@@intellecta2686 From Vieta we have x_{1}+x_{2}=-b/a Suppose that x_{1}-x_{2}=d (x_{1}-x_{2})^2 is symmetric function and it is called discriminant (x_{1}-x_{2})^2 is symmetric function so it can be expressed in terms of elementary symmetric functions (x_{1}-x_{2})^2=(x_{1}+x_{2})^2-4x_{1}x_{2} (x_{1}-x_{2})^2=(-b/a)^2-4(c/a) x_{1}+x_{2}=-b/a x_{1}-x_{2}=d 2x_{1}=-b/a+d 2x_{2}=-b/a-d d^2=(b^2-4ac)/a^2 so x_{1} = (-b+sqrt(b^2-4ac))/(2a) x_{2} = (-b+sqrt(b^2-4ac))/(2a) Symmetric functions can be also used for solving cubics and quarics
I have a problem (in fact I designed it) and I am struggling to solve it, other than by graphs. I am not a mathematician, I only play sometimes with math... when I am in the mood to... I try W Lambert and a lot of things but it didn't work. The equation is (4^x)/x + (4^(1/x))*x=12. The real solutions are 0.5 and 2 and many other complex solutions. I try to make substitutions and I get a/ln(a) + b/ln(b)=12/ln4, where a=4^x, and b=4^(1/x), and lna*lnb=(ln4)^2, but analytically I wasn't able to solve it... Maybe you have some suggestions?
This is a popular Golden Ratio problem where Φ = (√5 + 1)/2 = (3/2)^x. To see this, divide the given equation by 6^x to obtain (3/2)^x - 1 = 1/(3/2)^x, which is Φ - 1 = 1/Φ or Φ^2 - Φ - 1 = 0 with the above claimed solution. Thus, x*ln(3/2) = lnΦ or x = lnΦ/ln(3/2) = 1.18681...
In the second method you come up with d²=cd+c² It is indeed a quadratic equation in d, and you get one solution. But you forget that it is also a quadratic equation in c which will give you another solution.
The quadratic formula
The denominator is 2a but not 4a
You forgot
Please correct it
Yes, of course I will make another video just about it :)
Nooot , but will pin you!
@@intellecta2686
From Vieta we have
x_{1}+x_{2}=-b/a
Suppose that
x_{1}-x_{2}=d
(x_{1}-x_{2})^2 is symmetric function and it is called discriminant
(x_{1}-x_{2})^2 is symmetric function so it can be expressed in terms of elementary symmetric functions
(x_{1}-x_{2})^2=(x_{1}+x_{2})^2-4x_{1}x_{2}
(x_{1}-x_{2})^2=(-b/a)^2-4(c/a)
x_{1}+x_{2}=-b/a
x_{1}-x_{2}=d
2x_{1}=-b/a+d
2x_{2}=-b/a-d
d^2=(b^2-4ac)/a^2
so
x_{1} = (-b+sqrt(b^2-4ac))/(2a)
x_{2} = (-b+sqrt(b^2-4ac))/(2a)
Symmetric functions can be also used for solving cubics and quarics
@@intellecta2686 thanks for the demonstration
I have a problem (in fact I designed it) and I am struggling to solve it, other than by graphs. I am not a mathematician, I only play sometimes with math... when I am in the mood to...
I try W Lambert and a lot of things but it didn't work.
The equation is (4^x)/x + (4^(1/x))*x=12. The real solutions are 0.5 and 2 and many other complex solutions.
I try to make substitutions and I get a/ln(a) + b/ln(b)=12/ln4, where a=4^x, and b=4^(1/x), and lna*lnb=(ln4)^2, but analytically I wasn't able to solve it...
Maybe you have some suggestions?
Please use one method in a video and make another video for second method to keep the length of video short, Thanks a lot❤❤
it's (1+√5)/2 and x=log3/2((1+√5)/2)≈ 1.18 thank you for the video and btw i loved the posters
Yes, true! Thanks, I like them too:)
I believe it is not (1+sqrt(5))/4 but (1+sqrt(5))/2, isn't it?
If my teachers were like this, I would top my exams
I used the Lambert W function to solve it.
It's really interesting listening ur lecture..
Subscribed
Hi 👋
This is a popular Golden Ratio problem where Φ = (√5 + 1)/2 = (3/2)^x.
To see this, divide the given equation by 6^x to obtain (3/2)^x - 1 = 1/(3/2)^x, which is
Φ - 1 = 1/Φ or Φ^2 - Φ - 1 = 0 with the above claimed solution.
Thus, x*ln(3/2) = lnΦ or x = lnΦ/ln(3/2) = 1.18681...
Identifying the quadratic equation for a solution seemed to be the most important part. Thanks Ivana and Simba(?)
Over 2a
the answr is not static at all...lets replace x with 1 ...it became 9
In the second method you come up with d²=cd+c²
It is indeed a quadratic equation in d, and you get one solution. But you forget that it is also a quadratic equation in c which will give you another solution.
your 9 looks like the letter g. Great video.
Can we use natural logarithms (e) instead of log?
Yes
Thanks Ivana, a really interesting little problem requiring a bit of thought!
Thank you :)
Please solve a question from india JEE Advance paper
Probably will make, but not soon . I have so much to do, that's why I am making some simple examples. But, thanks for the suggestion 👍
New video featuring Koko and Simba :)
Hahahh maybe 🤔😅
Hello Ivana,
Very enjoyable video. 😀
Peter. Dublin.
Thank you Peter, thanks for stopping by:)
I am interested
The denominator should have been 2a instead of 4a
U forget!
Shouldn't 4a be 2a?
True. My mistake🙈
Miaoooo
😅
Ah my lover you are fouls
Cat doesn’t like math.
Haha for sure. 😁
@@intellecta2686 what is her name? I assume it is female
No, it is male. He is Simba and he looks like this 😼 I have another one, also male and the name is Koko, looks like this 🐈⬛ heheh
@@intellecta2686 very good. Maybe in the next video you put them on the table watching you with baby yoda:)
😅
On ne sait pas si l'on doit suivre vos explications sur le tableau ou regarder votre nudité et vos quartiers de viande que vous avez exposés.