Solving a Cubic Polynomial in Two Ways

This video is about solving a Polynomial Equation
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[seshnarayan]

I love how you explain more than one method for the same problem

[pmccarthy]

Wow! I don't think I've ever seen someone approach the cubic from quite that perspective before. You display quite a bit of depth of understanding of this material.

[snejpu]

Random, not connected observation: if you start from 1 and then add 1 and reverse the fraction, you will get 1/2. If you continue adding 1 and reversing a fraction, you will generate Fibonacci numbers in the denominators. That surprised me, because I've never known this.

[holyshit]

This can be solved using trigonometry using triple angle formula for cosine or sine
cos(3t)=4cos^3t-3cost

[n_digit_flow_]

Awesome sir, I really impressed by this channel

[mxsjncv]

From the subject "relationship between a function and its derivetives" I know that a three degree polinom f(x)=ax^3 .... if a ia positive, the graph is or only increasing. In this case it has two values that zero the derivetive. So -2 is minimum and 2 is max. So the roots are above 2, below -2 and between 0 and 1. But since the roots are not "nice" numbers, the suggested way of solving is very good.

[danielsittner]

I've got a cool question you (and anyone who reads this for that matter) can solve. There is an equilateral triangle with side lengths of 10 cm. On each vertex there is a turtle such that turtle 1 always looks at turtle 2, turtle 2 always looks at 3 and so on. They all start to move simultaneously in a speed of 10 cm per second while maintaining their "target" turtle in their sight. The turtles are dots.
1. How long will it take for them to meet and how long will they have to walk?
2. How many rounds will they go around the center of the equilateral triangle they started on?

[hsjkdsgd]

I, by my intuition, proceeded with the 2nd method. Finished it in style!

[timetraveller]

funny how i just learned cardano's method this morning

[happiness]

i just found your channel yesterday and i am loving the content :D
By the way one question for you

*Find the sum of all integers n, such that 1 ≤ n ≤ 1998 and*

*that 60 divides*

*n³ – 30n² + 100n*

[UttamKumar-ktjp]

Both the methods were equally good .just loved it.

[CaradhrasAiguo]

9:25 or note that the angles from 5 * pi/9 + n * 2pi/3 are 2*pi-compliments of the 3 angles found from pi/9 + n*2*pi/3, e.g. 5 * pi / 9 + 13 * pi / 9 = 2 * pi and so on, and cos(2 * pi - theta) = cos(theta) for all real theta

[usernamehere]

Very interesting, where did you discover this method?

[LOL-gnkv]

Great video !
Another way could be using cardan's method of solving cubics as it is already in the form of a reduced cubic

[tonyhaddad]

Good job !!! Very helpful video thanks !!!

[vanshr.sachan]

For the second method don't we have to go with the assumption that the value of x at which the function 8x³ - 6x -1 is zero is between -1 and 1 (as cos fucntions range is from -1 to 1)
Am I missing something... Pls let me know...

[caeoranger]

Hi at 2:38 you gave a+b=2x, if that is so then a^3 +b^3 should be in terms of x as well (should not equal 1)

[vishalmishra]

*General Formula* Transform equation to x^3 = 3mx + 2n. If m > 0 and m^3 > n^2 then x = 2√m Cos T/3 where Cos T = n/m^(3/2). Let's apply the general formula to this problem. x^3 = 3(1/4)x + 2(1/16). So, m = 1/4 and n = 1/16. Applying pre-condition, m > 0 and m^3 = 1/64 > 1/256 = n^2, so the formula applies in this equation. Cos T = n / m^(3/2) = (1/16) / (1/8) = 1/2 = Cos 60. Therefore, x = 2√m Cos T/3 = 2 √(1/4) Cos 60/3 = 1 x Cos 20 = Cos 20. Add/Subtract 120 for the other 2 real but negative solutions, -Cos 40 and -Cos 80, or simply x = Sin (70, -10, -50).

[owhycvb]

I actually did it with the second method you showed and seconds before the end of the video I was ready to comment about the other solutions from the trigonometric equation. It's important to wait till the end and listen carefully 😂

[mxsjncv]

Obviously, if it had rational roots there were 1, -1, 1/8, -1/8, 1/4, -1/4, 1/2, -1/2