Interesting and well presented until the last few seconds when a promotion for another of JJ's videos obscures the portion of the whiteboard where she is concluding her work.
A lot of people down here thinking the answer is just '-2'. I think there are clues in the title with the phrase 'Most Fail' indicating that the teacher is not looking for just the REAL solution. If you gave just -2 as your answer you would FAIL to get into Harvard and end up going to a community college instead. :)There are 3 solutions.
Depends on what the question asked was - was it "give all possible answers" or was it "find a solution for x". "Most fail" was the caption on the video.
This is very good approach - in advanced mathematics I would use the Rational Roots Method to find x = -2, then divide the cubic by x+2 to find the quadratic,
Thanks for the video ! Great thinking 💜 I did it slightly differently : after I realised that -2 is a root (by just looking at the polynomial), which means that (x³ - x² + 12) is divisible by (x + 2), I just performed the division and found (x² - 3x + 6), and then found the two complex solutions in the same way. Not sure I would have thought of using the identities for (a³ + b³) and (a² - b²), but I’ll try to remember that 😃
First, we will divide the expression by minus to get -X^2+X^3=-12 and X^3-X^2+12=0, (X+2)(X^2-3X+6)=0. One of the roots equals -2, and the other two are determined from the expression X^2-3X+6=0 by using the quadratic formula.
You can do it in your head. At a glance, you can figure out that an abvious answer is x = -2. If you divide (-x^3 + x^2 + 0x - 12) with (x+2), you get (-x^2 + 3x - 6). Problem solved in 2 steps.
I did this in about 30 seconds in my head. Harvard must be a cakewalk. The trick was that the answer had to MINUS due to any POSITIVE answer would not work since x cubed is > x squared, so the answer would always be < 0. -2 came instantly in my head as a guess and it solves the problem. Seriously? JK about Harvard - total respect for that institution.
X=-2 by inspection. So x+2 is a factor (factor theorem). Divide the expression =0 by x+2. This will yield 2 complex roots. Basic differentiation will allow curve sketching to confirm only 1 real root. Really no need for loads of algebra. Especially if it’s a Harvard entrance problem, which I seriously doubt.
Is the question. Find the real root? or is it find all roots? and perhaps teaching this helps people learn a tool for dealing with more complicated formulae where the real root is not found by simple inspection?
Intuitively the answer is x = -2 The procedure might seem long but if done mentally will give you the solution within seconds. x² - x³ is always negative if the value of x is more than 1. It's always positive if the value of x is less than 1. Now for anything between 0 and 1, we'll have x² - x³ also between 0 and 1. That would mean the value of x must be a negative number. If we write x as -k, then we get k² + k³ = 12, where k is a positive real number. Now all we have to do is take two or three guesses. Taking k as 1 will give us less than 12. Taking k as 2 gives us exactly 12. Anything more than that will be more than 12. This means k can have only one value, that is 2. This gives us x = -2. By our reasoning, there can't be any other real value of x. Now a real cubic equation will either have 3 real solutions or 1 real and a pair of complex solutions. This means the other roots are complex conjugates.
It takes about three seconds to see that x = -2 solves this. It is a cubic equation, of course, so it will have three solutions. The other two will be complex, though.
Why so much calculations ,by factor/remainder theorem we get immediate answer x=-2 then by synthetic division we get another quadratic factor which can be easily solved and get two complex roots
Thank you, that was fun. The best part of it is if it was on a college board test where you did not need to show Work, the guess and check method would be very simple because you would know that the imaginary part would not work as an answer. So just take the -2 and that would be the only one that does work. Always loved doing college algebra
X must be negative because whilst the square of a negative number is positive, the cube is negative. Futhermore, the cube is subtracted from the square to equal 12. 12 is a small number, so x must be smaller. In this case it's 2. 2 -(-8) = 12. It takes longer to type than to analyse and calculate mentally.
