I Solved A Cubic Equation in Two Ways
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- Опубліковано 28 гру 2023
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While I very much like your second method, it will only work in special situations. Your first method should work for any cubic polynomial: Define a shift transformation y = x - k that eliminates the quadratic term. This results in a reduced cubic equation involving only the cubic, linear, and constant terms. Then use your favorite cubic expansion for (a+b)ˆ3, which also involves only the cubic, linear, and constant terms in y = a+b. Easily identify the product ab and the sum aˆ3 + bˆ3 in that expression. Use the product expression to eliminate either a or b and insert into the latter sum expression. This results in a 6th-order polynomial in the selected variable, say a, that also involves only the cubic term and a constant term. This equation is easily solved using the quadratic formula for aˆ3. bˆ3 is then easily found via the product formula. While there are 6 solutions for this 6th-order equation, they result in only 3 unique solutions for y due to the equivalent roles that a and b play in the problem. For example, I found a_1ˆ3 = 16/27 and a_2ˆ3 = 4/27 from the quadratic formula. (The corresponding b solutions are b_1ˆ3 = 4/27 and b_2ˆ3 = 16/27 so that b_1ˆ3 = a_2ˆ3 and b_2ˆ3 = a_1ˆ3. ) A real solution for (a, b) is (4ˆ(2/3)/3, 4ˆ(1/3)/3). Use y = a+b and then x = y + 1/3 to get x = 1/(4ˆ(1/3) - 1).
I solved your equation another way. Your equation becomes
x³ − x² − x − ⅓ = 0 (thanks to @NadieFan for the text here and later)
We can make things easier by avoiding fractions by multiplying both sides by 3:
3x³ − 3x² − 3x − 1 = 0
Note it would be easier to reverse the coefficients, which we can do by letting x = 1/y, multiplying both sides by y³, then rearranging and getting
y³ + 3y² + 3y - 3 = 0
Now we do your substitution to remove the quadratic term: let y = t - 1, substitute and simplify. In this case, not only the quadratic term is eliminated but also the linear term, and we get
t³ - 4 = 0
Thus t = ∛4; y = t - 1 = ∛4 - 1; and x = 1/y = 1/(∛4 - 1). That last can have its denominator rationalized to become (∛16 + ∛4 + 1)/3 as in your video.
The result of your special equation was to have the final cubic equation without quadratic and linear terms, which is very easy to solve.
Nice. Your equation
t³ − 4 = 0
has three solutions
t₁ = ∛4, t₂ = ω₁∛4, t₃ = ω₂∛4
where ω₁ = −½ + i·½√3 and ω₂ = −½ − i·½√3 are the complex cube roots of unity, so the three solutions of the cubic in x are
x₁ = 1/(∛4 - 1), x₂ = 1/(ω₁∛4 - 1), x₃ = 1/(ω₂∛4 - 1)
which can have their denominators rationalized to become
x₁ = (∛16 + ∛4 + 1)/3, x₂ = (ω₁²∛16 + ω₁∛4 + 1)/3, x₃ = (ω₂²∛16 + ω₂∛4 + 1)/3
or
x₁ = (∛16 + ∛4 + 1)/3, x₂ = (ω₂∛16 + ω₁∛4 + 1)/3, x₃ = (ω₁∛16 + ω₂∛4 + 1)/3
since ω₁² = ω₂, ω₂² = ω₁. Substituting ω₁ = −½ + i·½√3, ω₂ = −½ − i·½√3 the two complex solutions can be rewritten as
x₂ = ⅓ − ⅙(∛16 + ∛4) − i·⅙√3(∛16 − ∛4)
x₃ = ⅓ − ⅙(∛16 + ∛4) + i·⅙√3(∛16 − ∛4)
Thanks for stepping through rationalization of the denominator in that first method.
My pleasure!
Nice!
Thanks
:) really like your videos
Glad you like them! 😍
Bro can u help me solve
(Ln (x²) / x) +x =0 without a graph? Like not from the graph, i need to solve it mathematically but I'm struggling
(Ln (x²) / x) +x = 0
Ln (x²) = -x²
x² + Ln (x²) = 0
e^(x² + Ln (x²) ) = 1
x²e^x² = 1
W(x²e^x²) = W(1)
x² = W(1)
x = +- sqrt(W(1))
W = Lambert's W Function (aka Product Log)
I need it asap pls
I don't agree that your first method (using the cubic formulas, yes, plural) is painful here, it is not if done properly. The equation to solve with the right hand side reduced to zero is
(1) x³ − x² − x − ¹⁄₃ = 0
As you noted, substituting
(2) x = y + ¹⁄₃
will turn this into the depressed cubic equation
(3) y³ − ⁴⁄₃y − ²⁰⁄₂₇ = 0
By the way, it is easy to find the correct substitution to depress (1) if you note that the sum of the roots of (1) is 1. Therefore, if we decrement each of the three roots of (1) by ¹⁄₃ we will get a new cubic equation where the sum of the roots is _zero_ which means that this new cubic equation will lack a quadratic term. If the variable of this new equation is y, we will therefore have y = x − ¹⁄₃ and so x = y + ¹⁄₃ is the substitution we need.
