I Solved A Homemade Quartic in Two Ways
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- Опубліковано 30 вер 2024
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Expanding the original equation you get x^4 - 12*(x^2) -x +30. I attempted to factor this as the product of two quadratics. I came up with
x^4 - 12*x^2 -x -30 = (x^2 -x - 6)(x^2 +x -5) So the four real roots are x=-2 x=3 x=.5*(-1+sqrt(21)) x=.5*(-1-sqrt(21)) The graph looks something like
a "W" and hits the x-axis four times.
Substitution is always coooool😎
I thought of another method of completing the square. Since there is only one x term in the whole equation 1/4, is a potential candidate for constant on RHS. Then the equation becomes
x^4-11x^2+121/4=x2+x+1/4 and it worked.
I also tried making RHS (x/2+1)^2 but it didn’t work.
x^4-12x^2-x=-30
x = -2 -> 16-48+2=-30
x = 3 -> 81-108-3=-30
(x^2-x-6)(x^2+x-5)=0
x = [-1+-{3^(1/2)*7^(1/2)}]/2
Let’s substitute a=6 and expand. You will get a quadratic equation of a.
Solve for a in terms of x then back substitute 6 in place of a.
You will get the same two quadratic equations of x.
nice
your tips can be collected in a book.😊
A book about what? Just a professor who likes mathematics brainwork without the use of computer days (before the 1960s)!!
Since he has a polynomial quartic x^4 - 12x^2 - x + 30 he can computationally evaluate all roots real by using Newton-Raphson Iterations, or the Secant Method ... or more computational negligible now time with today's computers the Muller's Method (capable of finding complex pair roots or the product of all real coefficient quadratic polynomials). He won't be so great formulating easy root finding for nth order polynomials, in general, where computations by computers roots are found by: for each real root the n-1 polynomial resulting in that root removed and for complex roots found to the k-2 reduced polynomials with the complex pair roots removed after finding them by Muller's Method for Iteratively computing to finding all polynomial roots of n-order polynomials.
I have a computer calculator ready for the index of coefficients of a0 + a1x + a2x^2 + a3x^3 + a4x4 + ... Where the "ai"s stop the request for more when subsequent "ai"s are no longer needed.
In my program the user just by index inputs a0 = +30.0, a1 = -1.0, a2 = -12.0, a3 = 0 and a4 = 1.0 where a4 became the last request by the computer program after the initial input data answer of the last power of the polynomial was input of polynomial size was determined.
Thank you!
Thank you for the kind words and the idea! 😍❤️
We can write the equation as x^2-6=sqrt(x+6) . If we consider f(x)= x^2-6, we have that its inverse is f^(-1)(y)=sqrt(y+6). Hence, we will solve the equation when we find x such that f(x)=f^(-1)(x). The points where a function equals its inverse are the intersection between the graph of the function and the straight line y=x (in fact, the graph of f and f^(-1) are symmetric with respect the straight line y=x).
Hence we solve x^2-6=x (or sqrt(x+6)=x) and we get x^2-x-6=0, which gives x=3, x=-2.
I hope I was clear, I am Italian . Buon Natale, SyberMath=Merry Christmas, SyberMath!
Ciao Silvia. Ho risolto l'equazione esattamente come te. Dimentichi però le altre due soluzioni: (-1+sqrt(21))/2 e (-1-sqrt(21))/2
yes ,but f^(-1)(y)=+\- sqrt (y+6) and we have 2 equations with 3 and -2 as asolutions then we keep the positive solution for the first equaton and the negative for the second.thank you from morrocco
@@hafedabrar4333 thank you, from italy.
Thank you for the replies and the remarks!! Syber has always challenging problems! Ciao!
a recursion of another variable
the rational root thm plus synthetic division take aboutfive lines of work (or 7 if you don't feel like trying 1 and -1 in your head)
Only for polynomials having real roots and roots that are integers or quotients (not real numbers). His way is more universal. However he cannot efficiently teach a computer in a program of algorithms to solve it not only for certain subset of all the infinite possible polynomials quartic and higher for finding the roots. Nor can his math. Nor can Synthetic Division. So in MAT 462 courses of Computer Science studying all numerical analysis of mathematics to be programmed into computers for computer solutions only at modular level computers are fast enough to flip through Iterative program algorithms to find real and complex pair roots (namely each quadratic polynomial paired of still real coefficients) by using the at least three different main Iterative techniques for root finding in any nth order polynomial size: Newton-Raphson, Secant and Muller's Iterative methods to find all polynomial roots programming algorithms.