The is one Real root that is -2, and two Complex roots that need to be calculated. So the answer x=-2 is wrong. And generally speaking a polynomial equation in x³ will always have 3 roots. You don't need to go to Harvard to know that.
@@JosephConcerned I understand her just fine and English isn't even my first language. While you're watching Maths videos, maybe you should improve your English skills too! :)
Je suis vraiment désolé mais ce n'est pas un exercice pour entrer à Harvard. C'est un exercice pour élève de 15 ou 16 ans qui découvre l'équation du second degré. Exercice fort simple. Bravo pour la factorisation mais face à une équation du 3e degré, il est préférable de chercher une racine ÉVIDENTE puis de FACTORISER. Une racine évidente est, par exemple, un entier compris entre - 5 et + 5. Puisque x²-x³ = 12, la racine x = -2 est évidente. On peut donc mettre (x+2) en facteur. On obtient x³ - x² + 12 = (x-2) (x² + ax -6) et une simple identification nous donne a = - 3. Il reste à résoudre (x-2) (x² - 3x -6)=0. Ce qui nous donne rapidement x = -2 ou x = (3 + i sqr(15))/2 ou x = (3 - i sqr(15))/2. Une dernière remarque. Dans un concours le temps joue un rôle primordial. Il est donc important d'apprendre à chercher des solutions simples et courtes, ou bien longue mais automatique. Sinon, la vidéo est claire et bien présentée. Dommage que les solutions proposées ne soient pas optimales...pour un concours. 😉 DEUX RÈGLES SIMPLES pour aider à faire les factorisations rapidement. Quand on fait le produit de deux polynômes, le TERME DE PLUS HAUT DEGRÉ est égal au PRODUIT DES TERMES DE PLUS HAUT DEGRÉ et le terme de plus bas degré est égal au produit des termes de plus bas degré.
100% d'accord. Avec la racine evidente trs facile a devinerer, j'ai trouvé les deux autres racines en factorisant en 10 fois moins de temps que la video. Je doute vraiment qu'un exercice aussi simple puisse servir a rentrer a Harvard... sauf a eliminer ceux qui prennent 10 fois trop de temps???? De plus je n'adhere pas a la notation employee avec une racine d'un nombre negatif. La fonction racine carree est definie pour IR+ . Mon vieux profs de maths des annees 80 aurait hurlé apres 6minutes 20 secondes, et m'aurait certaienment collé un zero
Retired physics/maths teacher here. Looked at thumbnail, thought a moment and answered, “negative two”. These were my thought processes: 1. For all positive numbers >1, X³ > X² and there is no possible solution, therefore X must be negative. 2. For negative numbers the square is positive but the cube is negative. 3. Subtracting a negative number is the same as adding its positive opposite. 4. So the problem can now be expressed as |X|² + |X|³ = 12. 5. 1 is too small, 3 is too big. 2 works. But remember it’s negative. Read a few comments and decided not to bother watching the video.
Retired Physics/Chemistry teacher here, California. (I can see that you are from UK....I taught in Derbyshire for a while)....I followed your exact line of thought.
@@q.e.d.9112 Visited Auckland just once; wanted to find Mr. Leo Bult.....he saved my Dad after his B-24 was shot down over the Netherlands. Found him just days before he passed. He was 97.; I tahnked him. The big steel mill in Sheffield is now a shopping mall (!!) Born in America, American for Life.
You should not be celebrating ignorance. If it means nothing to you does not mean it would not mean something to high school students preparing for uni!
