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)
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.
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
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.
@@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.
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.
@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!
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!
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.
@Mathsfocus8610 because it’s incorrect, simple as that. We only take square roots of positive numbers. If we assume we could write i=sqrt(-1), it’s wrong, because i would also be equal to -sqrt(-1). The thing is that i is ONLY defined by i^2=-1, and that’s the only and simplest way to express i directly.
@@ribionakzalatoi Is it possible that you confuse two different things here? For real numbers r, sqrt(r) is always positive - simply because it is defined as the positive number that gives r when muliplied with itself. (-sqrt(r)) is the second solution. For any complex number c (including -1) it's perfectly fine to calculate a pair of numbers again that give c when multiplied with themselves. Just like for the two solutions for sqrt(r) one of these numbers is defined as the principal value in the following way: if c = r*e^(i*φ) then the principal square root sqrt(c) = sqrt(r) * e^(i*φ/2) in other words, both i² and (-i)² are -1 but only i is the square root of -1.
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)
1) for a > 1, the equations does not hold
2) for 0
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.
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?
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
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 !!!!
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}.
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.
-5 by inspection. (-5)^2 - (-5)^3 = 25 - (-125) = 25 + 125 =150
Thank u
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.
-5. (-5)^2 - (-5)^3 25 - (-125) = 150
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.
-5 by inspection and the complex ones someone else can work out.
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.
a^2(a-1)=-5*5*3*2 thus a=-6
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 😂
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？
😊
Такие "сложные" задачи решаются устно😂, пока Европа расписывает микрошаги😂
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.
Very simple.
a = -5 (minus five)
Ans: -5
a must be negative
And think about a^2 + a^3 = 150
Complex Inapropiado. Imaginario!
A=-5
It required only 2 minutes to solve..
a=-5 not in 10 seconds but in 20 seconds cos I am not mathematician, I am just ordinary person.
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
Решила устно : - 5
I exposant 2= -1
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.
a =-5.
25 - (-125) = 150.
a is a negative integer. a= -5 jumped out at me.
-5
cubic equation. . .
Это не математика, это ребусы пол ютуба завалено Поиском разных оснований суммы квадрата и куба.
a=-5
After solving & applying inspection method we get 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!
Я сразу увидела -5.😂
a =5
@@rkadkar noooo
-5.
マイナス5
a=-5.
a=-5.,,, 5*5-((-5)*(-5)*(-5))=25-(-125)=25+125=150
a = AI
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
Po cholerę marnowanie czasu na głupią algebrę, skoro na początku intuicyjnie wiadomo, że a= -5.;((
푸는것이 속터지네요?
a = +25
La racine carré de -1n'existe pas la fonction racine carré est définie dans R+
inspection is the best, complex roots is a waste of time !
C'est faut
The algebra is fun to learn, but the answer does pop out intuivly. As 5
Please check very well. Thanks
-5, by inspection, less than 10 second
磨磨唧唧的
Just a remark: we should NEVER write square root of -1 !! that is NOT i !!
Can you explain this very well?
@Mathsfocus8610 because it’s incorrect, simple as that. We only take square roots of positive numbers. If we assume we could write i=sqrt(-1), it’s wrong, because i would also be equal to -sqrt(-1). The thing is that i is ONLY defined by i^2=-1, and that’s the only and simplest way to express i directly.
@@ribionakzalatoi Is it possible that you confuse two different things here? For real numbers r, sqrt(r) is always positive - simply because it is defined as the positive number that gives r when muliplied with itself. (-sqrt(r)) is the second solution.
For any complex number c (including -1) it's perfectly fine to calculate a pair of numbers again that give c when multiplied with themselves. Just like for the two solutions for sqrt(r) one of these numbers is defined as the principal value in the following way:
if c = r*e^(i*φ) then the principal square root sqrt(c) = sqrt(r) * e^(i*φ/2)
in other words, both i² and (-i)² are -1 but only i is the square root of -1.
😂😂😂😂😂😂😂😂😂
Reloved in 15 seconds.
一看就知道答案。-5
a=-5
-5
-5
-5
-5
-5