Germany | Can You Solve This ? | A Nice Math Olympiad Problem | Best Trick

In this Math Olympiad problem, we tackle the equation and find the value of with a clear, step-by-step solution. Perfect for math enthusiasts looking for a challenge!
#MathOlympiad #MathChallenge #CubicEquation #Algebra #ProblemSolving #STEM #MathLovers

[@diegoyotta]

a^2(1-a)=150. a^2(1-a)=[prime factors of 150:2, 3, 5, 6, 10, 15, 25, 30, 50, 75] a^2(1-a)=25*6=150 a^2=25 and (1-a)=6 thus a^2=25 -> a=[5, -5] (1-a)=6 -> a=-5

[@Leonhard-Euler]

1) for a > 1, the equations does not hold 2) for 0

[@LoveTonsure]

a²(1-a)=5*5*6. So I expected that 1-a=6 and a²=25, then found a=-5. The rest is just a simple factorization.

[@driterxholi6570]

The simplest solution.

[@mmagic5753]

well done

[@gatchangatchan5593]

The 2 you write is similar to 3. I tend to look at the screen without listening to the sound, I get confused between 2 and 3. The horizontal bar of 2 should be written straight.

[@Abdullah-wolf-syria]

a²-a³=150 a²(1-a)=150. ,will trying (±1,±2,±3,±4,±5) (±1,±2,±3,±4, and +5) rejected (-5)²× (1 - (-5)) = 25(1+5)=25×6=150 ( a = -5 )

[@ΒασιληςΑρετακης-μ2ψ]

Расписать 150=125+25.перенести в левую часть и разложить на множители сумму кубов и разность квадратов,а дальше понятно.

[@yvindaamodt4445]

there is way easier don't put it to wen u work with ^2 ^3 add minus equal number both sides that works with ^2 so easy make it to (a^2- a^2)+0 = (150)+0 and then change 0 with - 125 both sides to (a^2-a^3)-125 = 150 -125 (to show have split into 25 and 125) Is same if u have x^2-6x+5 =21 u can +4 both side to make it x^2-6x+9 =25 so make it to (x-3)^2 =25 so to attack this see if a number plus or minus both side make it easier to squire it:):)

[@chine9778]

Repeated use of input errors reach convergence

[@praporserg]

Так как а^3 всегда больше, чем а^2,очевидно,что ответ будет отрицательным числом, так как а^3 тогда даст плюс. А далее простым подбором получаем, что это не -1,-2,-3 и -4, а - 5. Решение в уме моментальное.

[@НинаИнсарова-э2м]

Я также решала

[@nenaud]

writing correctly your signs, your letters and your numbers will ease the understanding of your demonstration thanks for this help anyway

[@اقرء]

Tanks for your video

[@ressouguerrier6043]

a²-a³=150; solution; a²-a³-150=0; a²-b³-125-25=0; a²-a³-5³-5²=0; a²-5²-a³-5³=0; (a-5)(a+5)-[(a-5)(a²+5a+25)]=0; (a-5)[a+5-(a²+5a+25)]=0; (a-5)[a+5-a²-5a-25]=0; a-5=0; a=5; -a²-4a-20=0;(-1); a²+4a+20=0; S={5}.

[@LeukMACLEDIO]

Thanks you very much

[@BSrour-lv1co]

Why bother finding the complex roots exact values if they are to be discarded? A lot of wasted time. In addition, it was clear from the get-go that the solution is negative because otherwise the cube would be greater than the square and hence the difference would be negative!

[@satrajitghosh8162]

a^3 - a^2 + 150 = 0 a^3 - a^2 - ( -5) ^3 + ( -5) ^2 = 0 ( a + 5)( a ^2 - 5 a + 25) - ( a + 5)( a - 5) = 0 ( a + 5)( a^2 - 5 a + 25 - a + 5) = 0 ( a + 5)( a ^2 - 6 a + 30) = 0 Hereby a = -5, 3 + i √ (21), 3 - i √(21)

[@MukontwaJay2pin]

In Zambian syllabus we don't have such... 😂😂

[@AnhNguyen-hz1rw]

Same here just substitute. There are too Many of the problems like this on the internet that a middle school kid can do

[@IgorTs79843]

А вы не заметили что у вас во второй строке уже готовый ответ? А как вы его нашли? Просто угадали? Ну и толку от такого решения, чему оно учит?