Oh! I see another channel where the host is a genius who, thanks to supernatural powers, knows exactly what to factor out to solve the problem. I prefer teaching that shows how to solve problems without relying on supernatural powers. x^2-x^3=12 x^2(1-x)=12 After factoring, you can see that both x^2 and (1−x) must be divisors of 12. Since 12=2*2*3, then 12=2^2*3 so x^2=4; 1-x=3 From the first equation, we get x^2=4 (, and from the second equation, we find x=2, which we have to reject, but x=−2 is the solution. Now we know which factor to factor out. to find complex solutions,... but we do not need it complex solutions are x2 = a+b*i and x3=a-bi x2*x3=a^2+b^2 x2+x3=2a And we know the sum and product of the roots. x^3-x^2+12=0 sum is 1 product is -12 now we can add and multiply solutions and find a and b: -2+a+a=1 so a=3/2 -2*(a^2+b^2)=- 12 (a^2+b^2)=6 9/4+b^2=6 b^2=(24-9)/4=15/4 b=sqrt(15)/2
Why must x^2 and (1-x) be integer divisors of 12? In theory, x^2 could be say pi, and (1-x) 12/pi which obviously doesn't work but I don't think it is valid to pair these factors with integer divisors of 12, given your correct distaste for supernatural powers!
@@silver6054 Because we asked ourselves whether there are solutions in integers. Neither the method presented in the video nor the one I presented are universal methods. Can you imagine trying to factor a cubic equation if its solution was π and the coefficients were integers? There are universal methods for solving cubic equations, and from them, it is known that a cubic equation with integer coefficients cannot have π as a solution because every solution can be expressed using addition, subtraction, multiplication, division, and taking square or cube roots of equation coefficents. These methods often give solutions where it can be extremely challenging to prove that these operations result in an integer. The solution I provided is indeed a solution by "guessing," but its value lies in the fact that we only need to check two values using the method specified. Moreover, the solution I provided allows us to determine the other two solutions, whatever they may be. You can use this method to construct an example of an equation with coefficients for x^3, x^2, and the constant term as in the problem from the video, and determine those solutions when the real solution is ππ. However, in that case, the coefficient of xx would have to be transcendental, not zero, as in the equation being solved. If you have any further questions, please feel free to ask.
there is a much faster way. Just do X2 (1 - x ) = 12 One of the factors has to be 3 the other one 4. Take the 1st one. x could be either 2 or -2. Making an educated guess is allowed when polynomials are around. Even Newton allows it.
@@tinkerman9525 I have tested it.. .there are three roots and you have only listed one. Meaning that you didn't really understand the question and you would fail this question in an interview. But at least you would win at emojis
@@tinkerman9525 LOL 2 is definitely not another factor... come on try again. You were the one shouting No NO. Now try and answer the question properly. I have given you a clue (it is also in the video). There are three roots. -2 is one of them. what are the other two??
Je reprends à mon compte un commentaire judicieux qui a été fait à propos de SQR (-1). C'est une notation qui est souvent utilisée par nos amis anglo-saxons, malheureusement cette notation est MATHÉMATIQUEMENT FAUSSE . En effet l'équation x²=-1 admet DEUX SOLUTIONS complexes qui sont DIFFÉRENTES , elles ne peuvent donc pas être représentées par un seul et même symbole sqr(-1). C'est donc un abus de langage bien pratique mais faux. 😉
the correct answer for x is - 2
you mean ONE of the roots is -2. You would fail if -2 was your answer
Interesting and well presented until the last few seconds when a promotion for another of JJ's videos obscures the portion of the whiteboard where she is concluding her work.
Oh! So sorry about that.
Right😊
In the answer she arrived -2 is not there at all. So whatever she did is in correct.
@@rajamnaidu1962Look to the far left. It is there!
Thank you for taking the time to make this video. Much appreciated.
A lot of people down here thinking the answer is just '-2'. I think there are clues in the title with the phrase 'Most Fail' indicating that the teacher is not looking for just the REAL solution. If you gave just -2 as your answer you would FAIL to get into Harvard and end up going to a community college instead. :)There are 3 solutions.
Depends on what the question asked was - was it "give all possible answers" or was it "find a solution for x". "Most fail" was the caption on the video.
Who is a lot of people. It's only you who thinks so.
How brilliant you are ! Your experience & explanation is very very best.
