x^3 -3x=sqrt(x+2)
HTML-код
- Опубликовано: 6 май 2024
- This is the hardest algebra video I have ever done. It is so hard that I could not finish the video. I chose to use basic algebra skills for an Olympiad level problem. It was long and stressful. The method used this video is an alternative to what the RUclipsr @drpkmath1234 used to solve the same problem. Check out his solution here
• A Very Interesting Rad...
@drpkmath1234
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.
Brilliant 👌
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)
Sir please make a series of videos on methods to solve calculus problems
Email me all the problems or topics I have not covered. Make sure to search my channel first.
My way:
The conditions talked about at the beginning of the video give me:
-sqr3
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.
This is my next series of videos.
@@PrimeNewtons Can you tell me why you always wear a cap?
@Orillians Is this an investigation, inquisition, interrogation, inhibition, intrusion, invigilation, or instigation?
@@PrimeNewtons Do I get answer if I say all of them??
@@Orillians nope!
I hope to meet u someday, you've helped me a lot with my engineering entrance preparation!
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...
Wow, thank you!
When he turns to the camera and nods.
The dishes have been done.
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
Love u bro u r very good
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.
Thanks. It was hard!!!
@lawrencejelsma8118 : No, all three solutions of the cubic x³ + x² − 2x − 1 = 0 are real.
@@NadiehFan ... That is why Newton's Method of Iterative slope root finding hitting the x-axis is better than one looking for possible complex number solutions. The Iterative "Secant Method" is better than the "Newton Method" to find Polynomial real roots only by being a cleverly algorithm design to reduce the steps to find the roots within error tolerances. I've seen the cubic and quartic derived formulas (beyond having the quadratic equation existing for second order or x^2 not 0). Unless a programmer stores each of those mathematics operations in line codes they will have a neat less math operations program calculating the results by computer. The Newton's Method, Secant Method and Muller's Method were designed by Computing calculator mathematicians to let the an = f(an-1) iteration mathematics compute the solutions in feedback steps.
Where iteration is most important from mathematics are in Electrical Circuit Filters Design coursework studied in Junior and Senior electrical Engineering where there is no way a design other than electrical feedback produces the same signal transfer function signals of input signals to output signals without circuit feedback loops. That is because there are no electrical components that can be made to have the transfer function of all internal component hardware that can simulate a feedback of outputs of electrical signals back into an input of another electrical signal. Within the many components in electrical filter circuits designs an electrical hardware design in electronics would be nearly impossible without iteration mathematics. Because of this, iterative electrical signals mathematics will never go away. Ideal mathematics in presented nice Cubic and Quartic formulas have no neat electrical component equivalence in electrical entry needing. Using feedback of outputs to inputs within electrical hardware designs iteratively make designing electronics hardware an iterative simple no need to think about an electrical hardware component a must need, especially in electronic hardware "electric filter circuit designs!"
@@PrimeNewtons ... The reason why iterative mathematics of Newton's Method, Secant Method and Muller's Method is important is by reason of a "newest state" is described by a function of past states described by mathematics iteration models. The straightforward cubic and quartic polynomial formulas come from scientists historical difficulties to come up with design hardware components whose transfer function signal inputs to outputs would have been impossible without any feedback of signal theories in electronics hardware designs. Even for third and fourth order polynomial transfer function characteristics of output to input signals in an electronic circuit analysis in electrical engineering the first feedback electronics closed loops had to eliminate a hardware design impractical to make in electronics without some feedback of some part of output signal added back into an input signal.
One big example are simple radio frequency *mixer feedback signals" used to extract from a broad range of input signal frequencies a particular output frequency (a design the allies designed in World War II). Iteration mathematics is therefore, to be consistent with electronic component designs in electrical engineering, a must in considering polynomials of third order and higher. Note that a quadratic polynomial transfer function of output to input signals can result from resistors, capacitors and inductors components with no need to design another electronics engineering component. In fact, after those three passive components of resistors, capacitors and inductors, even with transistors components in circuitry there became the need to feed back some output signals somehow back into certain input signals. In Electrical Engineering this becomes "Iterative signals" combining to an input signal to in summation create a proper output signal. Iterative mathematics cannot be neglected if an Electrical Engineer plans to design any electronic circuitry with more components other than the circuitry of the three main resistor, capacitor and inductor components used in the designs.