[@yhira2010]

first examine you can find two integers x and y making a^3-a^2=a^2(1-a)=150=x^2 × y and y=1-x true. I could find x=-5, y= 6 so, this one solution enables me to compose an equation (a+5)(a^2-6a+30) Therefore I could find the solution a=-5, 3+i×Rt21, 3-i×Rt21

[@JPTaquari]

a =- 5 a² * ( 1 _ a ) = 150 Possibility: 5 * 30 or 30 * 5 10 * 15 or15 * 30 3 * 50 or 50 * 3 25 * 6 or 6 * 25 ( This!!!!!!) 25 * [ 1 - (-5) ] = 125 Bingo from Brazil !!!!

[@ngocdo5687]

a^2 - a^3 = 150. * 150 = 25 + 125 = 5x5 + 5x5x5 = (-5x-5) - (-5x-5x-5) ** a = -5./.

[@jim2376]

-5 by inspection. (-5)^2 - (-5)^3 = 25 - (-125) = 25 + 125 =150

[@Mathsfocus8610]

Thank u

[@tatjanagapesina5599]

-5. Методом логических рассуждений. Первое - число должно быть отрицательным, чтобы получился положительный ответ. Дальше простой подбор. Начала с 5, сразу получила нужный ответ. Зачем это безумное решение?

[@richardamullens]

The given solution is laborious. Instead: First find root -5 by inspection. Second divide a^3 - a^2 + 150 by (a+5) to get (a^2 - 6a + 30) This has roots (3 + i √ 21) & (3 - i √ 21)

[@egitimg]

a=-5 not in 10 seconds but in 20 seconds cos I am not mathematician, I am just ordinary person.

[@07Pietruszka1957]

Integer solutions can be found as follows a^2*(1-a)=150 But 150=5*5*3*2 or 150=(-5)*(-5)*(-3)*(-2) ......Half of the factors can be negative But a^2 is positive then a must be negative By checking negative numbers like -2, -3, -5, -6, -10...We quickly find the correct answer a=-5

[@wavelio7370]

the equation is cubic, it mean there should be 3 answers here, that is the basic, so the important thing is not only show a=-5 but show another two answer and also show two answer is not suitiable in real number/not make sense; if you only show a=-5; actually it is possible that you cannot find out another 2 answer(that only lucky that 2 answer is not make sense in this case) anyway, your way to find out a=-5 fastly can very helpful for find out the equation -a^3+a^2-150 must be rewrite to (a+5)(.......) is real helpful for find out another two answer

[@07Pietruszka1957]

@@wavelio7370 I understand the problem, but many times we do not always to look for complex solutions, but for simple integer.

[@bitboy17]

Можно короче: а^2 ( 1 - a ) = 150 = (25)(6), то есть 1 - а = 6 и а = - 5 . А комплексные корни можно найти потом из квадратного уравнения.

[@에스피-z2g]

a^2-a^3=150 a^2(1-a)=150 25×6=150 (-5)^2(1+5)=150 a=-5

[@Mathsfocus8610]

There are three solutions

[@AmédéeGerest]

I used this method that seemed much more simple to me 😉

[@christianborgelt8318]

