several have asked “where is c?” . The second equation yields (b/2)^2 =c -2d/b a simplification of the final result using that equation yields..u= -g+/-sqrt(5g^2-c) , where g=b/4
Thank You for your great job. in (4:30) x2+x3=-b/2 x2*x3=2d/b we can try my favorite substitution: x2=m+r x3=m-r g^2=x2*x3 ( this is because the favorite substitution is for common use g is the geometric mean, m is the arithmetic mean, r is the common difference forming the arithmetic sequence x2, m, x3) so we get: x2+x3=2m m^2-r^2=g^2 r^2=m^2-g^2 r=(+/-)sqrt(m^2-g^2) x2=m-r = m-sqrt(m^2-g^2) x2=m-r = m+sqrt(m^2-g^2) m=(x2+x3)/2=-b/4 m^2=b^2/16 g^2= 2d/b x2=m-r = m-sqrt(m^2-g^2)=-b/4-sqrt(b^2/16-2*(d/b)) = (-b-sqrt(b^2-32*(d/b)))/4 x3=m+r = m+sqrt(m^2-g^2)=-b/4+sqrt(b^2/16-2*(d/b))= (-b+sqrt(b^2-32*(d/b)))/4
Here sum of the roots is x_1+x_2 +x_3 = 2x_1 so x_1 = -b/2 again c = x_1*(x_2+x_3) + x_2*x_3 or c =(b/2)^2 + x_2*x_3 again d = x_1*x_2*x_3 = -b*x_2*x_3 /2 or x_2*x_3= 2d/b Hereby c= 2d/b + b*b/4 This essentially means x+b/2 is a factor of x^3 + bx^2 + cx + d = 0 or x+ b/2 is a factor of x^3 + bx^2 +( -2d/b + b*b/4)x + d = 0 or (x+b/2) (x"x + b*x/2 +2d/b).= 0 Now roots of x"x + b*x/2 +2d/b= 0 are -b/4 +√( (b/4)^2 -2d/b), -b/4 -√( (b/4)^2 -2d/b) or -b/4 (1 + √(1 - 32d/b^3)), -b/4 (1 - √(1 - 32d/b^3)),
I think you can't choose any value of c, as the given argument locks a relation between b, c, and d. If you plug the roots we've found into the second vieta formula we got, you would get what c has to be in relation to b and d.
The condition on the solutions can be turned into a constraint on the coefficients. In other words, there is a particular form that c must have (in terms of b and d) for the relationship between the three solutions to hold.
The thing that i find most interesting about this particular math problem is the way that the concept of systems of equations and the concept of the quadratic formula have "grown up" and now use things like x to the 4th power instead of the regular quadratic coefficients terms. Also i have a question: Is using the substitution variable "u" a concept in itself in order to go up into higher mathematics? I notice lots of math problems can be solved by introducing that letter "u".
I wish more people knew about this channel because you deserve millions of subscribers
Oh, thank you for the kind words! 💖
several have asked “where is c?” . The second equation yields (b/2)^2 =c -2d/b
a simplification of the final result using that equation yields..u= -g+/-sqrt(5g^2-c) , where g=b/4
Thank You for your great job.
in (4:30)
x2+x3=-b/2
x2*x3=2d/b
we can try my favorite substitution:
x2=m+r
x3=m-r
g^2=x2*x3
( this is because the favorite substitution is for common use
g is the geometric mean,
m is the arithmetic mean,
r is the common difference forming the arithmetic sequence x2, m, x3)
so we get:
x2+x3=2m
m^2-r^2=g^2
r^2=m^2-g^2
r=(+/-)sqrt(m^2-g^2)
x2=m-r = m-sqrt(m^2-g^2)
x2=m-r = m+sqrt(m^2-g^2)
m=(x2+x3)/2=-b/4
m^2=b^2/16
g^2= 2d/b
x2=m-r = m-sqrt(m^2-g^2)=-b/4-sqrt(b^2/16-2*(d/b)) = (-b-sqrt(b^2-32*(d/b)))/4
x3=m+r = m+sqrt(m^2-g^2)=-b/4+sqrt(b^2/16-2*(d/b))= (-b+sqrt(b^2-32*(d/b)))/4
Another great explanation, SyberMath!
