A Super Special Polynomial Equation
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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!
2(2x+2)+4x+3=8x+7
Wow! That's cool
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" th-cam.com/video/GKut5EJXnCs/w-d-xo.html ? I think you'll enjoy the GeoGebra applet I made for this, link in my comment.
It's worth setting z=-0.3.
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? 😁
It's the same thing but distributed differently
Realoze -0.5 is a solution. Also, the left hand side is an increasing function since -0.75. -0.5 is only real solution.
Got 'em all!
You are awesome! 😍
👍
The movie is u and SyberMath or You and SyberMath!
😊😍
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)+i√2..x=((-9/2)+i√2)/4
Nice method. At a certain point you lost a factor t which transformed the quartic into a cubic. :(
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
Solved it like in the 2nd method but got it wrong as two times four is NOT four. 🙄 It was a hard day thou.
👍😎👍🎉👍😎👍
still dont saw the video, but my answers to real values of x are
x = ( -7 + √7) ÷ 8 e x = ( -7 - √7) ÷ 8
unfurtunately i miss in √289 = 13, but still i solved by the right method
im quite happy with the result
It would be nice if you didn't mention science fiction roots...
science fiction roots?
@@SyberMath Imaginary. Same thing.
Trivial
The 2nd method is always the best method.
False. That is not true. Do not make this wrong generalization.
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, y⁴ - 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(2√2)]², 8x + 7 = ± i(2√2), x = [- 7 ± i(2√2)]/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 ± i(2√2)]/8: (8x + 7)² = - 8
(8x + 7)²(4x + 3)(2x + 2) = (- 8)[- 7 ± i(2√2)]/2 + 3}{2[- 7 ± i(2√2)]/8 + 1}
= (- 1)[- 1 ± i(2√2)][1 ± i(2√2)] = (- 1)(- 1 + 8i²) = (- 1)(- 9) = 9; Confirmed
Final answer:
x = - 1/2; x = - 5/4; Two complex value roots;
x = [- 7 + i(2√2)]/8 or x = [- 7 - i(2√2)]/8