Solving a Nice Radical Equation in Two Ways

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  • Опубликовано: 23 дек 2024

Комментарии •

  • @Rbmukthegreat
    @Rbmukthegreat 2 года назад +2

    I finally got the 1 year badge!!! I love your videos, and I hope to get to 2 years!! Keep making AMAZING videos!

    • @SyberMath
      @SyberMath  2 года назад

      Wow!!! Congratulations and thank you for the support! 🤩💖

    • @Drk950
      @Drk950 2 года назад

      How did yo get the badge?

  • @karimjemel7405
    @karimjemel7405 2 года назад +9

    you can raise to the power of 6 it will cancel the cube root and the square root giving you after simplifying the sum of two positive terms being equal to 0 which means they're both 0 hence x=0

  • @SuperYoonHo
    @SuperYoonHo 2 года назад +1

    wow i like these problems that includes radicals

  • @hans-rudigerdrzimmermann
    @hans-rudigerdrzimmermann 2 года назад +1

    Only to congratulate you for your math channel.
    I love many of your problems.

  • @hornet-e9g
    @hornet-e9g 2 года назад +1

    you always surprize even though i watched this video one time

  • @elmachevera
    @elmachevera 2 года назад

    73 / 5.000
    what hardware and software do you use to write on this digital blackboard ?

  • @GirishManjunathMusic
    @GirishManjunathMusic 2 года назад +1

    It's the Perry the Platypus colour.

  • @Vitzyk
    @Vitzyk 2 года назад

    A visual example, as of a simple thing to make a difficult, and nothing to learn, but firmware

  • @Drk950
    @Drk950 2 года назад

    Before to watch the video: My first method was guessing ('simple inspection') xd Second method: x=z^6; solve for z {idem than yours :) }

  • @Alians0108
    @Alians0108 2 года назад

    Do the compex solutions from that equation work?

    • @XJWill1
      @XJWill1 2 года назад

      There are no other solutions than x=0, complex or not.

    • @XJWill1
      @XJWill1 2 года назад

      @@souvik9632 No, there are NOT two solutions. The only solution is x = 0.

    • @XJWill1
      @XJWill1 2 года назад

      @@souvik9632 There is only one solution to the given equation, x=0. There are no other solutions, real or complex.

    • @tontonbeber4555
      @tontonbeber4555 2 года назад

      If you take the 6th power of both sides, you have a quadric equation, 3z4-2z3+3z2=0 with 0 as double solution and 2 complex conjugate solutions.
      However by definition of the complex roots (real part always positive), one can check that the two complex solutions are not solutions of the original equation.
      By the way ... a = b^(1/2) and a^2 = b are not always equivalent ...and also a = b^(1/3) and a^3 = b
      For example ... (-1)^3 = -1 ... everybody knows ... but if you calculate (-1)^(1/3) in a complex calculator, you obtain ... 0.5 + 0.866j

  • @wesleysuen4140
    @wesleysuen4140 2 года назад

    I modified the equations a bit… and graphed y=(x^3-1)^2 and y=(x^2-1)^3, kinda cute… confirming the same intersection btw…

  • @Thomasoacademiiia
    @Thomasoacademiiia 2 года назад

    Perfect solving 👌

  • @gigi4874-w3w
    @gigi4874-w3w 2 года назад

    Thanks!!

  • @oenrn
    @oenrn 2 года назад +1

    Slight mistake at the end: the intersection is actually at (0,1).

  • @InnocentNeuron
    @InnocentNeuron 2 года назад

    It is mildly interesting that I enjoy your PJ's (don't ask 'y', '2b' or not '2b', '2u', and so on...) 🤷‍♂

  • @Jalina69
    @Jalina69 2 года назад

    Not me finding complex solutions and using moivres formula 😭😭

  • @scottleung9587
    @scottleung9587 2 года назад

    From a first glance, it's obvious x=0 is the only real solution. However, for the complex solutions I got x=((-7+-2*i*sqrt(2))/9)^3. Does that look correct?

    • @XJWill1
      @XJWill1 2 года назад

      No, x=0 is the only solution, complex or otherwise.

  • @NXT_LVL_DVL
    @NXT_LVL_DVL Год назад

    ❤❤❤❤

  • @Vitzyk
    @Vitzyk 2 года назад +1

    просто возведи обе части в шестую степень, сократи подобные члены и сделай замену корень шестой степени из x =y. Вновь сократи и получи квадратное уравнение без действительных корней. Зато найдешь мнимые.

  • @zhugzhuangzury
    @zhugzhuangzury 2 года назад

    i got some random strange answers
    which is x = (2114±√2114^2-4) /2

  • @tobymurs6495
    @tobymurs6495 2 года назад

    You have a Turkish accent, are you Turkish by any chance ?

  • @wesleydeng71
    @wesleydeng71 2 года назад

    Both methods are too sluggish. Simply raise both sides to the power of 6 and see how it goes.

    • @tontonbeber4555
      @tontonbeber4555 2 года назад

      Exactly ! That's exactly what I did ... and by elevating to the power of 6 you obtain a quadric with 0 as double solution and 2 complex conjugate. However you have to check the solutions to the original equation, and so you should reject the non-0 solutions that are not solutions of the original equation.

    • @forcelifeforce
      @forcelifeforce 2 года назад

      @@tontonbeber4555 "quadric?" You mean "quadratic," correct?

    • @MichaelRothwell1
      @MichaelRothwell1 2 года назад

      @@forcelifeforce I think he meant quartic.