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,
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.
Factorising the cubic elements and by using the difference of two squares she's broken it down into one linear multiplied by one quadratic equation equals zero.In order to get zero one of them must be zero. She's pretty good. Next up completing the square to prove the quadratic formula- easy sums!Good 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 😃
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.
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.
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.
Thanks for the content you are really good at math. I understood how you resolved it even though I didn't do math since a long time, but it would be great for others if you explain more slowly, that would make you a great mathematician and a great teacher.
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
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.
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
Any number squared is a positive, a number cubed can be either polarity. By inspection, x must be negative. Let's try -2. (-2)^2 - (-2)^3 = 12 4 - (-8) = 12. It fits.
From the comments, it is clear that at least about 2/3 of them come from people who never heard about imaginary numbers, neither understand that there must be three solutions to the equation. An interesting aspect is the psychology behind these, mostly cocksure, responses - these individuals don't know what they don't know.
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 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.
Is this Harvard or Hardward? Answer of -2 in a couple of seconds. You know it has to be negative. It is not -1, and -2 fits in nicely. Cubic equation and hence 3 solutions for X. You figure X=-2. Factorize and solve tge quadratic to get the imaginary numbers. This will be a 1 min video.
I guess harvard wants to see you regurgitate and apply basic math formulas to determine you atleast know the basics. Im also guessing there are many of these questions on the exam with very little time to aee if you can handle stress not being able to finish. This question called for no real thinking,,its like riding a bike...
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.
Start with a negative number, as sqare is positive and cube is negative. So both terms are positive and add up. Thus, it is x sqare plus x cube is 12. So x is 2.
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.
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.
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
My sister, my head was turning, turning, turning! All those yardsticks I have no idea why and what it is ooo! I think you mathematicians are using those people from out of this world's brains!
Straight forward; good explanation. You explained everything you did, and you backed it up with the formulas that you would use.
Thank you for taking the time to make this video. Much appreciated.
How brilliant you are ! Your experience & explanation is very very best.
the correct answer for x is - 2
There are two, well three answers, 2 is the observe one if you have any brains.
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.
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,
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
Factorising the cubic elements and by using the difference of two squares she's broken it down into one linear multiplied by one quadratic equation equals zero.In order to get zero one of them must be zero. She's pretty good. Next up completing the square to prove the quadratic formula- easy sums!Good 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 😃
👍👍
12 = (-2)^2 -(-2^3)
X^2-X^3 = (-2^2) - (-2^3)
X = -2
Only one root?
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.
Thank you so much for explaining it so well and patiently.
Regards 🙏
U are welcome
This is painful
-2 in a few seconds
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.
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😊
@rajamnaidu1962Look to the far left. It is there!
@@JJONLINEMATHSCLASSchannel:
It's the same all over this platform, don't worry.
@@JJONLINEMATHSCLASSchannelyou know your stuff.
I did it in my head. No pen, no paper. Answer is -2 . ( -x- = +)(-×+=-) x=-2
you are a smart lady!!!
Thanks for the content you are really good at math. I understood how you resolved it even though I didn't do math since a long time, but it would be great for others if you explain more slowly, that would make you a great mathematician and a great teacher.
She is demonstrating methodology. It becomes more relevant when solving more complex problems which cannot be solved at a glance.
Thank you
Harvard doesn't have an "entrance exam".
x=-2 or x = (3 +/- sqrt(15)i)/2 where i^2=-1. Hint: use synthetic division.
Great presentation
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
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.
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.
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.
x^2-x^3=12 x=-2 x=(3±Sqrt[15])/2i=1.5±0.5Sqrt[15]i final answer
nice. It has been over 45 years since I did that. I now remember.
For instances where the answer is not as simple as -2, this more elaborate approach is probably applicable :-)
What the hell I just witnessed? One glance is enough to see that x=-2.
😂
One value of x is -2. So x+2 is a factor. By dividing I would find the other quadratic factor. Will factorize ......
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.
one root is -2. not sure about the two others, but i am sure you will explain in video
if x
x = -2 satisfies the equation
X2(1_x)=4×3.
1_x=3.
_x=3-1=2.
X=-2
Great! I was wondering how to solve it, thanks for a clear explanation.
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
Great job, please ignore all those who use this platform to act like they know better than you do, they can go make their own videos
Any number squared is a positive, a number cubed can be either polarity.