I used long division after implementing RRT - I believe synthetic division can only be used for quadratics.
@@scottleung9587 Synthetic division works for any polynomial by a linear polynomial. It can be modified to divide by nonlinear polynomials.
x=-2.
(x-3)(x+2)(x^2+x-5)
The cubic equation in k you have at 3:48 with the first method looks rather daunting. However, most of the time when you are solving quartics from competitions which usually have nice factorizations into two quadratics with integer coefficients you do not need to solve the cubic resolvent formally. In this case we have
(x² + k)² = (12 + 2k)x² + x + (k² − 30)
where the quadratic in x at the right hand side is a perfect square if and only if k satisfies
1 − 4(12 + 2k)(k² − 30) = 0
But if the quartic equation is to factor into two quadratics with _integer_ coefficients, then the quadratic term (12 + 2k)x² must be the square of an integer multiple of x, meaning that the coefficient 12 + 2k of this term must then be the square of an integer.
So, for factorizations into two quadratics with _integer_ coefficients we really only need to test values of k which make 12 + 2k the square of an integer, i.e. 1, 4, 9 ... If we start with 2k + 12 = 1, then we must have k = −¹¹⁄₂ and with this value for k the quadratic on the right hand side of our equation has a constant term k² − 30 = (−¹¹⁄₂)² − 30 = ¼ and we have 1 − 4(12 + 2k)(k² − 30) = 1 − 4·1·¼ = 0 as required. Thus we have found what we were looking for almost immediately and without having to solve the cubic in k. With k = −¹¹⁄₂ the equation becomes
(x² − ¹¹⁄₂)² = (x + ½)²
or
(x² − ¹¹⁄₂)² − (x + ½)² = 0
and using the difference of two squares identity this factors as
(x² − ¹¹⁄₂ − x − ½)(x² − ¹¹⁄₂ + x + ½) = 0
which is
(x² − x − 6)(x² + x − 5) = 0
As you can see, k itself does not need to be an integer but can also be an integer multiple of ½ to have a factorization into two quadratics with integer coefficients.
Pretty good!
It can easily be transformed to (x^2-8)(X+2)(x-2)=x+2 from where x=-2; Then (x^2-8)(x-2)=1. First guess is to try x-2=1 and x^2-8=1 simultaneously . And it works - x=3. The rest is trivial.
nice. Did you subtract 4 from both sides and then factored?
@@SyberMath Indeed :)
@@vladimirkaplun5774 nice
saw the similar question on an other yt channel. that guy would solve this equation by expanding LHS: x⁴ - 2x² * 6 + 6² = x + 6 which actually gives us a quadratic equation in terms of 6! and so 6 is expressed as one of two really nice functions of x (quadratic polynomials) so from there we need to solve 2 quadratic equations in terms of x
Yes, you can do that, but if you don't want to lose sight of what you are doing and want to keep track of constants and variables properly you should first replace 6 in the equation
(x² − 6)² = x + 6
with a second variable, then solve the equation for this second variable, and finally put 6 back in place of this second variable. If we replace 6 with y, we have
(x² − y)² = x + y
and expanding the left hand side this gives
x⁴ − 2x²y + y² = x + y
which can be written as
y² − (2x² + 1)y + (x⁴ − x) = 0
and this can be considered as a quadratic equation ay² + by + c = 0 with a = 1, b = −(2x² + 1), c = x⁴ − x. To solve this quadratic equation in y we first calculate the discriminant D = b² − 4ac = (2x² + 1)² − 4(x⁴ − x) = 4x⁴ + 4x² + 1 − 4x⁴ + 4x = 4x² + 4x + 1 = (2x + 1)². Using the quadratic formula y = (−b ± √D)/2a we then get
y = ((2x² + 1) + (2x + 1))/2 ⋁ y = ((2x² + 1) − (2x + 1))/2
which is
y = x² + x + 1 ⋁ y = x² − x
We started by replacing 6 in our equation with the second variable y, and we can now again put 6 back in place of y to get
6 = x² + x + 1 ⋁ 6 = x² − x
which is
x² + x − 5 = 0 ⋁ x² − x − 6 = 0
and now we only need to solve these two quadratic equations in x to find all four solutions of our original equation.
@@NadiehFan exactly. thanks for clarifying tho :)
Nice!