@kobethebeefinmathworld953 if lambda is rational and positive, then the solutions to the quartic involve the following possibilities: integers, fractions, and expressions with square roots.
Good question. Lambda does not need to be rational in order to solve the quartic equation. In fact, since lambda is determined by the corresponding cubic equation, as long as we substitute the coefficients into the cubic formula, we can find out what lambda is.
zero is the second iteration, step, to solve the equation variable. too determine them. it is actually not zero. we add equal to zero to solve an unknown varable. it starts without. the = 0. wich is the first step. first literation. the whole formula is not equal to zero. but has a different answer. but this is probably not a formula. but points on a graphic out of a table. wich means their are individual coordinates. wich are unique units.
x means x. until x is a number. x is a variable. a yet to defind number. otherwise it is just a... formula without an answer. just as is. it only make sense if you replace the numbers. with figures, numbers.
I'm in third grade Middle school and this is complex meybe .. so i want to know , why we don't have ( ∆ ) like ax²+bx+c=0 ∆=b²-4ac X1=(-b+√∆)/2a ou X2=(-b-√∆)/2a , if ∆ = 0 we have only one solution if he negative we have complex number but why and why we don't have this in ax⁴+...+e=0
That's a very good question. The quick answer is: Yes, there is (sort of), which is the lambda that's used as a parameter of finding the roots of the cubic equation that yielded from the original quartic equation.
As a parameter, you can use any 1 of the 3 solutions of λ and yield the same 4 solutions of x, so you can try to use the nonzero λ to get the results. Now, if all 3 solutions of λ are equal to 0, it provides that all p, q, and r are equal to 0, implying that there are some kind of conditions for a, b, c, and d su h that you can simplify the original equation.
more specifically, λ can only be 0 if the absolute term is 0. The absolute term is -q^2, but if q=0, that means we have a biquadratic equation which can be solved easily. When q≠0, we never get any zero-solutions for λ, so you never have to explicitly choose.
@kobethebeefinmathworld953 if p, q, and r all equal 0, the original equation has four roots all equal to -a/4, since this implies that a, b, c, and d are all 0.
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Quartic polynomials are of degree 4, this video's duration is 44:44. Perfection.
Finally someone noticed it.
@@kobethebeefinmathworld953it seems you also have 4 markers
bro hesitated to write the constant as 'e' in the thumbnail, relax nobody thought it was euler's number
Well done one 250 subsxribers! Keep up the good work!
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If you can obtain a rational root for the resolvent cubic, it can simplify work.
Not exactly. It needs some kind of symmetry with conditions that only degrees 4 and below fit, but not 5 or above
@kobethebeefinmathworld953 if lambda is rational and positive, then the solutions to the quartic involve the following possibilities: integers, fractions, and expressions with square roots.
@bobbyheffley4955 I thought you were referring to quintic equation. My mistake.
Yes, that can reduce the solution into something nicer.
Just get rid of the 3degree term by using a linear transformation like one did for cubic.
Yes, that's correct
is there a video for cubic formula?
Yes, there is
♥️
underated
Thanks
Hi! I have a question. How did you come up with 3/8a^2x^2? I got 21/16a^2x^2 instead. Thanks for helping out in advance😊
Good question! It is based on the expansion of (x +a/4)⁴, the x² term of it would then be 6×x²×(a/4)²=(6/16)a²x²=(3/8)a²x²
@@kobethebeefinmathworld953 Great! I mislooked the 4^2 of a^2/4^2 when I multiplied it by x^2
Hi, I have a question, if the third degree equation in lambda, has no rational solutions; how should we proceed?
Good question. Lambda does not need to be rational in order to solve the quartic equation. In fact, since lambda is determined by the corresponding cubic equation, as long as we substitute the coefficients into the cubic formula, we can find out what lambda is.
@kobethebeefinmathworld953 of course, lambda must be nonzero
zero is the second iteration, step, to solve the equation variable. too determine them. it is actually not zero.
we add equal to zero to solve an unknown varable. it starts without. the = 0. wich is the first step. first literation. the whole formula is not equal to zero. but has a different answer. but this is probably not a formula. but points on a graphic out of a table. wich means their are individual coordinates. wich are unique units.
x means x. until x is a number. x is a variable. a yet to defind number. otherwise it is just a... formula without an answer. just as is. it only make sense if you replace the numbers. with figures, numbers.
At 500 subs you should derive the quintic formula :D
Galois proved there can be no such thing.
@@carultchAt 500 subs he should prove that there is no formula
Correct!
I'll wait until 2500
Niels Henrik Abel proved that the quartic is the highest degree equation that has a general solution in radicals.
I'm in third grade Middle school and this is complex meybe .. so i want to know , why we don't have ( ∆ ) like ax²+bx+c=0
∆=b²-4ac
X1=(-b+√∆)/2a ou X2=(-b-√∆)/2a , if ∆ = 0 we have only one solution if he negative we have complex number but why and why we don't have this in ax⁴+...+e=0
That's a very good question. The quick answer is: Yes, there is (sort of), which is the lambda that's used as a parameter of finding the roots of the cubic equation that yielded from the original quartic equation.
@@kobethebeefinmathworld953 can you make videos for the solution of equation 3 , 4 , 5 degree or 3 , 4 please with this method ?
@@kobethebeefinmathworld953 and thank you brother for reponse xD
I'm your 444 liker,
my phone's battery is at 44%. perfect
what do you do if λ = 0
As a parameter, you can use any 1 of the 3 solutions of λ and yield the same 4 solutions of x, so you can try to use the nonzero λ to get the results. Now, if all 3 solutions of λ are equal to 0, it provides that all p, q, and r are equal to 0, implying that there are some kind of conditions for a, b, c, and d su h that you can simplify the original equation.
more specifically, λ can only be 0 if the absolute term is 0. The absolute term is -q^2, but if q=0, that means we have a biquadratic equation which can be solved easily. When q≠0, we never get any zero-solutions for λ, so you never have to explicitly choose.
@kobethebeefinmathworld953 if p, q, and r all equal 0, the original equation has four roots all equal to -a/4, since this implies that a, b, c, and d are all 0.
@@bobbyheffley4955 Yes, that is correct.
@Fluorineer if q is nonzero and you get a rational value for lambda, use it to simplify calculations.
44 minutes 💀💀💀
Difficult to solve.
Add Arabic subtitles to your videos please
I will try to find a way to add the option, but I don't speak Arabic, so I need someone to type the subtitles for me.
@@kobethebeefinmathworld953 deer sir you do not need to type subtitles youtube provides subtitles for videos .just activate it
👍👍🫡❤