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.
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.
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
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
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!
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)
-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?
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.
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
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!
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.
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
@@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.
@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.
@@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!
Интересно, в какой прикладной науке применяются недействительные числа, что американским школьникам засоряют мозги нахождением этих недействительных корней уравнения?
Дурака валяет. Вынес за скобки а в квадрате и сразу видно , что -5 есть корень. Потом все перенес в одну сторону и поделил на а+5 и получил квадратное уравнение.
Me shikimin e pare vertetoj qe a eshte nje numer negativ. Pastaj zgj ushtrimin. Qe tashme ju po e beni vete. Megjtheate un e ndjek zgjidhjen. Argetohem me matematiken dhe me mban pak a shume ne nje nivel normal llogjiken time prej 78 vjecareje. Flm.
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.
The simplest solution.
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.
1) for a > 1, the equations does not hold
2) for 0
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.
Я также решала
Расписать 150=125+25.перенести в левую часть и разложить на множители сумму кубов и разность квадратов,а дальше понятно.
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
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
@@wavelio7370 I understand the problem, but many times we do not always to look for complex solutions, but for simple integer.
a^2-a^3=150
a^2-a^3=(-5)^2-(-5)^3
Hence a= -5
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!
Это уравнение быстрее решается без нахождения корней! Зачем так заморачиваться!🤔☝️
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.
Такие "сложные" задачи решаются устно😂, пока Европа расписывает микрошаги😂
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)
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}.
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?
a=-5 not in 10 seconds but in 20 seconds cos I am not mathematician, I am just ordinary person.
-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?
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.
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)
In Zambian syllabus we don't have such... 😂😂
Same here just substitute. There are too Many of the problems like this on the internet that a middle school kid can do
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?
К чему столько писанины, пример для устного счёта, 10 сек. не более
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 !!!!
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
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？
-5 by inspection. (-5)^2 - (-5)^3 = 25 - (-125) = 25 + 125 =150
Thank u
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!
Какое практическое применение имеют два комплексных корня? Правильный ответ никакого. А вот при поступлении в ВУЗ имеют.
Repeated use of input errors reach convergence
Блин, прямым перебором простых целых чисел находим, что а=-5
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.
This is a juicy nightmarish Math Olympiad problem....
Thanks you very much
-150 squared minus -150 cubed should also = 150 without all that fuss
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
Love your Nigerian pronunciation 😀😀😀
Thank you
a^2-a^3=150
a^2(1-a)=150
25×6=150
(-5)^2(1+5)=150
a=-5
There are three solutions
I used this method that seemed much more simple to me 😉
@@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.
الرياضيات لغة عالمية، شكرا لك على هذا الشرح المميز.
зачем предлагать столько длинное решение
Tanks for your video
De tête, a=-5
Теперь я поняла,почему сейчас основная масса школьников ненавидит математику...
Да что тут это ,,пудрят мозг,,
-5 by inspection and the complex ones someone else can work out.
a=-5. 10 seconds.
You Forgot the complex roots, bro
Yeah.. but this is a native German solution...
Il faut justifier
Jajajaja te faltan 2 soluciones. Fallaste el examen
Клаус браво-брависимо.
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
If you can guess to split 150 into 25 and 125, than you should get the answer a= -5 without any calculation 😂
सीधी सी बात है। इतना बड़ा स्टेटमेंट क्यों?
You write 2 like 3. Please change how you write 2. Thanks.
a2 and a3 (×) a2 or a3 (o)
a=-5 ( -5)2-(-5)3=25-(-125)=25+125=150
Sir ; dimak ka dahi bana diya
a must be negative
And think about a^2 + a^3 = 150
-5. (-5)^2 - (-5)^3 25 - (-125) = 150
Did it in 15 seconds. What's the age group of this Olympiad?
How did you calculate the complex solutions in 15s?
@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.
@@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!
@@apoorvchauhan9275 A cubic equation has 3 roots and you are expected to find them all.
딱 보니 답이 나왔다.
a=-5
25+125= 150
После того как 150 разбил на 5в3 и 5во2 стал ясен ответ -5.
Посмотрела условие и сразу сказала ответ: -5
보자마자 5초만에 -5 대입하니 맞던데
a =-5.
25 - (-125) = 150.
It required only 2 minutes to solve..
cubic equation. . .
A=-5
Complex Inapropiado. Imaginario!
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
certainly a= 05
Il faut suivre la formule, il ne faut pas inventé
After solving & applying inspection method we get a=-5
-5 = a=
Интересно, в какой прикладной науке применяются недействительные числа, что американским школьникам засоряют мозги нахождением этих недействительных корней уравнения?
😊
Это не математика, это ребусы пол ютуба завалено Поиском разных оснований суммы квадрата и куба.
Я сразу увидела -5.😂
god give me a = -5, then (a + 5)(a^2 + B a + 30) = 150, then B = -6
A= - 1/50
-5
How much IQ is needed to understand that? :( i am to stupid for this!
~6
A= - 0.015
お国柄なのか、うｐ主さんの癖なのかは判りませんが、
２乗の「２」と３乗の「３」とがぱっと見た感じでは
区別がつけ辛いですね。
a is a negative integer. a= -5 jumped out at me.
а= -5
Дурака валяет. Вынес за скобки а в квадрате и сразу видно , что -5 есть корень. Потом все перенес в одну сторону и поделил на а+5 и получил квадратное уравнение.
I exposant 2= -1
Для олимпиады слабовато. Решается устно. а=-5.
Решила устно : - 5
1
a=-5.
А=-5
a=-5
Легко.
А вот почему так, тот кто задал это не ответит правильно сть этого примера пусть даст объяснить мне как кажется это все не стоит выеденного яйца. 😅
_5
Elegáns megoldás
Ta résolution n'est pas vérifiée ,vous n'avez pas eu de formule
Ой таки профессоры жуть просто. ,а ларчик просто открывался. а = 5
a =5
@@rkadkar noooo
-5.
Ans: -5
a^2(a-1)=-5*5*3*2 thus a=-6
マイナス5
푸는것이 속터지네요?
Me shikimin e pare vertetoj qe a eshte nje numer negativ. Pastaj zgj ushtrimin. Qe tashme ju po e beni vete. Megjtheate un e ndjek zgjidhjen. Argetohem me matematiken dhe me mban pak a shume ne nje nivel normal llogjiken time prej 78 vjecareje. Flm.
Waooo... That's great
I like Nigerian accent
😂😂😂😂😂😂😂😂😂
Its too long explanation solution!