To be constant to higher order polynomials greater than fourth order polynomials in math, third and fourth order polynomial mathematics rely on higher order iteration mathematics to consistently solve for roots by Iteration means. We know about Cubic and Quartic polynomial formula solutions as Electrical Engineers but there is no other component in electronics having properties of those equations. Note that even the next component (that isn't termed a passive component) of a transistor relies on feeding back some of its output signal in the circuit with a passive RLC or just short circuit wire back into a part of its input signal (as an iteratively mixed input signal in electronics designs).
This is quite informative. Thank you
Simple graphic sketching gives one immediate answer x=2
I don't think it can be zero because it won't be equal to each other , it be 0 = sqrt2
Just try numbers 🤠🤠 x=2
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
That's what I'm planning to explore soon. I know it would be another rabbit hole for me. Appreciate the suggestion.
Yes it can take 30 minutes to solve
If we want to get rid of complex radicals we need to use de Moivre theorem cube roots of unity and and complex mulitiplication is optional
@@PrimeNewtons general approach for the quintic needs hypergeometric functions or stuff like this
This is solvable quintic so maybe group theory or symmetric polynomials will help
As symmetric polynomial i mean polynomial usually in more than one variable such that any permutation of variables does not change the polynomial
@@holyshit922 That's the so called _casus irreducibilis_ but you can solve this equation trigonometrically. The intriguing thing here is that the roots of
x³ + x² − 2x − 1 = 0
are
x₁ = 2·cos(²⁄₇·π), x₂ = 2·cos(⁴⁄₇·π), x₃ = 2·cos(⁶⁄₇·π)
but _you will not find these expressions_ using the standard trigonometric method to solve a cubic equation with three real roots. Substituting
x = (z − 1)/3
you will get
z³ − 21z − 7 = 0
and solving this in the conventional way we can express the roots 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 have
x₁ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7))
x₂ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) − ²⁄₃·π)
x₃ = −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
x₁ = 2·cos(²⁄₇·π)
x₂ = 2·cos(⁴⁄₇·π)
x₃ = 2·cos(⁶⁄₇·π)
Graphing both sides y1(x)=x^3-3x and y2(x)=sqrt(x+2) can bring you more easily to the solution
Yeah, curve sketching might help here, but for now we have to decide what we are to do with the cubic.
Dr. PK used trig substitution in this video.
Oh that was painful 😓
i think it is called Horner
X³ +4x = 24 Do you have a solution for this ?
By inspection x=2
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)
The equation
4x³ − 3x = √(1 − x²)
is _much easier_ than the equation discussed in this video because, unlike the equation in the video, _this_ equation can be solved using nothing more than elementary algebra. Of course, a trigonometric solution of this equation is also possible, because if we substitute
x = cos α
this equation reduces to
cos 3α = sin α
_provided_ we restrict α to an interval where sin α does not become negative. You have to be careful here because the square root on the right hand side of the original equation is nonnegative but sin α can take on negative values. But if we restrict α to the interval [0, π] then sin α will be guaranteed to be nonnegative without imposing restrictions on the possible values of cos α because cos α can then still have any value on the interval [−1, 1]. We can now solve the trigonometric equation as follows
cos 3α = sin α
cos 3α = cos(½π − α)
3α = ½π − α + 2kπ ⋁ 3α = −½π + α + 2kπ, k ∈ ℤ
