Nice Olympiad Math | x^2–x^3=12 | Nice Math Olympiad Solution

This mathematical challenge is an Olympiad question. In solving this Olympiad polynomial equation I will guide you on some unique steps and techniques to apply.
I will introduce to you two special algebraic identities that will give a short cut to solving this in a short period of time.

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[todaaki]

” 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

[LarsEllerhorst]

It's pretty clear right from the start that x has to be negative, otherwise the right term would be <1. I needed no calculation to find the result x=-2.

[JAMESYUN-et]

Excellent math question and smart expanation. Muchas gracias

[markdagley]

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!!!!

[bobwineland]

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

[bencheng]

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.

[maurotivolesi]

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

[luisalbertogonzalez]

WOW! I LIKED THIS VIDEO...GOD BLESS YOU...

[JPTaquari]

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!!!!

[ichdu]

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.

[leylaaghazadeh]

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

[pasixty]

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<0. Going further, -1 can’t be the solution, for 1-1=0. Next step is -2. Bingo 4-(-8) in other words 4+8 equals 12. That’s all. You can call it an educated guess, but there’s a simple way to do it in „real math“ for sure. Your solution is respectable (even though you did a sign error in it, what I account for you being nervous diving into a complex solution) but it is truly a bit too complicated. All the other commenters: find the sign error in this comment 😊

[jan-willemreens]

... 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

[fernandomembrilacortez]

bommmbaaaa!!!! que fantástico, muy bien profesor. su didáctica es muy clara y enseñadora. muchas gracias

[vyacheslav-yarmak]

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 !!

[hgilbert]

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

[kevincozens]

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 +.

[lopezpablo]

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.

[jovankovacevic]

Explained in very good way. Thank you so much!

[volasvolas]

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? :)