
Glad it was helpful!
you made a mistake. to jump from x^2-2^2-x^3-2^3=0 to (x^2-2^2)-(x^3-2^3)=0 is very wrong.
it should have been (x^2-2^2)+(-x^3-2^3)=0
and then only you could have proceeded to (x^2-2^2)-(x^3+2^3)=0
yes, a mistake... the intention seems to be "we will group" but the parens are wrong. The way, say, SyberMath shows the intention to group is by underlining first.
There was a mistake but he recovered it
i noticed it too.
With due respect Sir,
Just by seeing the equation we can find the answer.. or hit and trial method can also be used..
Of course we will not be able to find complex answers mentally, but they are as it is rejected for the solution unless asked for in the exam.
Thank you for sharing this video.
Regards 🙏🙏
You must be Indian.
Trial and error method is not real Mathematics.
But you guys don't study or care about real Mathematics, nor appreciate it's beauty. To you, finding the answer faster than the other guy in an exam by any means possible is Maths. Higher score is victory. It is not methodical or exhaustive, it's just trick play. Like the difference between a cheap thrill seasonal action film and a timeless classic.
@@daakudaddy5453
Do something for your frustration, I wish you peace
@@daakudaddy5453The method discussed in the video is also nit methodical, in the sense that it would only work for this cases.
I guess the most reliable is the cubic formula here or numerical methods, everything else is tricks play.
What's the point on insulting every student here when most of them are not even responsible for it? Besides, your criticism of students not being exposed to the beauty of math isn't only common in india but everywhere.
I love it. Brilliant ❤
X^2-X^3=12 X^2(1-X)=4×3
X는 음수가 되어야 함으로
X=-2 이렇게 간단한 문제를
너무 어렵게 푸네요.
나랑 푼 방식이 같네. 이렇게 하면 간단하게 암산되는데 ㅋ
Согласен с @gorbachevaol. Имплементация алгоритма Горнера, подчеркнула бы структурную инвариантность, обеспечивает внедрение дифференциальных операторов в рамках алгебраического контекста. Так получаемые корни уравнения имеют больший математический смысл, по моему
J'ai eu de la chance de trouver plus rapidement ! On remarque que 12 = 3 x 4. Écrivons l'expression x**2 - x**3 = x**2 * (1 - x). Supposons xx*2 = 4 et (1 - x ) = 1 - (-2) , du coup on a bien x = -2 comme solution. (NB '**' signifie ici puissance d'un nombre et * multiplication)
FANTASTIC😍😍😍😍😍😍😍😍😍😍
this guy did 15 min video just to proof that x equals -2 when everyone guessed it in half a minute
true legend.
他是在教思考的““方式””
所以得用簡單的數字帶您思考
如果今天的常數從原本的12改成64160000或是更多大的數字那就很難30秒解答出來
@@nelsoneason5822 fair point. knowledge of the algorithm is always the most powerful weapon
@@nonsencephilosophy wow
you are a nice guy+9999999
Thank you, You are an excellent!
3:42 я не понял, как он превратил х^3-2^3 в х^3+2^3
он ошибся и исправился в следующей строчке.
Very beatiful explained and correctly solved problem. Because of the third degree term,there must be at least three roots for the equation. İf you draw the diagram of this equation in the analytical plane you will see the roots. I personally was not aware of the equation for root calculation and now learned it. Thank you very much for this informative video
Plus wolfram alfa gives the exact same roots for the equation, for ones information,who does not belive the solution.
You made silly error with the sign of 2 to the power of 3 (line 6 of the solution). Then you made another error in line 7. This is ridiculous ...
Plus the question is extremely simple for an Olympiad
Revise su video antes de mostrarlo....
Good catch, but his 2 wrongs turned out to be right
Bro there's nothing wrong in those steps.. Every step is just awesome
@@AhmedsNjie-hg5du dude, it's an easy question, 4+8 us an intuitive solution so you immediately know that one root is -2, then you divide by (x+2) and you solve a quadratic. I don't know why the weird steps..
x² - x³ = 12
Let's see, a square and a cube that add/subtract to 12? Well 2 would work if it were a sum, not a difference: 4 + 8 = 12.
But wait, we can make that happen just by changing the sign on the 2.
x = -2
x² - x³ = (-2)² - (-2)³ = 4 + 8 = 12
Having found a root, the next step is to factor it out and solve the resulting quadratic.
x³ - x² + 12 = (x + 2)(x² - 3x + 6) = 0
x = -2 or x² - 3x + 6 = 0
x = 3/2 ± ½i√15
So if you're looking for real solutions, x = -2 is the only one.
Fred
You made mistake. When you facror out -1, you should have -(x*3+2*2) not (x*3-2*2)
No need to use a long pocedure to solve that the value of x is -2. Just use simple arithmatic to get the correct answer
I was looking for a comment like yours, I couldn't be the only one who saw this error.
Yup there was an error on factorisation.