This is very good approach - in advanced mathematics I would use the Rational Roots Method to find x = -2, then divide the cubic by x+2 to find the quadratic,
Thank you so much for explaining it so well and patiently.
Regards 🙏
U are welcome
one root is -2. not sure about the two others, but i am sure you will explain in video
Thanks for the video ! Great thinking 💜 I did it slightly differently : after I realised that -2 is a root (by just looking at the polynomial), which means that (x³ - x² + 12) is divisible by (x + 2), I just performed the division and found (x² - 3x + 6), and then found the two complex solutions in the same way. Not sure I would have thought of using the identities for (a³ + b³) and (a² - b²), but I’ll try to remember that 😃
👍👍
First, we will divide the expression by minus to get
-X^2+X^3=-12 and X^3-X^2+12=0,
(X+2)(X^2-3X+6)=0. One of the roots equals -2, and the other two are determined from the expression X^2-3X+6=0 by using the quadratic formula.
x = - 2, (3 + - √- 15) /2
This method is shorter
-2 in a few seconds
You can do it in your head. At a glance, you can figure out that an abvious answer is x = -2. If you divide (-x^3 + x^2 + 0x - 12) with (x+2), you get (-x^2 + 3x - 6). Problem solved in 2 steps.
That’s what I did.
Harvard doesn't have an "entrance exam".
I did this in about 30 seconds in my head. Harvard must be a cakewalk. The trick was that the answer had to MINUS due to any POSITIVE answer would not work since x cubed is > x squared, so the answer would always be < 0. -2 came instantly in my head as a guess and it solves the problem. Seriously? JK about Harvard - total respect for that institution.
x = -2 by inspection
and what about the other two roots? did you inspect those as well
One value of x is -2. So x+2 is a factor. By dividing I would find the other quadratic factor. Will factorize ......
x = -2 satisfies the equation
X must be negative
And answer is -2 just by inspection.
X=-2 by inspection. So x+2 is a factor (factor theorem). Divide the expression =0 by x+2. This will yield 2 complex roots. Basic differentiation will allow curve sketching to confirm only 1 real root.
Really no need for loads of algebra. Especially if it’s a Harvard entrance problem, which I seriously doubt.
Is the question. Find the real root? or is it find all roots? and perhaps teaching this helps people learn a tool for dealing with more complicated formulae where the real root is not found by simple inspection?
Intuitively the answer is x = -2
The procedure might seem long but if done mentally will give you the solution within seconds.
x² - x³ is always negative if the value of x is more than 1. It's always positive if the value of x is less than 1.
Now for anything between 0 and 1, we'll have x² - x³ also between 0 and 1.
That would mean the value of x must be a negative number. If we write x as -k, then we get k² + k³ = 12, where k is a positive real number.
Now all we have to do is take two or three guesses.
Taking k as 1 will give us less than 12.
Taking k as 2 gives us exactly 12.
Anything more than that will be more than 12.
This means k can have only one value, that is 2.
This gives us x = -2.
By our reasoning, there can't be any other real value of x.
Now a real cubic equation will either have 3 real solutions or 1 real and a pair of complex solutions.
This means the other roots are complex conjugates.
The question was to find the roots of the equation. if you say the answer is -2, have you answered the question?
It takes about three seconds to see that x = -2 solves this. It is a cubic equation, of course, so it will have three solutions. The other two will be complex, though.
Because x^3 is bigger than x^2, x must be negative. The factors of 12 are 1, 2, 3, 4, 6 ,12 and their negatives. Try -1, -2 ,,,
-2works.
Why so much calculations ,by factor/remainder theorem we get immediate answer x=-2
then by synthetic division we get another quadratic factor which can be easily solved and get two complex roots
x^2-x^3=12 x=-2 x=(3±Sqrt[15])/2i=1.5±0.5Sqrt[15]i final answer
I find it hard to believe that JJ has subscribers. James Joyce however makes perfect sense.