​@@Mathsfocus8610 There is only one real-valued solution. Whether complex solutions may be counted, depends on what the problem actually is and should not be taken for granted. If the problem comes from some physical problem in which complex solutions do not correspond to anything physically meaningful or possible, they are meaningless and hence do not count. To make my point clearer: Depending on the situation, it may even be that a real-valued solution is meaningless and hence does not count. Suppose you describe the diagonal throw of a body in Galileian mechanics, say a ball thrown by a human. You want to know where the ball hits the ground, depending on the point from which it was thrown, the angle and the initial velocity. The trajectory is a parabola opening downward and with a proper coordinate system the point where the ball hits the ground is a root of that parabola. However, a parabola opening downward which has a part above the x-axis (which is the case in this example, because a human standing on the ground throws the ball) has two roots. Only one of these roots is meaningful, namely the one that lies in the direction in which the ball is thrown. The other one is "behind" the human throwing the ball and hence meaningless. Saying in this situation that there are two solutions because there are two roots is obviously not reasonable. Similarly, saying for the task discussed in the video that there are 3 solutions (which are the roots of a cubic equation, and any cubic equation has at least one real-valued root; the one considered here happens to have exactly one) has exactly the same taste to me. Without any further information whether complex solutions are meaningful, it is reasonable to restrict solutions to real-valued ones, and there is only one real-valued solution, which the original commenter found correctly. I always have serious problems with this sort of ivory tower mathematics where one moves in some ideal mental space without any relation to reality and based on that dismisses a perfectly correct answer, because in that ideal mental space something else can be considered, regardless of whether that is meaningful or not.

[@AbdoLaalou]

الرياضيات لغة عالمية، شكرا لك على هذا الشرح المميز.

[@gogaotinashvili2235]

Когда D

[@jesusjosemanuelsuarez5377]

Basta de complicar lo sencillo!!! Entre los divisores del término independiente está el (-5), que es raíz. Luego se reduce al producto entre (a+5) y una de segundo grado y ésta se resuelve por baskara, dando las otras dos raíces complejas conjugadas. ¡Menos de dos minutos!

[@mmagic5753]

ob，a is a0 b²(b+1)=150 = 2*3*5*5 only 5 be double, so b must 5 back to try equation, pass

[@ivobrazgodinho4878]

This calculation is equal 150 ÷3 = 75+150 ÷3=50= 75-50=25+75+50=150 All roads lead to Rome, which is in the middle of the crossroads, what's new?

[@FuWong-r4f]

做了那麼多高深的解程，我這大陸七十年代初的初中生是看不懂，但用初中的解程可簡單的就能解出來，就是用 最後的那三四層解程就完事了，為什麼要算得那麼久？

[@ЕленаТарасова-е8ю]

Это уравнение быстрее решается без нахождения корней! Зачем так заморачиваться!🤔☝️

[@HPNam1307]

Điều kiện xác định: 1-a>0 suy ra a

[@박창현-w1h]

a = -5 25 - ( -125) = 150

[@ПавелГерманов-ь4я]

Такие "сложные" задачи решаются устно😂, пока Европа расписывает микрошаги😂

[@arsaluisnubar1074]

a=-5 ( -5)2-(-5)3=25-(-125)=25+125=150

[@klaushoegerl1187]

a=-5. 10 seconds.

[@fredericimbert5969]

You Forgot the complex roots, bro

[@livnasta1333]

Yeah.. but this is a native German solution...

[@chantalvilar4001]

Il faut justifier

[@Renato-to9no]

Jajajaja te faltan 2 soluciones. Fallaste el examen

[@АндрейКожевников-о8й]

Клаус браво-брависимо.

[@lionelfischer8240]

Trivial try = -5. Finished in 30 s.

[@deneb80]

Love your Nigerian pronunciation 😀😀😀

[@Mathsfocus8610]

Thank you

[@IgorTs79843]

Когда он разделил 150 на 25 и 125, то фактически решение уже найдено и вся дальнейшая писанина не имеет смысла. Вот только откуда он знал, на какие цифры разделять число? Просто угадал... А если это будут не такие удобные цифры?... Я еще 30 лет тому назад придумал программку для решения уравнений. Программка работала на любом программируемом калькуляторе. У меня был БЗ-34... Тупо подставляем в формулу любое число, вычисляем результат... Еще раз то же самое. Сравниваем результаты и определяем в какую сторону развивается функция. Выбираем направление вычисления и подставляем следующее число... потом следующее пока функция не перейдет 0. Тогда умножаем шаг на -0,1 и продолжаем считать... вычисляем вторую цифру. Еще раз... и так находим нужное количество разрядов.