Felicitaciones por el formidable desarrollo analitico. Sin embargo la solución (x=-2) adviene simplemente por tanteo una vez que te percatas de que DEBE ser un número negativo. Y el proceso mental para hallarla no lleva ni un minuto, sin necesidad de ser una especie de Ramanujan, ni tener un infinitesimo de su talento, por asi hablar
I solved this in 5 seconds thus: 4 + 8 = 12, 4 - (-8) = 12, x = -2.
(If only real roots are wanted, long division shows there are no other)
x=-2; (3-V15*i)/2; (3+V15*i)/2. Найдём корень -2 по схеме Горнера и придём к уравнению x^2 + 4x +12 =0.
I like your explication 💭❤️
I think this way :
1. X must be negative, because if X positive, X^2 < X^3 => X^2 - X^3 never equals 12.
2. Negative but range ? X should be from 0 to -3 (I only consider integer), because -X^3 must < 12 itself to carry X^2.
3. Make a try with integer and easily find X = -2. Base on this to form the other equation to find another 2 X.
Посчитал в уме ответ "-2". Но мне повезло. что "12" - маленькое число))
Wow! Factor: x2(1-x)=12. x2 must be 4, and 1-x must be 3. If x=-2, it works!
This awesome, u just open my understanding to solving this using this unique method. Thanks for this video sir.
Glad to hear that
x^3-x^2+12=0
因数分解すると、
(x+2)(x^2-3x+6)=0
x^2-3x+6=(x-3/2)^2+15/4≠0
∴x=-2
Let's solve the equation X2−X3=12X^2 - X^3 = 12X2−X3=12.
Rearrange the equation to standard polynomial form:
−X3+X2−12=0-X^3 + X^2 - 12 = 0−X3+X2−12=0
Multiply through by −1-1−1 to simplify:
X3−X2+12=0X^3 - X^2 + 12 = 0X3−X2+12=0
This is a cubic equation, and we need to find the roots. Let's try to find the roots using the Rational Root Theorem, which suggests that any rational solution, in the form of pq\frac{p}{q}qp, is a factor of the constant term divided by a factor of the leading coefficient.
Here, the constant term is 12 and the leading coefficient is 1, so the possible rational roots are the factors of 12:
±1,±2,±3,±4,±6,±12\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12±1,±2,±3,±4,±6,±12
We can test these possible roots by substitution to see if they satisfy the equation.
Testing X=1X = 1X=1:
13−12+12=1−1+12=12≠01^3 - 1^2 + 12 = 1 - 1 + 12 = 12
eq 013−12+12=1−1+12=12=0
Testing X=−1X = -1X=−1:
(−1)3−(−1)2+12=−1−1+12=10≠0(-1)^3 - (-1)^2 + 12 = -1 - 1 + 12 = 10
eq 0(−1)3−(−1)2+12=−1−1+12=10=0
Testing X=2X = 2X=2:
23−22+12=8−4+12=16≠02^3 - 2^2 + 12 = 8 - 4 + 12 = 16
eq 023−22+12=8−4+12=16=0
Testing X=−2X = -2X=−2:
(−2)3−(−2)2+12=−8−4+12=0(-2)^3 - (-2)^2 + 12 = -8 - 4 + 12 = 0(−2)3−(−2)2+12=−8−4+12=0
So, X=−2X = -2X=−2 is a root.
Now, we can factor (X+2)(X + 2)(X+2) out of the cubic polynomial:
X3−X2+12=(X+2)(X2+aX+b)X^3 - X^2 + 12 = (X + 2)(X^2 + aX + b)X3−X2+12=(X+2)(X2+aX+b)
To find aaa and bbb, we can perform polynomial division or use synthetic division. After factoring, we get:
(X+2)(X2−3X+6)=0(X + 2)(X^2 - 3X + 6) = 0(X+2)(X2−3X+6)=0
Now we solve the quadratic equation X2−3X+6=0X^2 - 3X + 6 = 0X2−3X+6=0 using the quadratic formula:
X=−b±b2−4ac2aX = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}X=2a−b±b2−4ac
Here, a=1a = 1a=1, b=−3b = -3b=−3, and c=6c = 6c=6:
X=3±(−3)2−4⋅1⋅62⋅1X = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}X=2⋅13±(−3)2−4⋅1⋅6 X=3±9−242X = \frac{3 \pm \sqrt{9 - 24}}{2}X=23±9−24 X=3±−152X = \frac{3 \pm \sqrt{-15}}{2}X=23±−15 X=3±i152X = \frac{3 \pm i\sqrt{15}}{2}X=23±i15
So, the solutions are:
X=−2,X=3+i152,X=3−i152X = -2, \quad X = \frac{3 + i\sqrt{15}}{2}, \quad X = \frac{3 - i\sqrt{15}}{2}X=−2,X=23+i15,X=23−i15
4o
Nice job
I like this topic!
X = -2 calculated in my head!
It is pretty obvious that the x³
can create an addition to the first term by raising a negative number to 3, thus getting an effective addition.