Well this one is easy. The first observation is that x
If it is I expect you are supposed to solve it by inspection, which shows a far better understanding than using a rote method.
x=-2 or x = (3 +/- sqrt(15)i)/2 where i^2=-1. Hint: use synthetic division.
Thank you, that was fun. The best part of it is if it was on a college board test where you did not need to show Work, the guess and check method would be very simple because you would know that the imaginary part would not work as an answer. So just take the -2 and that would be the only one that does work. Always loved doing college algebra
For instances where the answer is not as simple as -2, this more elaborate approach is probably applicable :-)
In the competetive exams, this type of questions are to be solved within 30 to 40 seconds. So, after guess and try, simple answer is -2. That's all.
If x = -2 then we have 4 --8 = 4 plus 8 equals 12. At least that's,a solution.
👍👍
X2(1_x)=4×3.
1_x=3.
_x=3-1=2.
X=-2
No need for long complicated work or theorems. Just some common sense and guess and check.
nice. It has been over 45 years since I did that. I now remember.
Life's too short.
Really too short!!! I'm none the better for her explanation 😢😢 jah jah!!
excellent
X must be negative because whilst the square of a negative number is positive, the cube is negative.
Futhermore, the cube is subtracted from the square to equal 12.
12 is a small number, so x must be smaller. In this case it's 2.
2 -(-8) = 12.
It takes longer to type than to analyse and calculate mentally.
Great presentation
The is one Real root that is -2, and two Complex roots that need to be calculated. So the answer x=-2 is wrong. And generally speaking a polynomial equation in x³ will always have 3 roots. You don't need to go to Harvard to know that.
Nonsense. Where did she say you have to go to Harvard to know that x^3 won't have 3 roots?
You would have to know all the theories,Etc.,to solve this type of problem.
Please don't use the quadratic equation. As thinking people, we should complete the square. Please teach us how to do that!
I have videos on completing the square method. Please check
This is painful
I'm 20 seconds in and I can already intuit that the answer is -2.
Imagine having a professor who you can't understand.
Guess what? That's a *you* problem. ^^
@@mori1bund definitely would be a problem for more than I. Try millions.
@@JosephConcerned I understand her just fine and English isn't even my first language.
While you're watching Maths videos, maybe you should improve your English skills too! :)
@@mori1bund wouldn't hurt lol
Welcome to America to the x(-3)
Je suis vraiment désolé mais ce n'est pas un exercice pour entrer à Harvard. C'est un exercice pour élève de 15 ou 16 ans qui découvre l'équation du second degré. Exercice fort simple.
Bravo pour la factorisation mais face à une équation du 3e degré, il est préférable de chercher une racine ÉVIDENTE puis de FACTORISER. Une racine évidente est, par exemple, un entier compris entre - 5 et + 5. Puisque x²-x³ = 12, la racine x = -2 est évidente. On peut donc mettre (x+2) en facteur.
On obtient x³ - x² + 12 = (x-2) (x² + ax -6)
et une simple identification nous donne a = - 3.
Il reste à résoudre (x-2) (x² - 3x -6)=0.
Ce qui nous donne rapidement x = -2 ou x = (3 + i sqr(15))/2 ou x = (3 - i sqr(15))/2.
Une dernière remarque. Dans un concours le temps joue un rôle primordial. Il est donc important d'apprendre à chercher des solutions simples et courtes, ou bien longue mais automatique.
Sinon, la vidéo est claire et bien présentée. Dommage que les solutions proposées ne soient pas optimales...pour un concours. 😉
DEUX RÈGLES SIMPLES pour aider à faire les factorisations rapidement. Quand on fait le produit de deux polynômes, le TERME DE PLUS HAUT DEGRÉ est égal au PRODUIT DES TERMES DE PLUS HAUT DEGRÉ et le terme de plus bas degré est égal au produit des termes de plus bas degré.