Here sum of the roots is
x_1+x_2 +x_3 = 2x_1
so x_1 = -b/2
again c = x_1*(x_2+x_3) + x_2*x_3
or c =(b/2)^2 + x_2*x_3
again
d = x_1*x_2*x_3 = -b*x_2*x_3 /2
or x_2*x_3= 2d/b
Hereby c= 2d/b + b*b/4
This essentially means x+b/2 is a factor of x^3 + bx^2 + cx + d = 0
or x+ b/2 is a factor of
x^3 + bx^2 +( -2d/b + b*b/4)x + d = 0
or (x+b/2) (x"x + b*x/2 +2d/b).= 0
Now roots of x"x + b*x/2 +2d/b= 0 are -b/4 +√( (b/4)^2 -2d/b),
-b/4 -√( (b/4)^2 -2d/b)
or -b/4 (1 + √(1 - 32d/b^3)),
-b/4 (1 - √(1 - 32d/b^3)),
SyberMath: Vieta Strikes Back
Wait. So x1 x2 x3 have no relation to c? Can i understand that as for whatever value of C is the solitions are the same?
I think you can't choose any value of c, as the given argument locks a relation between b, c, and d. If you plug the roots we've found into the second vieta formula we got, you would get what c has to be in relation to b and d.
The condition on the solutions can be turned into a constraint on the coefficients. In other words, there is a particular form that c must have (in terms of b and d) for the relationship between the three solutions to hold.
That's right!
8d=4bc-b^3 must hold
The thing that i find most interesting about this particular math problem is the way that the concept of systems of equations and the concept of the quadratic formula have "grown up" and now use things like x to the 4th power instead of the regular quadratic coefficients terms. Also i have a question: Is using the substitution variable "u" a concept in itself in order to go up into higher mathematics? I notice lots of math problems can be solved by introducing that letter "u".
Substitution is a powerful tool!
Lets go this is gonna be fun
First again! Super interesting content every single day
Thank you!
@@SyberMath No problem. Why people have a problem with me enjoying your content is not an answerable question
Great video... but I think you should substitute the solutions in x1x2 + x2x3 + x3x1 = c and ensure that there is indeed a solution...
That's right!
8d=4bc-b^3 must hold
Alternatively the first two equations could have been used which gives a result independent of d rather than your result which is independent of c.
That's right!
8d=4bc-b^3 must hold
It's like Po Shen Lo method to solve quadratic equations
6:24 to be or not to be nice one 😃
If x1=x2*x3, can Vieta be used also?
sure
Woww new one !!!
It's only possible when 4bc=b^3+8d
That's right!
@@SyberMath Alhamdulillah 😊
4:48 2u yayyy!!! 😁
😁
Wow i thought first thhat after finding one root i will just divide and find other twoo roots but ur method surpised me
If x1x2+x2x3+x3x1 does not equal C, then there are no solution.
That's right!
8d=4bc-b^3 must hold
We really need a trademark for you on the 2b or not 2b jokes 😆
😍
easy one this time😊😉
*kind of
Nice method but you need to put some information in the description due to the mistake you made and didn't realize
bruh he fixed it...
it was not a mistake.
@@SyberMath that's exactly why i missed it
"For the youtube algorithm"What it really means,I see it at many videos🤔
It means when people comment on a video and like it, RUclips recommends it to more people rewarding the interaction and people's interests.
@@SyberMath Why use Vieta and not solve it using algebra?
@@leif1075 Vieta is awesome! 🤩
For the algorithm!
we gotta beat that thing lol
We can solve the equation even we didn't know c.
sorry i missed the live stream
No worries!
16th
Genial! Vieta dos cops
I want to give you a problem, how shall I do it?
Without an image: forms.gle/A5bGhTyZqYw937W58
With an image: twitter.com/intent/tweet?text=@SyberMath
@@SyberMath Uploaded my problem @SyberMath, sorry for being late as I was a bit busy...
How to solve it without the given condition?
Cubic formula
So it doesn't even matter what the value of c is?
Your question has already been answered by previous commenters.
That's right!
8d=4bc-b^3 must hold
3rd