3:55 Hi Mr.Eddie, I have another solution to prove as the following: beta^2 - gamma^2 = (beta+gamma)^2 - 4beta.gamma = 0 by substituting in the second eq with beta= - gamma 4gamma^2= 0 gamma = 2i and then beta = -2i
Apart from using given distinctness, one could say that gamma = conjugate of beta, and since beta^2=gamma^2 (beta-conj. beta)(beta+conj beta)=0, now since beta has been proven to be imaginary, first bracket isnt going to be zero since its twice the imaginary part of beta, hence beta+conj beta =0 and beta+gamma=0
It's very common that you get complex conjugate roots in polynomials like this. For quadratics of real coefficients, if you have complex roots, it's guaranteed that they are conjugates. Same with cubics of real coefficients. Cubic polynomials with real coefficients have at least one real solution, and in the event that they have only one real solution, the complex roots are guaranteed to be conjugates.
You'd think so at first thought, but since beta is negative gamma and they're distinct from one another it can only be one or the other but which one doesn't really matter because we don't have any further assumptions.
@@AndresFirte Really!? Can you explain why the unit i is depicted to have same length as the x axis unit? Are they equal or is that another "intuitive" way to depict absurdness?
@@pelasgeuspelasgeus4634 i and 1 have the same magnitude (defining magnitude as √(Cr^2 + Ci^2), where Cr is the real part of a complex number and Ci is the imaginary part, which is a generalization of the notion of distance. So we put them at the same distance from the origin to depict that.
@@pelasgeuspelasgeus4634 I didn’t say i=1. I said the *magnitude* of i is equal to the *magnitude* of 1, I even wrote the formula to calculate the magnitude. We also denote magnitude using vertical bars |x|. So I didn’t say i = 1. I said |i| = |1|. Which is also equal to |-1| and to |-i| btw
No, as the sqrt of a negative number does not output a negative number, instead the result is an imaginary number. For example, what multiplied by itself equals -1? There is no real answer, instead, it is denoted by 'i', short for imaginary.
It has to do with how the roots of equations are related to the coefficients of the equation. For e.g, take a quadratic equation of the form ax^2 + bx + c = 0 Now, if you take the roots α and β one at a time (i.e separate from each other): α + β = - b/a If you take them two at a time, i.e alpha*beta: α*β = c/a Let us see cubic equations,if they are of the form ax^3 + bx^2 + cx + d = 0. and the roots are α, β and γ: One at a time: α + β + γ = -b/a Two at a time: α*β + β*γ + α*γ = c/a Three at a time: α*β*γ = -d/a Observe, the denominator always stays 'a' but the numerator are the coefficients cycled from 'b' through the constant term. And they go from negative to positive, i.e from -b to c to -d to e to -f and so on. The equation given in the video is of the form: (1)x^3 + ax^2 + bx + (-54) so taking the roots one at a time: α + β + γ = -b/a here a = 1 and b = a substituting, α + β + γ = -a/1 = -a
alpha, beta, and gamma are roots, therefore, we can write it in factored form. Since the factorization equals zero, we can equate it to the given polynomial. For two polynomials to be equal their constants must equal, so when you expand the factorization you can equate the coefficients to get a system of equations. The equation you cited is in that system. Its sort of simple but important nonetheless. He shouldn't have omitted it.
oh my, that question is crazy. I understood every step but I don't know how I'm going to think like that in the exam :(
Well, you don't. Just practice a bit ans steps will come in smoothly.
Great question, great video, great teacher!
Wow! Loved every second of the video.