4α = ½π + 2kπ ⋁ 2α = −½π + 2kπ, k ∈ ℤ
α = ⅛π + ½kπ ⋁ α = −¼π + kπ, k ∈ ℤ
Taking into account that α ∈ [0, π] this gives the solutions
α = ⅛π ⋁ α = ⅜π ⋁ α = ¾π
and since x = cos α this gives the solutions
x = cos(⅛π) ⋁ x = cos(⅜π) ⋁ x = cos(¾π)
for our original equation. Of course you would want to express these solutions in algebraic form if possible, and that can be done in this case. First of all, we have cos(¾π) = −cos(¼π) = −½√2. Then, using the double angle identity cos 2θ = 2·cos²θ − 1 we have cos(¼π) = 2·cos²(⅛π) − 1 so cos²(⅛π) = ½(1 + cos(¼π)) = ½(1 + ½√2) so cos(⅛π) = √(½(1 + ½√2)) = ½√(2 + √2)) since cos(⅛π) > 0. Finally, again using the double angle identity we have cos(¾π) = 2·cos²(⅜π) − 1 so cos²(⅜π) = ½(1 + cos(¾π)) = ½(1 − ½√2) so cos(⅜π) = √(½(1 − ½√2)) = ½√(2 − √2)) since cos(⅜π) > 0. So, for our original equation we have the solutions
x = ½√(2 + √2)) ⋁ x = −½√(2 − √2)) ⋁ x = −½√2
But, unlike the equation in the video, the equation 4x³ − 3x = √(1 − x²) can also be solved using nothing more than elementary algebra. Squaring both sides we have
16x⁶ − 24x⁴ + 9x² = 1 − x²
16x⁶ − 24x⁴ + 10x² − 1 = 0
which is a bicubic equation, i.e. a cubic equation in x². Substituting y = x² we have
16y³ − 24y² + 10y − 1 = 0
To turn this into a monic cubic equation, that is, a cubic equation where the coefficient of y³ is 1, we could divide both sides by 16, but that is not a smart thing to do because we then get fractional coefficients. Therefore, we first multiply both sides by 4, because we can then rewrite 4·16y³ = 64y³ as (4y)³ and 4·24y² = 96y² as 6·(4y)² and 4·10y as 10·(4y). This gives
(4y)³ − 6·(4y)² + 10·(4y) − 4 = 0
and substituting z = 4y this gives
z³ − 6z² + 10z − 4 = 0
which is a monic cubic equation. Using the rational root theorem we know that the only possible candidates for rational solutions of this equation are 1, −1, 2, −2, 4, −4. However, since the coefficients of the _odd_ powers of z are _positive_ and the coefficients of the _even_ powers of z are _negative_ we can also see that this equation can not have any negative solutions. Therefore, we only need to try 1, 2, 4 and then we quickly find that z = 2 is a solution. This means that the cubic polynomial in z at the left hand side contains a factor z − 2 which we can take out in order to solve the remaining quadratic equation in z to get the other solutions of the cubic equation. The easiest way to take out this factor is to subtract 2³ − 6·2² + 10·2 − 4 = 0 from both sides which gives
(z³ − 2³) − 6(z² − 2²) + 10(z − 2) = 0
(z − 2)((z² + 2z + 4) − 6(z + 2) + 10) = 0
(z − 2)(z² − 4z + 2) = 0
z − 2 = 0 ⋁ z² − 4z + 2 = 0
z = 2 ⋁ (z − 2)² = 2
z = 2 ⋁ z = 2 + √2 ⋁ z = 2 − √2
Since z = 4y and y = x² and therefore z = 4x² which implies x = ½√z ⋁ x = −½√z this gives
x = ½√2 ⋁ x = ½√(2 + √2) ⋁ x = ½√(2 − √2)
x = −½√2 ⋁ x = −½√(2 + √2) ⋁ x = −½√(2 − √2)
for the solutions of the equation 16x⁶ − 24x⁴ + 10x² − 1 = 0. However, since we obtained this 6th degree equation by squaring both sides of our original equation 4x³ − 3x = √(1 − x²) these solutions also include the solutions of 4x³ − 3x = −√(1 − x²). So, we need to check which of these solutions are the solutions of 4x³ − 3x = √(1 − x²). This is easy to do since the left hand side 4x³ − 3x = x(4x² − 3) = x(z − 3) needs to be positive. For z = 2, x needs to be negative so we have the solution x = −½√2. For z = 2 + √2, x needs to be positive so we have the solution x = ½√(2 + √2) and, finally, for z = 2 − √2, x again needs to be negative so we have the solution x = −½√(2 − √2). So, the solutions of 4x³ − 3x = √(1 − x²) are
x = −½√2 ⋁ x = ½√(2 + √2) ⋁ x = −½√(2 − √2)
and this of course agrees with the solutions we found trigonometrically.