[@stephanebarbieri7254]

De tête, a=-5

[@jaggisaram4914]

-5. (-5)^2 - (-5)^3 25 - (-125) = 150

[@clancyhughes]

-150 squared minus -150 cubed should also = 150 without all that fuss

[@kayhchoo]

If you can guess to split 150 into 25 and 125, than you should get the answer a= -5 without any calculation 😂

[@shekharjoshi7292]

सीधी सी बात है। इतना बड़ा स्टेटमेंट क्यों?

[@cliffmo6707]

The thinking path should not be like this video. We should think about the circumstances under which this equation is true. Then we know that "a" must be a negative number. Then we think about what makes the cube of “a” close to 150, and the answer is obviously -5. Then we know the answer.

[@武中]

A^2-A^3 = A^2(1-A) = 5＊５＊２＊３＝(ー５)＊（ー５）＊２＊３＝１５０ 1－A ＝2＊3 ーA=2＊３－１＝５ A = ー５

[@mbmsafir4835]

a^2-a^3=150 a^2-a^3=(-5)^2-(-5)^3 Hence a= -5

[@AchtungBaby77]

Wouldn't it be easier to just apply the Rational Root Theorem, then use synthetic division to obtain a linear factor times a quadratic factor. From there, use the quadratic formula. Problem solved.

[@이정원-b8e]

a2 and a3 (×) a2 or a3 (o)

[@acojapacze]

>> 5

[@jackieking1522]

-5 by inspection and the complex ones someone else can work out.

[@nnfefe9451]

You write 2 like 3. Please change how you write 2. Thanks.

[@mafe52and17]

Для начала следует определиться с областью значения. Во множестве реальных чисел корень из отрицательного числа не извлекается. Решение растянуто. Следует также применять рq формулу. Она оптимальнее

[@chunyulo7605]

a must be negative And think about a^2 + a^3 = 150

[@ОлександрПрадеус]

-5. 25-(-125)=150

[@Mario-y8v]

Is this type of mathematics applicable to our work on field of computation.ewan ko ha.nag pakahirap aq mag aral para sa mga ganitong computation pero Indi ko magamit sa trabaho ko.para saan ba talaga ang mga eto..kita nu etong nag tuturo Indi maintindihan .hahaha.sir saan ko Po gagamitin to an computation na Yan sa field of work .kindly site one w/c applicable to use .bulol bulol pa pag tuturo.

[@rajailyasaman]

😊

[@SergeyTikhomirov-xd4ox]

Блин, прямым перебором простых целых чисел находим, что а=-5

[@peterhawran146]

cubic equation. . .

[@ОльгаЛебедева-ю6ы]

Ответ - 5. Практически сразу.

[@lilianaprina5991]

This is a juicy nightmarish Math Olympiad problem....

[@小瑋-p7y]

a=-150😊

[@Роман-ь3н5з]

Какое практическое применение имеют два комплексных корня? Правильный ответ никакого. А вот при поступлении в ВУЗ имеют.

[@عادلامام-م3ف]

A=-5

[@ИгнатВеселов-у2ы]

К чему столько писанины, пример для устного счёта, 10 сек. не более

[@rajeshtiwari1856]

Sir ; dimak ka dahi bana diya

[@НадеждаПанкова-к6й]

Теперь я поняла,почему сейчас основная масса школьников ненавидит математику...

[@ЛеонидИльницкий-д3ъ]

Да что тут это ,,пудрят мозг,,

[@heribertoayalareyes3628]

Complex Inapropiado. Imaginario!

[@박사진]

딱 보니 답이 나왔다. a=-5 25+125= 150

[@ryanchiang9587]

a =-5. 25 - (-125) = 150.

