Solving a cubic equation with a parameter

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  • Опубликовано: 25 янв 2025

Комментарии • 20

  • @a.raghavan9934
    @a.raghavan9934 3 года назад +2

    Isn’t there an error when we take sqrt 3 into cube root, you can’t multiply by 3.

    • @SyberMath
      @SyberMath  3 года назад +1

      That's an error!
      😂

  • @neuralwarp
    @neuralwarp 3 года назад +1

    You: okay, right, good, so, well, what we're gonna do is, hopefully you can already see, right, no, actually, I'll tell you that later, awesome, ...
    Me: JUST GET ON WITH IT

  • @neuralwarp
    @neuralwarp 3 года назад +1

    What happened to the sqrt 3? Did you distribute it into the cube root?

    • @SyberMath
      @SyberMath  3 года назад +1

      I think I did!
      😂

  • @eliasmazhukin2009
    @eliasmazhukin2009 3 года назад

    A smart Italian mathematician: *discovers the cubic formula so that people could solve cubic equations quickly and easily*
    People: Screw it, let's solve it in a smart way:
    The guy who discovered the cubic formula: Am I a joke to you?

  • @lucanina8221
    @lucanina8221 Год назад

    One way I discovered it has only one solution is to take the derivative of the expression and the a parameter gets away finding is zero at x=1/2 and always positive otherwise, that means that the cubic expression is of the form (cx+b)^3 +(a -b^3) with c and b to be calculated. but you solution is more elegant and clever

  • @seegeeaye
    @seegeeaye 2 года назад

    The cubic equation has one real solution and two complex solutions, or three real solutions.
    We rewirte the equation as 2X^2(2x -3) = a - 3x, the both sides as graphs, they intersect at only one point.

  • @michaelempeigne3519
    @michaelempeigne3519 4 года назад +4

    sqrt ( 3 ) * cbrt ( 7 ) is not equal to cbrt ( 21 )

    • @SyberMath
      @SyberMath  4 года назад +4

      You're right! Thanks for the heads up! That escaped my attention. The correct way to combine would be to use the 6th root and write sqrt(3) times cube root of (2a-1) as 6th root 27(2a-1)^2

  • @kaslircribs5804
    @kaslircribs5804 3 года назад

    You have a very nice way of solving problems. I like your method sir. Thank you.

    • @SyberMath
      @SyberMath  3 года назад

      You’re welcome! Thanks for the kind words! 💖

  • @kamaljain5228
    @kamaljain5228 3 года назад +1

    when you got the first solution, it has cuberoot of a real number. there are three cuberoots of a real number. the other two you get by multiplying the real cuberoot with w and w^2, where w is a cuberoot of unity. so right there you had all 3 solutions.

  • @michaelempeigne3519
    @michaelempeigne3519 4 года назад +3

    you cannot combine a square root and cube root like you did.

  • @hp.basketball
    @hp.basketball Год назад

    How do you solve a depressed cubic equation with unknown parametres example 2x^3 +(2k-1)x -4h=0