Cubic equation from determinant
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- Опубликовано: 2 окт 2024
- This determinant is computed using cofactor expansion. The resulting cubic equation is then factored using synthetic division since one root is already given.
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Hi! The trick of this problem is not expanding the determinant to a cubic equation. If you sum the second line and third line to the first line, the value of the determinant remains the same and the first line is converted to a line of (x+9) values. Then, the equation can be solved without the synthetic division.
Aditionally, you can check determinant is multiple of (x-7) if you substract the third column to the first column. Maybe something similar can be done to know that (x-2) is multiple.
Wow! What a visualization
To make the matrix triangular:
Add row 1 to row 3
Add row 2 to row 3
Subtract column 3 from column 1
Subtract column 3 from column 2
Now the diagonal is x-7, x-2, x+9
Great ideae @@xavigb5991
The fact that it's a depressed cubic is actually really nice, it allows you to skip the synthetic division step. Let a and b be the other roots.
By Vietta's formulas, the sum of the roots is the co-efficient of the x^2 term. Since there is no x^2 term, that means that the sum of the roots is 0. So, a + b + -9 = 0, or a + b = 9.
Also by Vietta's formulas, the product of the roots is the negative of the constant term of the polynomial. So ab(-9) = -126, or ab = 14.
So, we want two numbers that add to 9 and multiply to 14. So, we get (a,b) = (2,7) or (7,2). Therefore, the roots are 2, 7 and -9.
Great channel. Thanks for sharing this problem! Just a friendly heads up, I think you want your thumbnail to say x = -9 is a root
Great video!❤ But in the thumbnail the given root is wrong (x=9 at the moment I'm writing this)
Yes this guy makes many mistakes in his thumbnails and doesn’t correct them… I don’t like that
From the determinant, we can see that 7 is the root because the 1st and 3rd columns are identical (and therefore the determinant is 0).
interesting problem, but the known root made the problem way too easy, as soon as you can calculate determinant 3*3 you sure would find the root x=2 quite fast
Hello sir can you help me I have my exam by next week and I know nothing about newton Raphson method and newton Gregory both foward and backward interpolation añd also how to solve errors in the interpolation table ...thanks
Since we can put the determent of matrix, with unknowns, into a polynomial then can we put a polynomial into a matrix?
What would be ˣx ? Integral and derivative?
So incredible
I been wishing to reach u out but no way is that necessary 😳
Kh is read approximately like χ (here i used Greek letter)
X=9 isgiven as one of the rootr and not X=-9
There's a really nice way to solve this with some linear algebra. Since the x values are always on the diagonal, this can be written as A - xI, where A has 0s on the diagonal and I is the identity matrix. Then the roots of this polynomial are (-1 times the) eigenvalues of A. We can directly compute that tr(A) = 0 and det(A) = 126, so the eigenvalues of A sum to 0 and multiply to 126, and we know one is -9, so we get -9, 2, and 7 (remembering the sign change).
Solving the depressed cubic will be more interesting 😉
Really enjoyed this video, I love linear algebra!!
Thanks Sir
I've enjoyed watching your "first principles" series.
Can you solve the cubic equation without using the given root? I want to see how to do that
Using the form of a general cubic y = Ax^3 + Bx^2 + Cx + D = 0, here A = 1, B = 0, C = -67, and D = 126.
Let F = (3C/A - (B^2/A^2))/3 and G = ((2B^3/A^3) - (9BC/A^2) + (27D/A))/27. So F = -67 and G = 126.
Let H = G^2/4 + F^3/27. So H = -7170.37037037. Since H is negative, all 3 cubic roots are real and we proceed as follows:
I = square root of (G^2/4 - H) = 105.54321565.
J = cube root of I = 4.72581563.
K = arccos of (-G/2I) in radians = 2.21044286.
L = -J = -4.72581563.
M = cos (K/3) in radians = 0.74061290.
N = square root of 3 * sin (K/3) in radians = 1.16382027.
P = -B/3A = 0.
Root 1 = 2JM + P = 7 (within rounding).
Root 2 = L(M+N) + P = -9 (within rounding).
Root 3 = L(M - N) + P = 2 (within rounding).
@@woodchuk1is there a quicker way to figure out if all roots are real? It is difficult to remember those formulas
@@bobross7473 Δ (or) D = B^2C^2 − 4AC^3 − 4B^3D− 27A^2D^2+ 18ABCD. If this is positive, all 3 roots are real and distinct.
@@bobross7473 Cubic discriminant D = B^2*C^2 − 4B^3*D− 27A^2D^2 + 18ABCD. Here D = 774400. Since it’s positive, the cubic has 3 real roots.
@@bobross7473 Discriminant for a cubic equation = b^2c^2 − 4ac^3 − 4b^3d − 27a^2d^2 + 18abcd. Since b = 0, D = -4(1)(-67)^3 - 27(1)^2*(126)^2. This is positive (the value is irrelevant), so it means that all three roots are real.
I seem to lack some skills here. 5:56
What is this about? Synthetic division? Never heard that.
😢 I'm puzzled. 🤔🤯
Look up Briot-Ruffini.
Horner's method : en.wikipedia.org/wiki/Horner%27s_method
It's just simpliif simplified polynomial division
@@niloneto1608
Thanks a lot!
I have never heard of him. I'll check it out.
@@PrimeNewtons
Thank you!
I haven't learned this before. You got me curious.
But, as you say: never stop learning.
❤️🙏
Wait, why did he divide by -9 during synthetic division? Seems incorrect