100% d'accord. Avec la racine evidente trs facile a devinerer, j'ai trouvé les deux autres racines en factorisant en 10 fois moins de temps que la video. Je doute vraiment qu'un exercice aussi simple puisse servir a rentrer a Harvard... sauf a eliminer ceux qui prennent 10 fois trop de temps???? De plus je n'adhere pas a la notation employee avec une racine d'un nombre negatif. La fonction racine carree est definie pour IR+ . Mon vieux profs de maths des annees 80 aurait hurlé apres 6minutes 20 secondes, et m'aurait certaienment collé un zero
X is -2
and what about the other roots?
Mental arithmetic! By initially splitting 12 into 2 x 2 + 2 x 2 x 2 you have arrived at the answer, the rest is beating about the bush!
But that ignores complex numbers
Why hide the answer by adverts to another video?
That is a bother she had address, otherwise good video
Why do I need this to study accounting at Harvard...or anywhere else???
Very well done. Thank you.
thank you
U are welcome
Retired physics/maths teacher here. Looked at thumbnail, thought a moment and answered, “negative two”.
These were my thought processes:
1. For all positive numbers >1, X³ > X² and there is no possible solution, therefore X must be negative.
2. For negative numbers the square is positive but the cube is negative.
3. Subtracting a negative number is the same as adding its positive opposite.
4. So the problem can now be expressed as |X|² + |X|³ = 12.
5. 1 is too small, 3 is too big. 2 works. But remember it’s negative.
Read a few comments and decided not to bother watching the video.
Retired Physics/Chemistry teacher here, California. (I can see that you are from UK....I taught in Derbyshire for a while)....I followed your exact line of thought.
@@DGill48
I grew up in Sheffield (technically Yorkshire, but the Peak District was our playground). But I’ve been a Kiwi for 46 of my 80 years.
@@q.e.d.9112 Visited Auckland just once; wanted to find Mr. Leo Bult.....he saved my Dad after his B-24 was shot down over the Netherlands. Found him just days before he passed. He was 97.; I tahnked him. The big steel mill in Sheffield is now a shopping mall (!!) Born in America, American for Life.
x^2*(1-x)=12=4*3 --- x^2=4 and 1-x=3 --- x=2 because 2^2=4
😂😂😂 how this pays the bills, I don't know
You should not be celebrating ignorance. If it means nothing to you does not mean it would not mean something to high school students preparing for uni!
x=-2
Thanks you so much for your incredible help in calculus
That's not calculus
X.1+X.1-X
X=-2
-2 by inspection
Oh! I see another channel where the host is a genius who, thanks to supernatural powers, knows exactly what to factor out to solve the problem.
I prefer teaching that shows how to solve problems without relying on supernatural powers.
x^2-x^3=12
x^2(1-x)=12
After factoring, you can see that both x^2 and (1−x) must be divisors of 12. Since 12=2*2*3, then 12=2^2*3
so x^2=4; 1-x=3
From the first equation, we get x^2=4 (, and from the second equation, we find x=2, which we have to reject, but x=−2 is the solution.
Now we know which factor to factor out. to find complex solutions,...
but we do not need it
complex solutions are x2 = a+b*i and x3=a-bi
x2*x3=a^2+b^2
x2+x3=2a
And we know the sum and product of the roots.
x^3-x^2+12=0
sum is 1 product is -12
now we can add and multiply solutions and find a and b:
-2+a+a=1 so a=3/2
-2*(a^2+b^2)=- 12
(a^2+b^2)=6
9/4+b^2=6
b^2=(24-9)/4=15/4
b=sqrt(15)/2
Why must x^2 and (1-x) be integer divisors of 12? In theory, x^2 could be say pi, and (1-x) 12/pi which obviously doesn't work but I don't think it is valid to pair these factors with integer divisors of 12, given your correct distaste for supernatural powers!
@@silver6054
Because we asked ourselves whether there are solutions in integers.
