x^3 -3x=sqrt(x+2)  

Bro really pulled a “the solution to this equation is left to the viewer as an exercise” at the end

Here's how I factored the 5th-degree polynomial: ASSUME that the polynomial can be factored into a quadratic and a cubic with whole-number coefficients. That means that the factored polynomial will take the form (x^2+Ax+B)(x^3+Cx^2+Dx+E). Now multiply those two polynomials. The result isn't pretty, but after you expand everything and then collect terms, you'll get: x^5 + (A + C)x^4 + (AC + B + D)x^3 + (AD + BC + E)x^2 + (AE + BD)x + BE Compare that with the original 5th-degree polynomial: x^5 + 2x^4 - 2x^3 - 4x^2 + x + 1 Thus: A + C = 2 AC + B + D = -2 AD + BC + E = -4 AE + BD = 1 BE = 1 And voila, you've got five equations and five unknowns. The algebra can be simplified further because, as you note in the video, B and E can only be 1 or -1. So do a Case 1 where B = E = 1, and do a Case 2 where B = E = -1. In both cases the problem reduces to three equations and three unknowns, and some basic algebra will reduce those three equations into a quadratic equation in terms of A. Use your favorite method to solve for A. Of the four possible results for A (two from Case 1, two from Case 2), only one is a whole number. From there, you can find values for C and D. And Bob's yer uncle.

Sir please make a series of videos on methods to solve calculus problems

I got your back bro, the cubic can be solved very fast. But in this specific problem before even solving the cubic you should use Calculus to see where’s both functions maximums and minimums if any. And see where they are decreasing or increasing because it could have been the case that this problem only had two real solutions and the cubic solution although guaranteed to have 1 real root it doesn’t imply that it will satisfy the original equation because it could be extraneous due to you squaring both sides in the beginning. Now to solve this cubic we can apply Scipione Del Ferro’s method of solving a cubic by depressing it. Which means we have to get rid of x^2 to complete the Cube. Which means we let x=y-b/3, where b is the coefficient of x^2. After this substitution we will have a depressed cubic and can solve it by using the Cardano/Tartaglia cubic formula. Even though I think Del Ferro was the one who gave it to Cardano.

The cubic at the end is fairly well-known. Its roots are 2cos(2pi/7), 2cos(4pi/7) and 2cos(6pi/7). The remaining solution is 2cos(6pi/7)

We just found a very rare 6th degree polynomial that all solutions are reals we got 2, -phi, 1/phi (such that phi=1+sqrt(5)/2) and 3 other real solution that we can see on a graph

Absolutely love your channel. You're a heck of a teacher! Great enthusiasm, great delivery, great explanations. And a wondefful motto. What a way to end the lessons. So great. When are you selling merch?? Mugs! T-shirts! My daughter wants one...

My way: The conditions talked about at the beginning of the video give me: -sqr3

Just try numbers 🤠🤠 x=2

When he turns to the camera and nods. The dishes have been done.

I don't think it can be zero because it won't be equal to each other , it be 0 = sqrt2

Let f(x)=x^3-3x, g(x)=sqrt(x+2). At first (you can prove it easily) for x>2 f(x)>g(x), so no one x>2 can be a solution. Thus, x=-2(Domain). Then, -2 -2sin(7t/4)sin(5t/4)=0... I think this way we can save time for solution.

I wrote a program to iteratively solve the equation. My initial guess was x=2. (I didn't watch the video).

Sir, kindly show the finding of 3 nos. imaginary roots for x^3+x^2 -2.x -1=0 .

my warm up in the morning i going through your videos.. it helps stimulates my brain plus my love for maths

I know there is a Cubic Root Formula but this is Mathematics overkill. You could use Wolfram Alpha for x^3 + x^2 - 2x - 1 = 0 for the complex conjugate pair and irrational real root (3 roots for that part). Wolfram Alpha will use Cubic Root solving (probably different than Muller's Method of Iterative Numerical Analysis beyond real roots finder Newton's Method). That part is highly advanced as the Cubic Roots formulas and any iterative method of Newton's Method (preferably if not Secant Method of Iterative Roots finding). Muller's Method finds the Complex Conjugate Imaginary Roots in addition. All this is Computer Algorithms Iterative formula coding to find irrational, real, and integer solutions of Polynomials. It was nice seeing your getting the linear (x + 2)(x^2 + x - 1)(x^3 + x^2 - 2x - 1) and first three possible xs of x=2, x= (-1 + √5)/2, x not equal to (-1 - √5)/2 parts before trying to find the Cubic real root not using Iterative Newton's, Secant or Muller's Methods in Numerical Analysis for Computer Programmer algorithms.

