Solving A Cubic Polynomial | Inspired By Math Olympiads

[@Blaqjaqshellaq]

If a=172, x=-3
If a=-63, x=-2
If a=-14, x=-1
If a=-1, x=0
If a=0, x=1
If a=13, x=2
If a=62, x=3
If a=171, x=4

[@dan-florinchereches4892]

I think the left hand side is almost a perfect cube. If weultiply by 2 and subtract 1 from both sides:
(2x)^2-3(2x)^2+3(2x)-1=2a+2-1
(2x-1)^3-(2a+1)=0
If y=2x-1 and t^3=2a+1
y^3-t^3=0
(y-t)(y^2+ty+t^2)=0
y1=t 2x1-1=(2a+1)^1/3
Delta=t^2-4t^2

[@robyzr7421]

y^2 + ty + t^2 = 0 ->
- t (+, -) [ t^2 - 4 t^2]^(1/2) / 2 =
- t (+, -) [ - 3 t^2]^(1/2) /2 =
- t (+, -) [ - 3]^(1/2)
[ t^2]^(1/2) /2
- t (+, -) [ - 3]^(1/2)[ t ] / 2
😮

[@leif1075]

I don't see anyone noticing the left hand is apmost a lerfect cube when it's not?

[@pietergeerkens6324]

Nice!
I spotted Simon's Favourite Factoring Trick right off the bat, and so jumped straight to method 2.

[@vaggelissmyrniotis2194]

Let f(x)=4x³-6x²+3x-1 which is increasing function for all reals and its domain is ( -inf,+inf) therefore for each real value of (a) there is only one solution.

[@Don-Ensley]

problem
4 x³-6 x² + 3 x = a + 1
Divide by 4.
x³-3/2 x² + 3/4 x = (a+1)/4
Remove the quadratic term by substitution.
x = y + 1/2
This results in
y³ + 1/8 = (a+1)/4
y³ = (2a+1)/8
One obvious root is
y = ∛(2a+1)/2
Denote this as r.
r = ∛(2a+1)/2
Our equation is
y³ - r³ = 0
Recalling the formula for factoring the difference of 2 cubes:
y³ - r³ = 0
= (y-r)(y² +yr +r²)
By the zero product property,
y² + yr +r² = 0
And by the quadratic formula,
y = [ -r ± √( r²- 4r² ) ] / 2
= -r / 2 ± √(-3 r²) / 2
= -r / 2 ± i √3 r / 2
= r { -1/2 ± i √3 / 2 }
So all the y solutions are
y = ½ ∛(2a+1)
y = ½ ∛(2a+1) [-½ - i ½ √3]
y = ½ ∛(2a+1)[-½ + i ½ √3]
Back substitution gives us
x = y +½
x = ½ + ½ ∛(2a+1)
x = ½ +
½ ∛(2a+1)[-½ - i ½ √3]
x = ½ +
½ ∛(2a+1)[-½ + i ½ √3]
answer
x ∈ {
½ + ½ ∛(2a+1),
½ +
½ ∛(2a+1)[-½ - i ½ √3],
½ +
½ ∛(2a+1)[-½ + i ½ √3]
}

[@giuseppemalaguti435]

Svolgendo i calcoli l'equazione diventa (x-1/2)^3=(a+1)/4-1/8...

[@scottleung9587]

Nice, but you should include the complex solutions.

[@ajitandyokothakur7191]

The second method is more elegant than the first one. Dr. Ajit Thakur (USA).

[@leif1075]

Is it not a matter of opinion to some extent?

[@jcfgykjtdk]

I used second method

[@WahranRai]

Your solution is incomplete, you had given the real one : 2 other complex conjugate solutions are missing

[@SyberMath]

This channel is not about complex numbers. I have another channel for that:
youtube.com/@aplusbi

[@WahranRai]

@@SyberMathYou present complex solutions well in your video:://ruclips.net/video/6GRA7zSEKDs/видео.html

[@TedHopp]

You didn't watch to the end of the video where he talks about obtaining the complex solutions with his second method.

[@leif1075]

​@@SyberMathbut wouldn't you agree no one would muktiply by 2 because it's not la perfect cube so isn't that method too contrived amd maybe even a cheat snd so shouldn't count? Why not factpr put the x minus 1 you can split it up into? Isnt that more valid? Hope you can respond