Neither the method presented in the video nor the one I presented are universal methods. Can you imagine trying to factor a cubic equation if its solution was π and the coefficients were integers?
There are universal methods for solving cubic equations, and from them, it is known that a cubic equation with integer coefficients cannot have π as a solution because every solution can be expressed using addition, subtraction, multiplication, division, and taking square or cube roots of equation coefficents.
These methods often give solutions where it can be extremely challenging to prove that these operations result in an integer.
The solution I provided is indeed a solution by "guessing," but its value lies in the fact that we only need to check two values using the method specified. Moreover, the solution I provided allows us to determine the other two solutions, whatever they may be. You can use this method to construct an example of an equation with coefficients for x^3, x^2, and the constant term as in the problem from the video, and determine those solutions when the real solution is ππ. However, in that case, the coefficient of xx would have to be transcendental, not zero, as in the equation being solved.
If you have any further questions, please feel free to ask.
It s obvious that -2 is a solution of this équation. So let factorize by X+2....
👍👍
can you show us?
It is -2 dear lady
That’s a lot of work when you can see that -2 is a solution and then use polynomial division. Don’t complicate things, math is hard enough as it is.
there is a much faster way. Just do X2 (1 - x ) = 12 One of the factors has to be 3 the other one 4. Take the 1st one. x could be either 2 or -2. Making an educated guess is allowed when polynomials are around. Even Newton allows it.
But then you wouldn’t have found the two imaginary solutions
@@vinuzo9548 that´s true!
then how do you get the other 2 roots?
Is the real solution! Not the prime ones.
Easy -2
Plus two other solutions.
The answer is -2
Are you kidding? This should be KG entrance
Wow it's interesting
Thanks
Easy/x^2(1--x)=2^2or(--2)^×3etc.
-2
Is it x=-2 ?
It’s -2 is it not?
Only way a square minus a cube can be positive is if x is a negative number.
Now I know why I never went to Harvard.
Stop the BS. Harvard does not have a specific entrance examination. Like all universities they work off SAT scores.
12= -2 × -2 x 3
So, we will get x= -2.
Later, I will watch exact solution.
I will have to give some break.
Thanks.
x² - x³ = 12
x³ - x² + 12 = 0 → let: x = z + (1/3)
[z + (1/3)]³ - [z + (1/3)]² + 12 = 0
[z + (1/3)]².[z + (1/3)] - [z² + (2/3).z + (1/9)] + 12 = 0
[z² + (2/3).z + (1/9)].[z + (1/3)] - z² - (2/3).z - (1/9) + 12 = 00
[z³ + (1/3).z² + (2/3).z² + (2/9).z + (1/9).z + (1/27)] - z² - (2/3).z - (1/9) + 12 = 0
[z³ + z² + (1/3).z + (1/27)] - z² - (2/3).z - (1/9) + 12 = 0
z³ + z² + (1/3).z + (1/27) - z² - (2/3).z - (1/9) + 12 = 0
z³ - (1/3).z + (322/27) = 0 → let: z = u + v
(u + v)³ - (1/3).(u + v) + (322/27) = 0
(u + v)².(u + v) - (1/3).(u + v) + (322/27) = 0
(u² + 2uv + v²).(u + v) - (1/3).(u + v) + (322/27) = 0
(u³ + u²v + 2u²v + 2uv² + uv² + v³) - (1/3).(u + v) + (322/27) = 0
u³ + v³ + 3u²v + 3uv² - (1/3).(u + v) + (322/27) = 0
u³ + v³ + 3uv.(u + v) - (1/3).(u + v) + (322/27) = 0
u³ + v³ + (u + v).[3uv - (1/3)] + (322/27) = 0 → suppose that: [3uv - (1/3)] = 0 ← equation (1)
u³ + v³ + (u + v).[0] + (322/27) = 0
u³ + v³ + (322/27) = 0 ← equation (2)
(1): [3uv - (1/3)] = 0
(1): 3uv = 1/3
(1): uv = 1/9
