A Nice Rational Equation |
Вставка
- Опубліковано 1 жов 2023
- 🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts).
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermathshorts
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/stores/sybermat...
Follow me → / sybermath
Subscribe → ua-cam.com/users/SyberMath?sub...
⭐ Suggest → forms.gle/A5bGhTyZqYw937W58
If you need to post a picture of your solution or idea:
intent/tweet?text...
#rationalequations #algebra
via @UA-cam @Apple @Desmos @NotabilityApp @googledocs @canva
An Exponential Equation | 2^{3^{5^x}} = 3^{2^{5^x}}: • An Exponential Equatio...
A Surprising Exponential Equation: • A Surprising Exponenti...
PLAYLISTS 🎵 :
Number Theory Problems: • Number Theory Problems
Challenging Math Problems: • Challenging Math Problems
Trigonometry Problems: • Trigonometry Problems
Diophantine Equations and Systems: • Diophantine Equations ...
Calculus: • Calculus
I did a variation of method 1: After cross multiplying you get 2*(x^4) - 7(x^3) -7*x +2 =0 I didn''t to see the trick used in the video but I
guessed there would be a bi-quadratic factoring and after some trial and error I tried [(x^2) +B*x +1)] [2*(x^2) +C*x +2] By equating coefficients with
the quartic you get B=-4 and C=1 thus the quartic factors as (x^2 - 4*x +1)(2*(x^2) +x +2) Solving these two quadratics gives the answers in the video.
I did what you did at the end, dividing the polynomial by x^2.
Set z=x+1/x and you get (z^2-2)/z=7/2.nice and very easy😃
For the life of me i couldn't find a clean way to solve this so i had to brute force through, it was very painful..
After simplifying, we get 2(x^4) - 7(x^3) - 7x + 2 = 0
Based of the 2, there's only so many ways this could be factored.
I initially tried [ 2(x^2) + Ax + 1 ] [ x^2 + Bx + 2 ], but after solving the system you get a contradiction.
Afterwards i did [ 2(x^2) + Ax + 2 ] [ x^2 + Bx + 1 ], it simplifies in to solving the system:
{ a + 2b = -7
{ ab = -4
We get two solution sets:
(a1, b1) = ( 1, - 4)
(a2, b2) = ( -8, 1/2)
After plugging back in we get the following :
[ 2(x^2) + x + 2] [ x^2 -4x + 1]
and
[ 2(x^2) - 8x + 2] [ x^2 + x/2 + 1]
Notice you can factor out a 2 from factor1 of the second equation and put it in factor2. Doing so will give us the first equation, meaning they are equivalent!
Now since we have it in factored form, using the quadratic formula twice we get two real and two complex solutions
X = 2 ± root(3)
X = ¼ [ -1 ± i * root(15)]
6^¼=x
(2-6^¼)^⅓=x
x^4+1 can be factored over the reals as x^2-sqrt(2)x+1 and x^2+sqrt(2)x+1
That's right!
x=(-1±√15i)/4, 2±√3
I learnt the method you used at the end for the symmetric quartic in 11th grade. Nobody seemed to know it in the US. Are you from abroad?
He was born and raised in Turkey, but now lives in the US.
thank you!
@@scottleung9587
you're right. not known widely in the US.
👍😁
how do you get rid of the 4th power if you substitute x with y? did you mean x^2 by y? probably not...
get rid of the 4th power?
@@SyberMath you said something about making the equation not of the 4th power, when you suggested replacing x with y, which does nothing to the power...
factoring after cross-multiplication = exercise in futility?!? :)
Missed joke opportunity: there are two t values - tea for two and two for tea
Wow! That's a good one 🤩
@@SyberMath yeah, except that probably almost no one remembers that song
You sound like Cypher from matrix 😅
😄😜
I used a "3rd method". When you cross multiply and divide by 2 you get
x⁴ - x³.7/2 - x.7/2 + 1 = 0
= (x² + ax + 1).(x² + bx + 1)
a = 1/2 and b = -4
You get two quadratic equations that can be solved to get the 4 roots (2 real and 2 complex) that you got.
Why do you use ∓ and not ±?
Especially where in this video you got the first value of t by using the "bottom" + and wrote it "on top" and vice versa...
I like ∓
@@SyberMath most people say "give or take", "plus or minus", and if not reading left to right, then top to bottom...