This is pure excellence
3:55
Hi Mr.Eddie, I have another solution to prove as the following:
beta^2 - gamma^2 = (beta+gamma)^2 - 4beta.gamma = 0
by substituting in the second eq with beta= - gamma
4gamma^2= 0
gamma = 2i and then beta = -2i
well explained
Apart from using given distinctness, one could say that gamma = conjugate of beta, and since beta^2=gamma^2 (beta-conj. beta)(beta+conj beta)=0, now since beta has been proven to be imaginary, first bracket isnt going to be zero since its twice the imaginary part of beta, hence beta+conj beta =0 and beta+gamma=0
It's very common that you get complex conjugate roots in polynomials like this. For quadratics of real coefficients, if you have complex roots, it's guaranteed that they are conjugates. Same with cubics of real coefficients. Cubic polynomials with real coefficients have at least one real solution, and in the event that they have only one real solution, the complex roots are guaranteed to be conjugates.
Me justing finished year 10 was like, what the heck is this?
For (iii) shouldn’t beta be plus or minus i times root a?
Doesn’t change much but just a thought
You'd think so at first thought, but since beta is negative gamma and they're distinct from one another it can only be one or the other but which one doesn't
really matter because we don't have any further assumptions.
I am doing IB AA HL but I COULD NOT UNDERSTAND THIS INSANE QUESTION RIP
Excuse me, can you show those roots on a graph???
Yes, we can show them in a graph. Watch the series of videos called “Imaginary numbers are real” by Welch Labs. It shows how to graph it
@@AndresFirte Really!? Can you explain why the unit i is depicted to have same length as the x axis unit? Are they equal or is that another "intuitive" way to depict absurdness?
@@pelasgeuspelasgeus4634 i and 1 have the same magnitude (defining magnitude as √(Cr^2 + Ci^2), where Cr is the real part of a complex number and Ci is the imaginary part, which is a generalization of the notion of distance. So we put them at the same distance from the origin to depict that.
@@AndresFirte You really don't see it? You said that i=1. But i is also equal to sqrt(-1). So, sqrt(-1)=1=sqrt(1). So - 1=1. Right?
@@pelasgeuspelasgeus4634 I didn’t say i=1. I said the *magnitude* of i is equal to the *magnitude* of 1, I even wrote the formula to calculate the magnitude. We also denote magnitude using vertical bars |x|.
So I didn’t say i = 1. I said |i| = |1|. Which is also equal to |-1| and to |-i| btw
If we sqrt both sides at won’t the sqrt and square cancel and we get -a at 5:15?
No, as the sqrt of a negative number does not output a negative number, instead the result is an imaginary number. For example, what multiplied by itself equals -1? There is no real answer, instead, it is denoted by 'i', short for imaginary.
why do we take 54 while it is -54 in 7:11
you gotta take -d/a if its product of roots in a cubic
MITSAKE
Why is Alpha + Beta + Gamma = -a 🤔
It has to do with how the roots of equations are related to the coefficients of the equation.
For e.g, take a quadratic equation of the form ax^2 + bx + c = 0
Now, if you take the roots α and β one at a time (i.e separate from each other):
α + β = - b/a
If you take them two at a time, i.e alpha*beta:
α*β = c/a
Let us see cubic equations,if they are of the form ax^3 + bx^2 + cx + d = 0. and the roots are α, β and γ:
One at a time:
α + β + γ = -b/a
Two at a time:
α*β + β*γ + α*γ = c/a
Three at a time:
α*β*γ = -d/a
Observe, the denominator always stays 'a' but the numerator are the coefficients cycled from 'b' through the constant term. And they go from negative to positive, i.e
from -b to c to -d to e to -f and so on.
The equation given in the video is of the form:
(1)x^3 + ax^2 + bx + (-54)
so taking the roots one at a time:
α + β + γ = -b/a
here a = 1
and b = a
substituting,
α + β + γ = -a/1 = -a
alpha, beta, and gamma are roots, therefore, we can write it in factored form. Since the factorization equals zero, we can equate it to the given polynomial. For two polynomials to be equal their constants must equal, so when you expand the factorization you can equate the coefficients to get a system of equations. The equation you cited is in that system. Its sort of simple but important nonetheless. He shouldn't have omitted it.
Okay thank you very much !
@@Гио-ж5ю I like how you're asking a question about Gamma, and your name starts with a Gamma.