(1): u³v³ = 1/27² ← this is the product P
(2): u³ + v³ + (322/27) = 0
(2): u³ + v³ = - 322/27 ← this is the sum S
u³ & v³ are the solution of the equation:
x² - Sx + P = 0 ← do not confuse with x in the original equation
x² + (322/27).x + (1/27²) = 0
Δ = (322/27)² - [4 * (1/27²)] = (322² - 4)/27² = 103680/27² = (27 * 256 * 15)/27² = (256/9) * 5 = (16/3)² * 5
x = [- (322/27) ± (16/3).√5]/2
x = - (161/27) ± (8/3).√5
u³ = - (161/27) + 8√5 → u = [- (161/27) + 8√5]^(1/3)
v³ = - (161/27) - 8√5 → v = [- (161/27) - 8√5]^(1/3)
Recall:
x = z + (1/3) → recall: z = u + v
x = u + v + (1/3)
x = [- (161/27) + 8√5]^(1/3) + [- (161/27) - 8√5]^(1/3) + (1/3)
→ x = - 2
I did in 10sec. No need this complex method. Minus 2 is the ans
cube is greater than square so x has to be negative. so what adds to 12? hmm, computer grad knows that 2^2 = 4 and 2 ^ 3 = 8 which is basic binary.
X=-2 is correct answer
No, it's only half the answer.
Join me bro. That's why we (you and I) did not enter Physics and Chemistry class.
Abeg commot me dey shaa...wetin you dey come shakara person y head here!
Lol. Puts the pop-ups over the final answers.
why all this fuzz... minus two is the right answer. (x = -2)
X=-2
Please stop saying "minus" for "negative" numbers!
Answer is -2, I did this in my head. (-2)^2 - (-2)^3 = 4 + 8 = 12.
What about the other two solutions?
Harvard doesn’t have an entrance exam. You just have to be rich and connected
👍
The answer is -2 and I did it in my head in under 10 seconds.
I also.
Una soluzione immediata è x=-2
The answer is -2. Thus isn't difficult at all.
Too difficult for you, since you only found half the answer. ^^
Pls where are you based. My mathematics is a bit rusty now!
No. No. X= -2
LOL, did you even watch the video? and what are the two other roots?
Test your answer😂😂@@noomade
@@tinkerman9525 I have tested it.. .there are three roots and you have only listed one. Meaning that you didn't really understand the question and you would fail this question in an interview.
But at least you would win at emojis
@@noomade or 2 is another factor
@@tinkerman9525 LOL 2 is definitely not another factor... come on try again. You were the one shouting No NO. Now try and answer the question properly.
I have given you a clue (it is also in the video). There are three roots. -2 is one of them. what are the other two??
Je reprends à mon compte un commentaire judicieux qui a été fait à propos de SQR (-1). C'est une notation qui est souvent utilisée par nos amis anglo-saxons, malheureusement cette notation est MATHÉMATIQUEMENT FAUSSE . En effet l'équation x²=-1 admet DEUX SOLUTIONS complexes qui sont DIFFÉRENTES , elles ne peuvent donc pas être représentées par un seul et même symbole sqr(-1). C'est donc un abus de langage bien pratique mais faux. 😉
Why can the one symbol ( for that is all it is) not represent both solutions?
Thank you so much.
U are welcome
(3±√15 i)/2 by inspection
Follow the KISS principle
By inspection it’s -2! No need for all this baloney
Well, you need to prove that's the only answer.
@@luarluarwick8304 Did the question in the exam state that?
@@karhukivi The Math Science says that!
@@karhukivi Obviously and sadly, you don't know the first ABC of math and science. School years wasted with gandjabus?
X=3
Good teacher.
I'm guessing -2 would probably do the trick, lol...
Is the real solution! Not the prime ones.
To find the other solutions, lol, one can use the sum of roots, product of roots formulas, etc, I guess...
Interesting ok sister