By inspection, x must be negative.
Let's try -2.
(-2)^2 - (-2)^3 = 12
4 - (-8) = 12. It fits.
Life's too short.
Really too short!!! I'm none the better for her explanation 😢😢 jah jah!!
x= -2 in my head < 5 seconds
From the comments, it is clear that at least about 2/3 of them come from people who never heard about imaginary numbers, neither understand that there must be three solutions to the equation. An interesting aspect is the psychology behind these, mostly cocksure, responses - these individuals don't know what they don't know.
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.
Very well done. Thank you.
Thanks you so much for your incredible help in calculus
That's not calculus
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.
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.
My dear guess the first root.
Divide to find the quotient
Use quadratic formula to solve the quotient.
X must be negative
And answer is -2 just by inspection.
It is a 3rd degree polynomial, so there are 3 roots. -2 is just one of them.
As long as you must have "i" in the "solution" it is not a real solution
Is this Harvard or Hardward?
Answer of -2 in a couple of seconds.
You know it has to be negative.
It is not -1, and -2 fits in nicely.
Cubic equation and hence 3 solutions for X. You figure X=-2. Factorize and solve tge quadratic to get the imaginary numbers. This will be a 1 min video.
Came to the same conclusion, 6th grade level math and that's harvard standard? 😂😂
I guess harvard wants to see you regurgitate and apply basic math formulas to determine you atleast know the basics. Im also guessing there are many of these questions on the exam with very little time to aee if you can handle stress not being able to finish. This question called for no real thinking,,its like riding a bike...
Hee Haw
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?
If x = -2 then we have 4 --8 = 4 plus 8 equals 12. At least that's,a solution.
👍👍
Start with a negative number, as sqare is positive and cube is negative. So both terms are positive and add up.
Thus, it is x sqare plus x cube is 12. So x is 2.
excellent
Wow, I forgot about the math entirely.
Excellent thank you...👍
U are welcome
You would have to know all the theories,Etc.,to solve this type of problem.
U r woman with beautiful face and mind too, genius!!!!!!
BRILLIANT!!!
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.
good stuff, I enjoyed it
thank you
U are welcome
Brilliant. Obviously used to the board, algebra and the marker. You belong to Harvard.
X is -2
I'm 20 seconds in and I can already intuit that the answer is -2.
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
❤❤ you are doing well , my colleague , A mathematician .
Thank you so much
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
The key is knowing that the cube can be either polarity, which makes -2 obvious....
I find it hard to believe that JJ has subscribers. James Joyce however makes perfect sense.
(1-x)x^2=3*4,
x^2=4, x=±2,
1-x=3. x=-2.
It s obvious that -2 is a solution of this équation. So let factorize by X+2....
👍👍
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.
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
-2 by inspection
Lol. Puts the pop-ups over the final answers.
x^2*(1-x)=12=4*3 --- x^2=4 and 1-x=3 --- x=2 because 2^2=4
Finally someone I can hear!
🤣
Why do I need this to study accounting at Harvard...or anywhere else???
Wow it's interesting
Thanks
very good
X.1+X.1-X
X=-2
Easy -2
Plus two other solutions.
Stop the BS. Harvard does not have a specific entrance examination. Like all universities they work off SAT scores.
-2, took me 3 seconds
Please remove those images that are covering the result. Don't you see we can't see?
Ok
@@JJONLINEMATHSCLASSchannel Thank you. Nice explanation, but we can't see the end.
Answer, X =-2, X= (3+sqrt15.i)/2 and X= (3-sqrt15.i)/2.
@@striker44 Thanks for your attention. Everything was brilliant up to the stage we couldn't see the result.
super easy. I could see it right away. -2.
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.
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.
And i wouldn't have gotten non-Real Roots.
I need them but I'm planning to be a Theorist.
Thanks !
No. No. X= -2
Test your answer😂😂@@noomade
@@noomade or 2 is another factor
It is -2 dear lady
Thank you so much.
U are welcome
Abeg commot me dey shaa...wetin you dey come shakara person y head here!
My sister, my head was turning, turning, turning! All those yardsticks I have no idea why and what it is ooo! I think you mathematicians are using those people from out of this world's brains!
🤣🤣🤣, we love you. Please your head should stop turning
@@JJONLINEMATHSCLASSchannel Hahahahahha!😀🤣😂
Do you privet tutoring??