China | A Nice Algebra Problem | Math Olympiad

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  • čas přidán 27. 08. 2024
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Komentáře • 15

  • @reguedebulle
    @reguedebulle Před měsícem +3

    I'm so happy. The more I look your video, the more I'm able to find the solution by myself. I'm 33 and it makes a long time I went to school and I continu to increase my level in math. You are not the only "Math CZcamsr" I follow, but I thank you a lot. It's because of you a lot ^^
    I hope my English is good lol

  • @greninjamariokartpokemonfan

    x⁴-12x³+54x²-108x+81=16
    (x²-6x+9)²=16
    (x-3)⁴=16
    x-3=2 or x-3=-2
    x=5 or x=1

  • @1976anands
    @1976anands Před měsícem +2

    Why can this be done through simple substitution ?
    (X-3) ^4 = 2^ 4
    So X-3 = 2
    Hence X = 5.

  • @andryvokubadra2644
    @andryvokubadra2644 Před měsícem

    Method I
    (x-3)⁴ = 16
    (x-3)⁴ = 2⁴
    x-3 = ±2
    x = ±2 + 3
    x = 5 & 1
    Methode 2
    (x-3)⁴ = 16
    (x-3)² = 4
    x² - 6x + 9 = 4
    x² - 6x +5 = 0
    (x-5)(x-1) = 0
    x = 1 & 5
    ==============
    But it shall have 4 roots. I've found 2, left 2 more. Coment below if found other methode beside above 😅😅😅

  • @doichecabano
    @doichecabano Před měsícem +1

    X=1/5/3+2i/3-2i

    • @YloGamerPro
      @YloGamerPro Před měsícem

      That's a big fraction. You could simplificate.
      (It's a joke) I know that you are separating the solutions XD.

  • @zlatabunjevac4094
    @zlatabunjevac4094 Před měsícem

    HVALA

  • @pelasgeuspelasgeus4634
    @pelasgeuspelasgeus4634 Před měsícem

    Tell me please. If equation was (x-3)^8 how many solutions would exist?

    • @mathshunter
      @mathshunter  Před měsícem

      8 solutions
      Number of solution will be equal to degree of equation (highest power of the variable)

    • @pelasgeuspelasgeus4634
      @pelasgeuspelasgeus4634 Před měsícem

      @@mathshunter How many real and how many imaginary?

    • @mathshunter
      @mathshunter  Před měsícem +1

      Depends on several factors
      You can read the chapter 'quadratic equation' to know the number of real and imaginary root of any complex equation. You can also get my ebook on quadratic equation in few days (link will be in the description)

    • @pelasgeuspelasgeus4634
      @pelasgeuspelasgeus4634 Před měsícem

      @@mathshunter I see. Well, as a help on your ebook, please provide a counter argument on the following:
      (1) the definition of a complex number contradicts to the laws of formal logic, because this definition is the union of two contradictory concepts: the concept of a real number and the concept of a non-real (imaginary) number-an image. The concepts of a real number and a non-real (imaginary) number are in logical relation of contradiction: the essential feature of one concept completely negates the essential feature of another concept. These concepts have no common feature (i.e. these concepts have nothing in common with each other), therefore one cannot compare these concepts with each other. Consequently, the concepts of a real number and a non-real (imaginary) number cannot be united and contained in the definition of a complex number. The concept of a complex number is a gross formal-logical error;
      (2) the real part of a complex number is the result of a measurement. But the non-real (imaginary) part of a complex number is not the result of a measurement. The non-real (imaginary) part is a meaningless symbol, because the mathematical (quantitative) operation of multiplication of a real number by a meaningless symbol is a meaningless operation. This means that the theory of complex number is not a correct method of calculation. Consequently, mathematical (quantitative) operations on meaningless symbols are a gross formal-logical error;
      (3) a complex number cannot be represented (interpreted) in the Cartesian geometric coordinate system, because the Cartesian coordinate system is a system of two identical scales (rulers). The standard geometric representation (interpretation) of a complex number leads to the logical contradictions if the scales (rulers) are not identical. This means that the scale of non-real (imaginary) numbers cannot exist in the Cartesian geometric coordinate system.

  • @deep_uu4335
    @deep_uu4335 Před měsícem +1

    Waste of time...why can't we equate L. H.S and R.H.S..in second step we will get answer...but you wasted so much of steps for solving this

    • @PedroBatista-lc8sj
      @PedroBatista-lc8sj Před měsícem

      In that way, he will find only one of the four possible solutions

    • @pelasgeuspelasgeus4634
      @pelasgeuspelasgeus4634 Před měsícem

      ​@@PedroBatista-lc8sjThere's no 4 possible. It's 2 real and 2 fake invented.