Having watched both your video and that of Dr PK Math on solving the equation (1) x³ − 3x = √(x + 2) I must conclude that this is clearly a contrived problem. Squaring both sides we obtain (2) x⁶ − 6x⁴ + 9x² − x − 2 = 0 This is a 6th degree equation which of course also has the roots of (3) x³ − 3x = −√(x + 2) since this gives (2) as well when both sides are squared. For anyone who is familiar with Chebyshev polynomials it is quite clear what we have here. If we substitute x = 2t in (2) we obtain (4) 64t⁶ − 96t⁴ + 36t² − 2t − 2 = 0 which can be written as (5) 32t⁶ − 48t⁴ + 18t² − 1 = t where the left hand side is a Chebyshev polynomial of the first kind. In fact the left hand side of (5) is T₆(t) and the right hand side is T₁(t) so we can rewrite (5) as (6) T₆(t) = T₁(t) Chebyshev polynomials of the first kind can be defined by the recurrence (7) T₀(x) = 1, T₁(x) = x, Tₙ₊₁(x) = 2x·Tₙ(x) − Tₙ₋₁(x) which means that they satisfy (8) Tₙ(cos θ) = cos nθ Therefore, if we substitute t = cos α in (5) or x = 2t = 2·cos α in (2) these equations reduce to (9) cos 6α = cos α and this trigonometric equation is easily solved. We have 6α = α + 2kπ ⋁ 6α = −α + 2kπ, k ∈ ℤ 5α = 2kπ ⋁ 7α = 2kπ, k ∈ ℤ α = ²⁄₅·kπ ⋁ α = ²⁄₇·kπ, k ∈ ℤ so for (2) we have the roots x₁ = 2·cos 0, x₂ = 2·cos(²⁄₅·π), x₃ = 2·cos(⁴⁄₅·π), x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) Here x₁ = 2·cos 0 = 2 so the left hand side of (2) has a factor x − 2 and x₂ = 2·cos(²⁄₅·π) = −¹⁄₂ + ¹⁄₂√5, x₃ = 2·cos(⁴⁄₅·π) = −¹⁄₂ − ¹⁄₂√5 are the zeros of the quadratic polynomial x² + x − 1 which therefore also is a factor of the left hand side of (2). The remaining roots x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) are the roots of the cubic equation (10) x³ + x² − 2x − 1 = 0 To prove this, we can substitute x = u + 1/u in (10) which results in (11) u⁶ + u⁵ + u⁴ + u³ + u² + u + 1 = 0 Since the left hand side of (11) is a geometric series with sum (u⁷ − 1)/(u − 1) we can also write this as (12) (u⁷ − 1)/(u − 1) = 0 Clearly, the roots of (12) and therefore of (11) are the seventh roots of unity except for u = 1 itself, so we have u = exp(k·²⁄₇·π·i), k = 1..6 and since x = u + 1/u this means that the roots of (10) are x = exp(k·²⁄₇·π·i) + exp(−k·²⁄₇·π·i) = 2·cos(k·²⁄₇·π), k = 1..6. Of course, for k = 4..6 we get the same values as for k = 1..3 so we indeed have the three roots x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for (10) and the left hand side of (10) is therefore also a factor of the left hand side of (2). WolframAlpha is incapable of coming up with the solutions x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for (10) and instead produces expressions for the solutions of (10) that contain cube roots of complex numbers even though all solutions of (10) are real. This is because the solutions of a cubic equation that has three distinct real but irrational roots cannot be expressed algebraically in radicals without using complex numbers. This is known as the _casus irreducibilis_ but there is a standard method to solve cubic equations with three distinct real but irrational roots trigonometrically, without using complex numbers. However, the intriguing thing here is that the expressions x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for the roots of (10) cannot be obtained either by using the standard trigonometric method to solve a cubic equation with three real roots. Substituting x = (z − 1)/3 in (10) results in (13) z³ − 21z − 7 = 0 and solving this depressed cubic trigonometrially in the conventional way we can express the roots of this equation as z₁ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7)) z₂ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) − ²⁄₃·π) z₃ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) + ²⁄₃·π) So, since x = (z − 1)/3, we can express the roots of (10) as x₄ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7)) x₅ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) − ²⁄₃·π) x₆ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) + ²⁄₃·π) but it is not evident at all how these expressions can be reduced to the much simpler expressions x₄ = 2·cos(²⁄₇·π) x₅ = 2·cos(⁴⁄₇·π) x₆ = 2·cos(⁶⁄₇·π) Since we have proved that x − 2, x² + x − 1 and x³ + x² − 2x − 1 are all factors of the left hand side of (2) and since there are no other factors since any root of (2) is a zero of one of these factors it follows that we can write (2) as (14) (x − 2)(x² + x − 1)(x³ + x² − 2x − 1) = 0 Since the set of roots of (2) consist of the union of the set of roots of (1) and the set of roots of (3) we still need to find out which roots of (2) are roots of (1) and which roots of (2) are roots of (3). To do this, we can apply the double angle identity for the cosine to (9) which gives (15) 2·cos² 3α − 1 = 2·cos² ½α − 1 so we have cos² 3α = cos² ½α and therefore (9) is equivalent with (16) cos 3α = cos ½α ⋁ cos 3α = −cos ½α where the first equation is also obtained by substituting x = 2·cos α in (1) and the second by substituting this in (3) subject to the condition 0 ≤ α ≤ π to ensure the proper sign of √(x + 2) = cos ½α. Solving cos 3α = cos ½α gives α = ⁴⁄₅·kπ ⋁ α = ⁴⁄₇·kπ, k ∈ ℤ so taking the condition 0 ≤ α ≤ π into account we have the roots x₁ = 2·cos 0 = 2, x₃ = 2·cos(⁴⁄₅·π) = −¹⁄₂ − ¹⁄₂√5, x₅ = 2·cos(⁴⁄₇·π) for (1). Similarly, solving cos 3α = −cos ½α gives α = ²⁄₇·(2k + 1)π ⋁ α = ²⁄₅·(2k − 1)π, k ∈ ℤ and taking the condition 0 ≤ α ≤ π into account this gives the roots x₂ = 2·cos(²⁄₅·π) = −¹⁄₂ + ¹⁄₂√5, x₄ = 2·cos(²⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for (3).

