Can You Solve 4x^3-3x-2=0 in Two Ways?

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[adandap]

I think you can make the second method work. cos(3t) = 2 doesn't have real solutions, but let's solve it over complex numbers by putting cos(3t) = (e^(i3t) +e(-3it) )/2 and then y = e^(3it) so y + 1/y = 2, or y^2 - 4y + 1 = 0, which gives y = 2 +- root(3). So going back to t we get t = - i/3 ln(2 +- root(3)) and then back to x = cos (-i/3 ln (2 +- sqrt(3) ) ). Now that is real because cos(i x) = cosh(x) = (e^x + e^(-x) )/2. Using e^(a ln(b) ) = b^a we finally get x = 1/2 ( cuberoot(2 +- sqrt(3)) + 1/cuberoot(2 +- sqrt(3) ) ) which simplifies to the answer given by Syber.

[randomjin]

The trig method not only works, but it is *the method* to get all roots - including complex ones. We set x = cosz, get cos3z=2 from where e³ⁱᶻ+e⁻³ⁱᶻ = 4 or (e³ⁱᶻ)²-4e³ⁱᶻ+1 = 0, so e³ⁱᶻ = 2±√3 . Now we want to solve for cosz which is (eⁱᶻ+e⁻ⁱᶻ)/2. If we take *principal value* of cube root for e³ⁱᶻ we get that eⁱᶻ = ∛(2±√3) and e⁻ⁱᶻ = ∛(2∓√3) - so regardless of which sign we choose, the sum is invariant: cosz = (∛(2+√3)+∛(2-√3))/2. Now the other complex roots are obtained from other branches of the cube root. This is far easier than trying to divide and solve the quadratic.
P.S. For Cardano (1st method) I find it easier to multiply by 2, set 2x = y and solve y³-3y-4 = 0

[NadiehFan]

You can do a hyperbolic substitution

x = cosh(t)

Then your equation becomes

cosh(3t) = 2

How about that?

The real solution of your equation then of course is

*x = cosh(⅓·arcosh(2))*

Now, we have

arcosh(2) = ln(2 + √3) = 3t

for a positive real value of t which also implies

ln(2 −√3) = −3t

since (2 + √3)(2 − √3) = 1, so

t = ln(∛(2 + √3)), −t = ln(∛(2 − √3))

and since

cosh(t) = ½(eᵗ + e⁻ᵗ)

it is clear that the real solution can be written algebraically as

*x = ½(∛( 2 + √3) + ∛(2 − √3))*

The complex solutions require a bit more work, but ultimately that is just a matter of multiplying each of these real cube roots by one of the two complex cube roots of unity −½+i·½√3 and −½−i·½√3 (in either order). It is easy to see why this is so. The exponential function has a period 2πi which implies that we can add any integer multiple of 2πi to 3t = ln(2 + √3) to satisfy cosh(3t) = 2 and therefore any integer multiple of ⅔πi to t = ln(∛(2 + √3)). So, all the solutions of the cubic can be represented as

x = cosh(ln(∛(2 + √3)) + ⅔kπi)

where we can take three consecutive integer values of k to obtain all solutions, e.g. k = −1, 0, 1 or k = 0, 1, 2. This can be written as

x = ½(exp(ln(∛(2 + √3)) + ⅔kπi) + exp(ln(∛(2 − √3)) − ⅔kπi))

That is

x = ½(∛(2 + √3)·exp(⅔kπi) + ∛(2 − √3)·exp(−⅔kπi))

Since exp(⅔πi) = −½+i·½√3 and exp(−⅔πi) = −½−i·½√3 this means that the two complex solutions are obtained by multiplying the real cube roots by −½+i·½√3 and −½−i·½√3 in either order. Therefore, the complex solutions can be written algebraically as

*x = −¼(∛( 2 + √3) + ∛(2 − √3)) + ¼√3(∛( 2 + √3) − ∛(2 − √3))i*

and

*x = −¼(∛( 2 + √3) + ∛(2 − √3)) − ¼√3(∛( 2 + √3) − ∛(2 − √3))i*

Alternatively, we can start from

x = cosh(t + ⅔kπi)

and use the identity

cosh(u + iv) = cosh(u)·cos(v) + i·sinh(u)·sin(v)

to get

x = cosh(t)·cos(⅔kπ) + i·sinh(t)·sin(⅔kπ)

and since t = ln(∛(2+√3)), −t = ln(∛(2−√3)) so cosh(t) = ½(eᵗ+e⁻ᵗ) = ½(∛( 2+√3)+∛(2−√3)), sinh(t) = ½(eᵗ−e⁻ᵗ) = ½(∛(2+√3)−∛(2−√3)) this gives us

*x = ½·cos(⅔kπ)·(∛( 2 + √3) + ∛(2 − √3)) + i·½·sin(⅔kπ)·(∛(2 + √3) − ∛(2 − √3))*

which comprises all roots in algebraic form.

I do not know why WolframAlpha spits out such complicated expressions for the exact complex roots, that is certainly not necessary.

[hedayaty]

I remember the trig method was replacing x = r cos(t) and it will give 3 roots (obviously can be hideous)

[cube]

Wow. I learned so many new things. Love your videos ♥️♥️

[Drk]

Thank you! Last night, i was trying to remember how to get the Cardano's formula. I understood with your method by comparison

[makara]

Hello sir . Can you solve ? 5^x.2^2x+1/x+1=50 😊

[HoSza]

Why are you all afraid of Cardano's formula? It gives the solutions directly, after sub y=2x: 8x^3-6x-4=0; y^3-3y-4=0;

[fabiog]

The cardano's formula gives all the real solution if they exist. The value under the square root would have been negative if the real solutions were more than 1. The fact that the radical were positive is a conclusive proof that the equation has only one real solution.