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).
Just wow
"I must conclude that this is clearly a contrived problem."
You don't say.
This is not really contrived. Just a bit challenging. We dont say math problem is contrived just because it is difficult.
@@highlyeducatedtruckerI agree with you
@@highlyeducatedtrucker I agree with you
Sensical 😂😂😂
You've been struggling hard, like a math soldier, and have done all you could! 🫡
Yooo love you two collaborating on a problem. BTW, that drpkmath1234 on the title is not clickable
X=2
You lost me e we With the remainder Theron
It is also called as a rational zero theorem
@@iqtrainer The rational root theorem for polynomials and the polynomial remainder theorem are two very different theorems, even though they can often be used together in solving an algebraic equation.
The _rational root theorem_ gives a (finite) list of potential _candidates_ for rational solutions of an algebraic equation with _integer_ coefficients. This theorem can be used to draw up a list of potential rational solutions which can then be tested. In this way you can find all rational solutions of an algebraic equation with integer coefficients. If none of the potential candidates check out, then you can be sure that the equation has no rational solutions, because if it had, these should have been on the list of candidates.
The _polynomial remainder theorem_ says that if you divide a polynomial P(x) by (x − a) then the quotient is some polynomial Q(x) and the _remainder_ of this division is P(a), that is, the value of P(x) for x = a.
Essentially, for any polynomial P(x) and any number a the polynomial remainder theorem guarantees the existence of a polynomial Q(x) and a number R such that
P(x) = (x − a)·Q(x) + R
and by setting x = a this gives P(a) = (a − a)·Q(a) + R = 0·Q(a) + R = R so we have
P(x) = (x − a)·Q(x) + P(a)
If x = a happens to be a zero of the polynomial P(x), that is if x = a satisfies the equation P(x) = 0, so P(a) = 0, then this reduces to
P(x) = (x − a)·Q(x)
So, if x = a is a zero of a polynomial P(x), then (x − a) is a _factor_ of this polynomial, that is, then there exists a polynomial Q(x) such that P(x) = (x − a)·Q(x) for all x. Of course the converse is also true and this is trivial because if we have P(x) = (x − a)·Q(x) then P(a) = (a − a)·Q(a) = 0·Q(a) = 0, regardless of the value of Q(a).
So, we also have the _polynomial factor theorem_ which says that (x − a) is a factor of a polynomial P(x) _if and only if_ x = a is a zero of P(x). Of course, the factor theorem is really just a special case of the polynomial remainder theorem.
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.
Only cheaters see something as cheating. That means you are a cheater. You said for a cubic look for a trig sub and call the exact same direction as a very unlikely direction? Prove that he already knows the answer before making a video. Prove in a logical way. Otherwise, watch your dirty mouth. If you present your ground for making such a negative comment (maybe out of jealousy), that would be helpful to make youtube more of a clean space rather than meaningless criticism like yours.
Prove he already knows the answer before making a video. Did you watch it? Did you see it? Are you jealous of anyone with higher math skills than you? Are you born with a DNA for negative comments? You are a life cheater so see something as cheating.
What's your reason to say cheating? Did you see him in person that he knows the answer before making a video? Did you hide a camera in his house? How can you prove he already knows the answer?
@@domedebali632 Well said. That Andrew guy seems to be very crooked and twisted mentality guy. He said trick works then said Dr PK already knows the answer. What a loser comment
@@domedebali632 Come on: I wrote "I think". The point is that Dr P K Math galloped off in a certain direction which turned out very luckily to be successful. Maybe he spent 3 hours before the video trying out lots of answers. Maybe he took this problem from a list of "algebraic problems solved by nifty trig substitution". If students are looking at this who need to tackle such problems in exams, then it would be valuable for Dr P K Math to motivate the direction taken. If you want a conjuring show, then there are plenty of RUclips channels for that. But this is meant to be educational