Вообще идиотическое решение идиотской задачи

Sensical 😂😂😂

What a lucid and dedicated explanation to the factorization part. May you be granted the best.

You've been struggling hard, like a math soldier, and have done all you could! 🫡

Simple graphic sketching gives one immediate answer x=2

I hope to meet u someday, you've helped me a lot with my engineering entrance preparation!

X=2

X³ +4x = 24 Do you have a solution for this ?

If we let y = sqrt(x+2) we will get polynomial equation (y^3-y^2-3y+2)*(y^3+y^2-2y-1) = 0 We have two cubic equations to solve y^3-y^2-3y+2 = 0 is a little bit easier because we can test candidates for integer roots and reduce deqree by division to get quadratic equation For equation y^3+y^2-2y-1 = 0 we probably need general approach to the cubic equation like Cardano formula or its derivation

Можно ли писать крупнее? Подписался.

i just study this problem today and my teacher gave me the same solution as drpkmath1234. Here is a smiliar problem. My teacher used it as a stepping stone for us to get the idea to solve this problem using trigonometry 4x^3-3x=sqrt(1-x^2)

i think it is called Horner

Graphing both sides y1(x)=x^3-3x and y2(x)=sqrt(x+2) can bring you more easily to the solution

By inspection x=2

Love u bro u r very good

Oh that was painful 😓

You lost me e we With the remainder Theron

Yooo love you two collaborating on a problem. BTW, that drpkmath1234 on the title is not clickable

Thanks I liked this. I think it's an honest approach to a hard problem. In contrast, I think Dr P K Math's approach is cheating. He already knows the answer before he heads off in what seems like a very unlikely direction. His is not a general approach, but a trick that only works in this specific case (cos(pa)=cos(qa)). If he has good reasons for exploring the avenue that he does, it would be helpful to students who need to budget time in exams if he can explain this motivation, rather than just pulling a rabbit out of a hat. However once you've come down to a cubic, it's believable to look for a trig substitution. Cubics are simple enough that one can probably look at the coefficients and see when that can work out. But that's for your second video.