[satyapalsingh]

Very good method, highly benefited, dear professor

[NadiehFan]

You are doing a lot of work to solve a³ + b³ = ½, a³b³ = ¹⁄₆₄ even though you mention Poh-Shen Loh. If we put

a³ = ¼ + u, b³ = ¼ − u

we get

¹⁄₁₆ − u² = ¹⁄₆₄

so

u² = ³⁄₆₄

and we immediately see that

a³ = (2 + √3)/8, b³ = (2 − √3)/8

or vice versa. This gives three values for a and b each, one of which is real while the other two are complex. So it may seem at first sight that we can get nine values for a + b, but only three solution pairs (a, b) satisfy ab = ¼.

Let a₁ and b₁ respresent the real solutions of a³ = (2 + √3)/8 and b³ = (2 − √3)/8 respectively, then

a₂ = ε₁a₁, a₃ = ε₂a₁, b₂ = ε₁b₁, b₃ = ε₂b₁

are the other solutions, where

ε₁ = −½ + i·½√3, ε₂ = −½ − i·½√3

represent the complex cube roots of unity. Since ε₁ε₂ = 1 it is clear the the solution pairs (a, b) satisfying ab = ¼ are

(a₁, b₁), (a₂, b₃), (a₃, b₂)

and since you set x = a + b, the three solutions of the cubic are therefore

x₁ = a₁ + b₁, x₂ = a₂ + b₃, x₃ = a₃ + b₂

or

x₁ = a₁ + b₁, x₂ = ε₁a₁ + ε₂b₁, x₃ = ε₂a₁ + ε₁b₁

or

x₁ = a₁ + b₁, x₂ = −½(a₁ + b₁) + ½√3(a₁ − b₁)i, x₃ = −½(a₁ + b₁) − ½√3(a₁ − b₁)i

So, it is clear that we have one real and two conjugate complex solutions. We can also see from this that it is easy to find all three solutions of the equation 4x³−3x−2 = 0 once we have the real solutions a₁ and b₁ of a³ + b³ = ½, ab = ¼.

[WolfgangKais]

Here’s how to proceed with method 2:
Clearly, cos(3α) = 2 has no solution for real α, but if we substitute α = iβ, we might be able to solve cos(3iβ) = 2, and then clearly x = cos(α) = cos(iβ) will solve our equation, and it will turn out that we will find the exact same real solution as from method 1 this way:
cos(iy) = cosh(y) = (e^y + e^(-y))/2, so substituting e^(3β) = u, we must solve (u + 1/u)/2 = 2. Multiplying both sides by 2u (oh no, please don’t say it…), the resulting equation is u^2 – 4u = 1, which is solved by u = 2 ± sqrt(3). The reciprocal value of 2 + sqrt(3) is its “conjugate” 2 – sqrt(3), and so -ln(2 + sqrt(3)) = ln(2 – sqrt(3), meaning that both ln(2 + sqrt(3)) and -ln(2 + sqrt(3)) = ln(2 – sqrt(3)) have the same cosh-value of 2 = cosh(3β) = cosh(-3β). It follows, that
x = cos(α) = cos(iβ) = cosh(β) = cosh((1/3)*ln(2 + sqrt(3))) =
= (e^((1/3)*ln(2 + sqrt(3))) + e^(-(1/3)*ln(2 + sqrt(3))))/2 =
= (e^((1/3)*ln(2 + sqrt(3))) + e^((1/3)*ln(2 - sqrt(3))))/2 =
= (2 + sqrt(3))^(1/3) + (2 – sqrt(3))^(1/3))/2 solves our equation (as seen in the 1st method).

[ardavanashkiani]

Using Newton-Raphson method (Numerical Method), the following solutions would be achieved:
X1 = 1.097911673
X2 = -0.548955836 + 0.392501632J
X3 = -0.548955836 - 0.392501632J

Bear in mind that the method is not similar to Baistow''s method

[ThAlEdison]

The second method directly gives you the real solution if you use the hyperbolic cosine.
4x^3-3x-2=0 let x=cosh t
4(cosh^3 t)-3cosh t-2=0
e^3t=cosh 3t + sinh 3t
e^3t=(e^t)^3=(cosh t + sinh t)^3
e^3t=cosh^3 t+3cosh^2 t sinh t+3cosh t sinh^2 t+sinh^3 t
cosh 3t is even
therefore cosh 3t = cosh^3 t+3cosh t sinh^2 t
=cosh^3 t+3cosh t(cosh^2 t - 1)
=4cosh^3 t - 3cosh t
cosh 3t=2

t=(1/3)ln(2+sqrt(3))
cosh

[rssl]

Can’t we do vieta’s substitution?
x=Z+1/4Z

[namunamu]

رائع بل اكثر من رائع
وممكن اسم التطبيق الذي تستخدمه في الكتابة

[largestcamil]

if we draw the function as y=4x^3 -3x-2 can we get the solution if we can someone give as a program that can graph all kinds of polynomials

[ronecisilva]

why is 0.5 not a solution of this function?

[HH-mwsq]

X = 1/2, and x = -1/2 are not the maxima or minima of this equation. That would be +infinity and -infinity.

[adamsmithson]

Pozdrawiam serdecznie i życzę miłego dnia