[@ghulamshabir3460]

It required only 2 minutes to solve..

[@Апполон-б6ш]

- 5 .

[@bne141]

動画見たら14分、コメント欄見たら2秒で解決。便利な世の中になった笑

[@kingmaker_984]

-5 without watching this vdo.....

[@Kcc-c5y]

~6

[@NhanDuongDuc-g5x]

Consider the expression : X2 - 6x +30  (x-3)2 +21 > 0 , positive for all values ​​of x , Therefore, this expression cannot be equal to zero, so why keep assigning it equal to zero to solve the quadratic equation . i do not know who you are teacher , professor or student ... i think you should relearn basic knowledge

[@sunnysharma5166]

After solving & applying inspection method we get a=-5

[@hulyaaycan4660]

♥Maşaallah

[@abdelrahmanaboukassem6222]

-5 = a=

[@hassanhrimach3784]

I exposant 2= -1

[@toddprifogle7]

1

[@ВладимирКарапетян-г7е]

зачем предлагать столько длинное решение

[@dineshatri2417]

certainly a= 05

[@cabasantbab]

-5

[@apoorvchauhan9275]

Did it in 15 seconds. What's the age group of this Olympiad?

[@lower_case_t]

How did you calculate the complex solutions in 15s?

[@apoorvchauhan9275]

@lower_case_t took a^2 common. a^2(1-a) = 150. This indicates that a is such a number whose square is a factor of 150 and also when subtracted by 1 is a factor of 150(hence the number would be negative). Then started looking at square numbers(4, 9, 16, 25) bec a^2 is a factor of 150. Rejected the first 3 bec they aren't factors of 150. When I checked for a=-5 it worked.

[@lower_case_t]

@@apoorvchauhan9275 OK, so you didn't calculate them at all and would have failed the test. There are two complex solutions to the problem, 3 + i * sqrt(21) and 3 - i * sqrt(21) here's a check for 3 + i * sqrt(21), it goes analogous for 3 - i * sqrt(21): a² : (3 + i * sqrt(21))² = 9 + 6 * i * sqrt(21) - 21 = -12 + 6 * i * sqrt(21) a³ : (-12 + 6 * i * sqrt(21)) * (3 + i * sqrt(21)) = -36 -12 * i * sqrt(21) + 18 * i * sqrt(21) - 6 * 21 = -36 -126 + 6 * i * sqrt(21) = -162 + 6 * i * sqrt(21) a² - a³ = -12 + 6 * i * sqrt(21) + 162 - 6 * i * sqrt(21) = 150 I guess most people would not think of that, but then the condescending tone in the question about the age group is a bit inappropriate. Anyway, thanks for responding!

[@richardamullens]

@@apoorvchauhan9275 A cubic equation has 3 roots and you are expected to find them all.

[@Darkz5w]

보자마자 5초만에 -5 대입하니 맞던데

[@uuwata]

god give me a = -5, then (a + 5)(a^2 + B a + 30) = 150, then B = -6

[@羁绊在可视宇宙之外]

2or3😂

[@prasadrasikawidanagamachch3932]

a=-5

[@zzzungnguyen]

A= - 0.015

[@zzzungnguyen]

A= - 1/50

[@賴驄仁]

Ans: -5

[@РифхатХакимов-к7м]

Легко.

[@d.bpatil6361]

Very simple. a = -5 (minus five)

[@diendoan467]

Đau hết cả đầu.

[@gewerkschaftderhonigschabe167]

How much IQ is needed to understand that? :( i am to stupid for this!

[@narenmenon6906]

a is a negative integer. a= -5 jumped out at me.

[@pas6295]

a=-5.

[@KirillZavrazhny]

Это не математика, это ребусы пол ютуба завалено Поиском разных оснований суммы квадрата и куба.

[@I.S-u7r]

После того как 150 разбил на 5в3 и 5во2 стал ясен ответ -5.

[@오말말]

결국 인수분해 문제네요.