Then it isn’t rocket science to figure out that (-2)² - (-2)³ = 4
- (-8) = 12
My two x:s
I solved this in 5 seconds in my mind with this simple logic:
"If x is positive, the RHS should be negative, so x must negative so that x^3 is also negative, and the ( - ) in front of x^3 makes it positive again, so it adds to x^2." Then I just thought of simple numbers, and 2, seemed to be a good |x| value, and so, the answer with the logic is: -2.
You the best, sir.
@@onlineMathsTV Thanks 😀. My logic works, but your method is really useful if the answer was not as simple. Because it was just -2, I could guess, but if it was a more difficult value, your method is very useful.
*Actually it was more simple than you did*
Sir, we can also solve it by factorising 12 to 4×3
Also factorising x²-x³ into x²(x-1)
Now,x²(1-x)=4×3
=>x-1=3
=>-x=2 so, x=-2
And,(-2)²=4 tooo
But, still there was soo much to learn in this video thanks for that 🙏🙏
I solve it in only 7 steps in One minute ❤❤❤
Really
Графически и ананалитически задача быстрее решается. - 2 сразу видно решение, простым подбором. Не в комплексных числах, конечно, там ещё могут быть решения
X^2(1-x)=4*3 so x^ 2= 4and 1-x=3 so x=-2 by substituting x=-2 the result will come out
X^2 - X^3 = 12
X^2(1-x) = 2^2*(1-(-2)) , or X^2(1-x) =(-2)^2*(1-(-2))
If we compare left and right sides of equations, It is obvious, that
true version of equation is X^2(1-x) = (-2)^2*(1-(-2))
That is x = -2
HOW CAN YOU CHANGE SIGN FROM (-) TO (+) ON THE SEVENTH LINE FROM THE SIXTH LINE WITHOUT OPENING THE FIRST BRACKET (X-3...........) ?
x² - x³ ＝ 12
x²(1 - x) = 12
x²(1-x) = 4*3
x²(1-x) = 2²*3
i) x² = 2²일 때, x = 2, -2
ii) 1-x = 3일 때, x = -2
따라서 x=-2
Thanks!
You prolonged the problem. You could have factored the problem out earlier without going through all those additional steps. But I understand you were trying to show the entire process of thought. Good job.
At 3:06, the second sign in the second bracket is wrong. It should be plus sign.
X^2(1-X)=4*3
X^2 et 1-X sont premiers entre eux ce qui donne, d'après Gauss, X^2=4 et 1-X=3
Donc X=-2, Puis on factorise par X+2
These type of problems in exams, usually have a simple solution that by simple looking can be obtained (for example (-2) is obviously a solution of the problem). Then you can divide the equation by x+2 and easily solve the obtained second order equation for other two roots.
Excelente video. Nuevo suscriptor a tu canal. like gran video
Preferably subtract 12 from both sides ( addition property of equality).
𝑥^2−𝑥^3=12
𝑥^2 (1−𝑥)=12
Factorization of 12 = 2x2x3
=(-2)x(-2)x3
(−2)^2 [1−(−2)]=12
Hence x= -2
x=0の場合とx=1の場合は0
1
Можно 12 записать как 3 умножить на 4. А в левой части х в квадрате вынести за скобки и решить уравнение х в квадрате равно 4, а 1-х = 3, отсюда х =-2
You the best sir.
Bravo!!!
I like you explanation,it is fantastic
Thanks sir
This is brilliant. However, at 3:16, when we introduce the second bracket. the sign should have changed to +. We did not need to wait until the next step..
Noted sir. Thanks for the observation sir.
Молодец, услышал, исправил!
I put X2 = Y so
I resolve Y2 - Y + 12
Y1 = -3 REJECTED
Y2 =[ -/+] 2
So the solution is -2
Уважаемый автор канала!
Задание очень простое, решение методом группировки.
Ответ минус два.
Ранее было размещено уравнение- перевёртыш, тоже самое, но со знаком плюс.
Это простейшая задача, она не олимпиадная, можно решить методом устного счёта.
(x+2)(3x-x^2+6)=0
x+2=0 or 3x-x^2+6=0
-> 3x-x^2+6=0 -> (x-3/2)^2+15/4 >0 -> x=-2
X is -2
X^2(1-x) =12
Since rhs is +tive meaning lhs is also positive
Hence 1-x is positve which means x is a negative number
Now by put -1, then -2 we get our answer
Trial and error method much faster than any methods in this case since you can tell x is a small negative number intuitively. So plug in x = -1,-2,-3. Therefore, x = (-2) is the solution
Thats what I did too. Figured it out in about 20 seconds. LOL
Bravo!!!
You guys are good at what you do and I love you both for that.
Respect sir. 💪💪💪👍👍
Let P(x)= x^3-x^2+12. One can easily verify that x=-2 is a solution of the equation, so (x+2) is a factor of p(x). Use long division we see that p(x)=(x+2)(x^2-3x+6) ....
Just cut ³ to ² using kuadratik.... that ez ,so that will be x²-x+12= 0 or just use calcul and you will get the answer
Wuau: "Nice EXERCISE" Lo resuelvo antes de ver el video: X²-X³=12=4+8 --> X²-X³=2²+2³ --> X²-X²=X³+2³