A Super Special Polynomial Equation
VloĹžit
- Äas pĹidĂĄn 7. 09. 2024
- 𤊠Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! đ§ĄđĽ°đđĽłđ§Ą
/ @sybermathshorts
/ @aplusbi
â¤ď¸ â¤ď¸ â¤ď¸ My Amazon Store: www.amazon.com...
When you purchase something from here, I will make a small percentage of commission that helps me continue making videos for you.
If you are preparing for Math Competitions and Math Olympiads, then this is the page for you!
CHECK IT OUT!!! â¤ď¸ â¤ď¸ â¤ď¸
â¤ď¸ A Differential Equation | The Result Will Surprise You! ⢠A Differential Equatio...
â¤ď¸ Crux Mathematicorum: cms.math.ca/pu...
â¤ď¸ A Problem From ARML-NYSML Math Contests: ⢠A Problem From ARML-NY...
â¤ď¸ (8x+7)^2(4x+3)(2x+2)=9
â¤ď¸ LOGARITHMIC/RADICAL EQUATION: ⢠LOGARITHMIC/RADICAL EQ...
â¤ď¸ Finding cos36¡cos72 | A Nice Trick: ⢠Finding cos36¡cos72 | ...
â Join this channel to get access to perks:â bit.ly/3cBgfR1
My merch â teespring.com/...
Follow me â / sybermath
Subscribe â www.youtube.co...
â Suggest â forms.gle/A5bG...
If you need to post a picture of your solution or idea:
in...
#radicals #radicalequations #algebra #calculus #differentialequations #polynomials #prealgebra #polynomialequations #numbertheory #diophantineequations #comparingnumbers #trigonometry #trigonometricequations #complexnumbers #math #mathcompetition #olympiad #matholympiad #mathematics #sybermath #aplusbi #shortsofsyber #iit #iitjee #iitjeepreparation #iitjeemaths #exponentialequations #exponents #exponential #exponent #systemsofequations #systems
#functionalequations #functions #function #maths #counting #sequencesandseries
#algebra #numbertheory #geometry #calculus #counting #mathcontests #mathcompetitions
via @CZcams @Apple @Desmos @NotabilityApp @googledocs @canva
PLAYLISTS đľ :
Number Theory Problems: ⢠Number Theory Problems
Challenging Math Problems: ⢠Challenging Math Problems
Trigonometry Problems: ⢠Trigonometry Problems
Diophantine Equations and Systems: ⢠Diophantine Equations ...
Calculus: ⢠Calculus
let y=2x+2 -> original becomes (4y-1)^2(2y-1)y=9
expand -> (16y^2-8y+1)(2y^2-y)=9
let z=2y^2-y
-> (8z+1)z=9
-> z=1 or z=-9/8
... substitute back for y and x.
Very nice problem and various substitution solutions!
Many thanks!
Realoze -0.5 is a solution. Also, the left hand side is an increasing function since -0.75. -0.5 is only real solution.
Nice problem & solution.
I was stuck until I saw you put spaces between the factors. Then I paused the video and found the same solution you presented. :)
(alas, I missed the u substitution clue in the thumbnail...)
Glad it helped!
@@SyberMath BTW did you see my comment on your video "Solving An Infinite Radical with i | Problem 280" czcams.com/video/GKut5EJXnCs/video.html ? I think you'll enjoy the GeoGebra applet I made for this, link in my comment.
It's worth setting z=-0.3.
2(2x+2)+4x+3=8x+7
Wow! That's cool
đ
Oh no, he's got his voice back!!!
Now I've got my headache back.
Multiplying one factor by 2, and another by 4, instead of multipying all of it by 8. That simple thing seems kinda new to me! I'm poorly educated, that's why I'm here.
Ahaha! Am I causing headaches? đ
It's the same thing but distributed differently
If you can get 3*3 = 9 you're in business. This can happen with 9*1 or 9*1*1.
8x + 7 = 3 => x = -1/2.
So (8x + 7)^2 = 9.
Further 4x + 3 = 1 =>
x = -1/2 with 2x + 2 = 1 => x = -1/2.
EUREKA I have found it!!!
At least one real root.
nice!
Pongo t=4x+3..risulta (2t+1)^2*t*((t+1)/2)=9...(2t+1)^2(t+1)=18...4t^3+8t^2+5t-17=0..(t-1)(4t^2+12t+17)=0...unica soluzione reale t=1..x=-1/2..t=(-3/2)+iâ2..x=((-9/2)+iâ2)/4
Nice method. At a certain point you lost a factor t which transformed the quartic into a cubic. :(
Got 'em all!
You are awesome! đ
still dont saw the video, but my answers to real values of x are
x = ( -7 + â7) á 8 e x = ( -7 - â7) á 8
unfurtunately i miss in â289 = 13, but still i solved by the right method
im quite happy with the result
Solved it like in the 2nd method but got it wrong as two times four is NOT four. đ It was a hard day thou.
The 2nd method is always the best method.
False. That is not true. Do not make this wrong generalization.
Trivial
It would be nice if you didn't mention science fiction roots...
science fiction roots?
@@SyberMath Imaginary. Same thing.
đđđđđđđ
The movie is u and SyberMath or You and SyberMath!
đđ
9*8=(8x+7)^2*(8x+6)(8x+8) set u=(8x+7)^2 so 72=u(u-1)
u=9,,or u=-8
u=9 x=((-+3)-7)/8 x=-1/2 x=5-5/4
u=-8 x=((-+2sqrt(2)i-7)/8
A Super Special Polynomial Equation: (8x + 7)²(4x + 3)(2x + 2) = 9; x = ?
(8x + 7)²[2(4x + 3)][4(2x + 2)] = (2)(4)(9), (8x + 7)²(8x + 6)(8x + 8) = 72
Let: y = 8x + 7, 8x + 6 = y - 1, 8x + 8 = y + 1; y²(y - 1)(y + 1) = 72
y²(y² - 1) - 72 = 0, yⴠ- y² - 72 = (y² - 9)(y² + 8) = 0; y² - 9 = 0 or y² + 8 = 0
y = 8x + 7; (8x + 7)² - 3² = (8x + 7 - 3)(8x + 7 + 3) = (8x + 4)(8x + 10) = 0
4(2x + 1) = 0, 2x = - 1, x = - 1/2; 2(4x + 5) = 0, 4x = - 5, x = - 5/4
(8x + 7)² + 8 = 0; (8x + 7)² = [i(2â2)]², 8x + 7 = Âą i(2â2), x = [- 7 Âą i(2â2)]/8
Answer check:
x = - 1/2: (8x + 7)²(4x + 3)(2x + 2) = (- 4 + 7)²(- 2 + 3)(- 1 + 2) = 9; Confirmed
x = - 5/4: (- 10 + 7)²(- 5 + 3)(- 5/2 + 2) = (- 3)²(- 2)(- 1/2) = 9; Confirmed
x = [- 7 Âą i(2â2)]/8: (8x + 7)² = - 8
(8x + 7)²(4x + 3)(2x + 2) = (- 8)[- 7 Âą i(2â2)]/2 + 3}{2[- 7 Âą i(2â2)]/8 + 1}
= (- 1)[- 1 Âą i(2â2)][1 Âą i(2â2)] = (- 1)(- 1 + 8i²) = (- 1)(- 9) = 9; Confirmed
Final answer:
x = - 1/2; x = - 5/4; Two complex value roots;
x = [- 7 + i(2â2)]/8 or x = [- 7 - i(2â2)]/8