You can use Horner scheme to get y^4+py^2+qy+r=0 equation
@dkiproch4 жыл бұрын
is it possible to solve any quartic equation using this method?
@holyshit9223 жыл бұрын
Yes, moreover reducing to the form y^4+py^2+qy+r=0 is not necessary Reducing quartic to the form y^4+py^2+qy+r=0 is helpful when you compare quartic to the product of two quadratics written using undetermined coefficients
@michaelrosen49167 жыл бұрын
Thank you for explaining the substitutions, really helpful.
@calvinjackson81103 жыл бұрын
All of this song and dance could be completely cut out if you used the substitution x = y - (a/4). It gives the desired form the FIRST time!
@mariomario-ih6mn3 жыл бұрын
Nice video I didn’t get really understand your video when I was 11 now I’m 13 and I fully understand it.
@heavennoes3 жыл бұрын
cool, I'm 10 and I'm here to see what I can do.
@mariomario-ih6mn3 жыл бұрын
@@heavennoes I turned 14 now
@holyshit9224 жыл бұрын
Try to reduce general quartic in x to quadratic in y^2 I tried and got sextic which is difficult to solve for me
@finnboltz4 ай бұрын
How did you do it?
@holyshit9224 ай бұрын
@@finnboltz a_{4}x^4+a_{3}x^3+a_{2}x^{2}+a_{1}x+a_{0} = 0 Let x = (pt+q)/(t+1) Equate coefficients of t^3 and t to zero You will have 10th degree polynomial but P(t) = a_{4}p^4+a_{3}p^3+a_{2}p^{2}+a_{1}p+a_{0} is factor of this polynomial so after division I reduced resolving equation to 6th degree equation but i am unable to solve this sextic equation Resolving equation is sextic in p
@jonjonmia3 жыл бұрын
Can someone help me find the general formula for q?????? given p=-3a^2)/8 + b
@bobbyheffley4955 Жыл бұрын
[(a^3)-4ab+8c]/8
@cameronspalding97924 жыл бұрын
I think your missing a +y^4
@grankoczsk3 жыл бұрын
He cancelled them out
@kawawashaibu68546 жыл бұрын
Sante
@kjchung92304 жыл бұрын
this get even complicated!!
@elliotnicholson51174 жыл бұрын
Sorry :) I hope it isn’t too difficult to follow.
@Anonymous-wj6bu4 жыл бұрын
Solving quartic equations involve a lot of algebraic manipulations so I don’t know see how it could be less complicated