Can we NOT solve x^4+64=0 like this?

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  • Опубликовано: 3 дек 2024

Комментарии • 48

  • @bprpmathbasics
    @bprpmathbasics  День назад +4

    Solve x^(2/3)=64: ruclips.net/video/JzfO9dI4aIk/видео.html

  • @PlzDoNotTheFade
    @PlzDoNotTheFade 6 часов назад +44

    *_"Believe in geometry, not 4throoting both sides"_* 🗣🗣🔥🔥🔥🔥

  • @MrConverse
    @MrConverse Час назад +5

    5:48, Why do you like to write down (A - B) first? I prefer to write (A + B) first because if A & B are perfect squares then the second factor factors further and the final answer ends up in the traditional order more naturally. For example, x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x + y)(x - y).

    • @tfg601
      @tfg601 3 минуты назад

      "Traditional order"?

  • @rogierownage
    @rogierownage 4 часа назад +7

    Before watching the video.
    The fourth roots of 1 are:
    1, -1, i, -i
    So the four solutions to our equation are those numbers times the fourth root of -64, respectively.
    The fourth root of -64 is the square root of its square root. This can be written as sqrt(sqrt(-64)) = sqrt(sqrt(64)*sqrt(-1)) = sqrt(8i). The square root of i is known as sqrt(0.5) + sqrt(0.5)*i. So the square root of 8i is sqrt(8)(sqrt(0.5) + sqrt(0.5)*i). Simplifying gets you sqrt(4) + sqrt(4)*i = 2 + 2i
    So the four solutions are:
    1(2 + 2i)
    -1(2 + 2i)
    i(2 + 2i)
    -i(2 + 2i)
    Or simplified:
    2 + 2i
    -2 - 2i
    -2 + 2i
    2 - 2i
    Using a calculator confirms that these are correct, and you expect exactly 4 answers for a x^4 equation.
    This can also be written as +-2 +-2i
    After watching the video:
    It's very interesting how many legitimate ways there are to solve this!

    • @Nikioko
      @Nikioko 32 минуты назад

      Or simply:
      i = cos(90°) + i sin(90°)
      And therefore:
      √i = cos(90°/2) + i sin(90°/2)

  • @Nikioko
    @Nikioko 29 минут назад

    6:55: This is a perfect opportunity to use the pq formula instead. You get x₁,₂ = 2 ± √(2² − 8) = 2 ± √−4 = 2 ± 2i

  • @excentrisitet7922
    @excentrisitet7922 3 часа назад +2

    Why not:
    x^4+64 = (x^2 -8i)(x^2+8i) = 0
    First bracket: x^2 = 8i 8i = 8e^(i*pi/2)
    Square root halves the angle and reduces the magnitude, so in becomes 2*sqrt(2) * (cos(45)+/-i*sin(45)) = 2*sqrt(2) * (sqrt(2)/2 +/-isqrt(2)/2) = 2+/-2i
    With second bracket we work analogously. And get the same result as in video. (mor imaginary units will appear but they should turn into minus ones)

  • @Chriib
    @Chriib 4 часа назад +3

    you could easily substitute x^2 with u and get two easy solutions with the quadratic formula and when you change back to x^2 you get two solutions from each of the first two.

    • @dogbreaththe3rd851
      @dogbreaththe3rd851 Час назад

      64 = -64i^2 substitute giving x^4 - 64i^2
      The factorization then becomes a difference of squares
      (x^2 -8i) * (x^2 + 8i)

  • @eganrabiee627
    @eganrabiee627 6 часов назад +3

    x^4= -64
    x^2=+-8i
    x=+-sqrt(+-8i)

    • @Nikioko
      @Nikioko 35 минут назад

      And since
      i = cos(90°) + i sin(90°),
      √i = cos(45°) + i sin(45°)

  • @ukdavepianoman
    @ukdavepianoman 6 часов назад +1

    I would do this as x^4 = -64 = 64 e^(k.i.pi) where k = 1,3,5,7. I know i need 4 values of k as there will be 4 roots.
    So x = 2sqrt(2) * e^(k/4.i.pi)
    = 2sqrt(2) * [1+i, -1+i, -1-i, 1-i]/sqrt(2) putting the various k options into [ ] brackets.
    Therefore x = [2+2i, -2+2i, -2-2i, 2-2i]

  • @trueriver1950
    @trueriver1950 8 часов назад +1

    i^n. 4throot 64 (for all n={1..4})
    Just as for squared equations you need +/- for a fourth power you need to take all possibilities if the fourth root of unity.

  • @RyanLewis-Johnson-wq6xs
    @RyanLewis-Johnson-wq6xs 9 часов назад +2

    Input
    (-2 - 2 i)^4 + 64 = 0
    Result
    True

  • @dlevi67
    @dlevi67 24 минуты назад

    Since you are going to talk about complex numbers anyway, it's a lot easier to factor a² + b² when noting that
    a² + b² = a² - (-b²) = a² - (i²b²) = (a + ib)(a - ib)
    With a fourth power binomial as here, you end up with two sums/differences of squares that can be factored again as differences/sums of squares.

  • @Hunni125
    @Hunni125 9 часов назад +1

    makes sense

  • @ironicstupidity7102
    @ironicstupidity7102 8 часов назад +2

    instead of taking the fourth root of both sides couldnt you instead take the square root twice and do plus or minus both times? You should get ±sqrt(±sqrt(-64)) which when worked out would give the same four answers?

  • @randomjin9392
    @randomjin9392 7 часов назад +3

    When I was in mid-school and started to learn more math, I was always wary of the "cool" tricks. It was always looming "what if" to me. In this case, the past me would've immediately asked: "hey, what if that 64 was 63?". So, with the amount of effort you put into it - I feel like the moment you talk "complex numbers" you could've briefly introduce the polar form and exponents and then talk Euler's formula. Come to think of it, some cool geometry would be there too - and it would actually solve it for any number, not just special ones.