Now, the first thing we want to do is to get rid of the fractional coefficients in (3). This can be done by substituting
(4) y = ¹⁄₃z
which gives
¹⁄₂₇z³ − ⁴⁄₉z − ²⁰⁄₂₇ = 0
and multiplying both sides by 27 then gives
(5) z³ − 12z − 20 = 0
A depressed cubic equation z³ + pz + q = 0 with real p and q has a _single_ real solution and two conjugate complex solutions if and only if its discriminant D = (½q)² + (⅓p)³ is _positive_ and then this unique real solution is
(6) z₁ = ∛(−½q + √D) + ∛(−½q − √D)
For (5) we have p = −12, q = −20, so D = (−10)² + (−4)³ = 100 − 64 = 36, so √D = 6, and from (6) we then immediately find that the only real solution of (5) is
(7) z₁ = ∛16 + ∛4
and in accordance with (4) we therefore have
(8) y₁ = (∛16 + ∛4)/3
and from (2) we then get
(9) x₁ = (∛16 + ∛4 + 1)/3
as the sole real solution of (1). As you can see this is all very simple and straightforward, no messing with fractions or rationalizing denominators.
But what about complex solutions? Well, this is also a lot simpler than you suggest in your video and indeed also a lot simpler than the untractable expressions WolframAlpha spits out for the complex solutions of this equation.
In fact all we need to do to get the complex solutions of z³ + pz + q = 0 with real p and q if D > 0 is to multiply each of the cube roots of the real solution (6) with the complex cube roots of unity ω₁ = −½ + i·½√3 and ω₂ = −½ − i·½√3 in either order, so we have
(10a) z₂ = ω₁∛(−½q + √D) + ω₂∛(−½q − √D)
(10b) z₃ = ω₂∛(−½q + √D) + ω₁∛(−½q − √D)
Separating the real and imaginary parts these expressions can be written as
(11a) z₂ = −½(∛(−½q + √D) + ∛(−½q − √D)) + i·½√3(∛(−½q + √D) − ∛(−½q − √D))
(11b) z₃ = −½(∛(−½q + √D) + ∛(−½q − √D)) − i·½√3(∛(−½q + √D) − ∛(−½q − √D))
So, the complex roots of (5) are
(12a) z₂ = −½(∛16 + ∛4) + i·½√3(∛16 − ∛4)
(12b) z₃ = −½(∛16 + ∛4) − i·½√3(∛16 − ∛4)
and in accordance with (4) the complex roots of (3) are therefore
(13a) y₂ = −⅙(∛16 + ∛4) + i·⅙√3(∛16 − ∛4)
(13b) y₃ = −⅙(∛16 + ∛4) − i·⅙√3(∛16 − ∛4)
and from (2) we then get
(14a) x₂ = ⅓ − ⅙(∛16 + ∛4) + i·⅙√3(∛16 − ∛4)
(14b) x₃ = ⅓ − ⅙(∛16 + ∛4) − i·⅙√3(∛16 − ∛4)
as the complex solutions of (1).
Incidentally, in this video you goofed big time, because none of the solutions from WolframAlpha you show are solutions of the equation you discuss. You can easily verify that (14a) and (14b) are the correct complex solutions of (1). If you enter these solutions in WolframAlpha it will recognize these as zeros of the polynomial 3x³ − 3x² − 3x − 1.
goofed?
That's what WA provides! 😮😁
@@SyberMath Check your video. The solutions from WolframAlpha you show are the solutions of
x³ − x² − x = 1
and not the solutions of the equation
x³ − x² − x = ¹⁄₃
you are discussing. At 9:57 you ask yourself why they didn't write the real solution in the way you did. Well, now you know.
For what its worth your first method, which ends in the system involving a^3 and b^3 isn't too bad. One can solve this system by introducucing and new and using a method due to the Mesopotamians.
Well, here (because it is not a cubic equation with three different real roots) you can just use Cardan's formulas directly and save yourself the trouble of solving a system in a³ and b³. SyberMath never does that, but it is really as easy as using the quadratic formula and will also get you the two complex roots of this equation easily. See my main comment on this video.
still easy
3x^3=3x^2+3x+1
3x^3-3x^2-3x=1
x = 1 -> 3-3-3=-3
x = 2 -> 24-12-6=6
1 < x < 2
1.701 < x < 1.702
x = 1.7015 (approx.)
(3x-1.7)(x^2-0.43x-0.57)=0
100x^2-43x-57=0
x = [43+-{(1849+22800)^(1/2)}]/200
x = [43+-{(24649)^(1/2)}]/200
sqrt = 157
x = (43+-157)/200
x = 1 or -57/100