  • @eisagdix6692
    @eisagdix6692 9 часов назад +12

    But why doesn't it work to just take the root, like shown at the start of the video? Is it because we have a function of the 4th degree and the first "method" only gives us 2 instead of 4 possible solutions?

    • @breqbs
      @breqbs 8 часов назад +1

      exactly

    • @ronaldking1054
      @ronaldking1054 7 часов назад +1

      i^4 = 1 rather than -1.

    • @jmr5125
      @jmr5125 3 часа назад +2

      It does work, but you have to apply rules that seem arbitrary to produce all 4 answers.
      This channel is intended for students that are learning _algebra_, and this is the correct approach for this group of people. If you are learning calculus then you can use more advanced techniques to properly calculate all solutions of the nth root and produce the same answers.

  • @dogbreaththe3rd851
    @dogbreaththe3rd851 Час назад

    64 = -64i^2 substitute giving x^4 - 64i^2
    The factorization then becomes a difference of squares
    (x^2 -8i) * (x^2 + 8i)

  • @lornacy
    @lornacy 7 часов назад

    Does the Po-Shen Loh method work with a quartic like this?

  • @JossWainwright
    @JossWainwright 5 часов назад

    Very nice.

  • @RyanLewis-Johnson-wq6xs
    @RyanLewis-Johnson-wq6xs 9 часов назад +7

    (-2-2i)^4+64=0 x=±2±2i It’s in my head.

  • @HarshvardhanMishra-ql2mj
    @HarshvardhanMishra-ql2mj 9 часов назад

    Good one❤

  • @stpat7614
    @stpat7614 24 минуты назад

    Does this work two, or am I off?
    x^4 + 64 = 0
    x^4 + 64 - 64 = 0 - 64
    x^4 = -64
    sqrt(x^4) = +/-sqrt(-64)
    x^2 = +/-sqrt(8*-1)
    x^2 = +/-sqrt(8)*sqrt(-1)
    x^2 = +/-sqrt(8)*i
    x^2 = sqrt(8)*i or x^2 = -sqrt(8)*i
    sqrt(x^2) = +/-sqrt(sqrt[8]*i), or sqrt(x^2) = +/-sqrt(-sqrt[8]*i)
    x = +/-sqrt(sqrt[8]*i), or x = +/-sqrt(-sqrt[8]*i)
    x = +/-sqrt(sqrt[8]*i), or x = +/-sqrt(sqrt[8]*i*[-1])
    x = +/-sqrt(sqrt[8])*sqrt(i), or x = +/-sqrt(sqrt[8])*sqrt(i)*sqrt(-1)
    x = +/-sqrt(sqrt[8])*sqrt(i), or x = +/-sqrt(sqrt[8])*sqrt(i)*i
    x = +/-(8^[1/2])^(1/2)*i^(1/2), or x = +/-(8^[1/2])^(1/2)*i^(1/2)*i^(2/2)
    x = +/-8^([1/2]*[1/2])*i^(1/2), or x = +/-8^([1/2]*[1/2])*i^(1/2+2/2)
    x = +/-8^(1/4)*i^(1/2), or x = +/-8^(1/4)*i^(3/2)
    x = +/-(2^3)^(1/4)*i^(1/2), or x = +/-(2^3)^(1/4)*i^(3/2)
    x = +/-2^(3*1/4)*i^(1/2), or x = +/-2^(3*1/4)*i^(3/2)
    x = +/-2^(3/4)*i^(1/2), or x = +/-2^(3/4)*i^(3/2)
    x = 2^(3/4)*i^(1/2), or x = -2^(3/4)*i^(1/2), or x = 2^(3/4)*i^(3/2), or x = -2^(3/4)*i^(3/2)

  • @Xlrupt
    @Xlrupt 9 часов назад

    Looks good

  • @hydo-7653
    @hydo-7653 8 часов назад +2

    Amazing video! this is love from the 8th graders haha

  • @APerson-ws4cw
    @APerson-ws4cw Час назад

    I'm usually a math wizz, this was wild to witness lol

  • @leonardobarrera2816
    @leonardobarrera2816 2 часа назад

    This video must to be called,
    How to demostrate the sqrt(i) to non caclulus studients!!!

  • @sina4948
    @sina4948 Час назад

    Why ±√-64 is not amongst the answers?

    • @NLGeebee
      @NLGeebee 53 минуты назад

      Because
      1) it is a 4th power
      2) calculating with ⁴√-64 to directly ±2 ±2i is sloppy maths ;)

  • @ChavoMysterio
    @ChavoMysterio 4 часа назад +1

    x⁴+64=0
    (x²)²+8²=0
    (x²)²+16x²+8²-16x²=0
    (x²+8)²-(4x)²=0
    (x²+4x+8)(x²-4x+8)=0
    x²+4x+8=0
    x²+4x+4=-4
    (x+2)²=-4
    |x+2|=2i
    x+2=±2i
    x=-2±2i ❤❤
    x²-4x+8=0
    x²-4x+4=-4
    (x-2)²=-4
    |x-2|=2i
    x-2=±2i
    x=2±2i ❤❤

    • @dogbreaththe3rd851
      @dogbreaththe3rd851 58 минут назад

      x^4 - 64i^2 = 0 is easier - difference of squares requires no geometry

  • @safestate8750
    @safestate8750 8 часов назад

    I have never seen the geometric approuch to the difference of squares before, thanks man

    • @pluto9000
      @pluto9000 7 часов назад

      Same here. My mind is blown.

  • @CasiMediocre
    @CasiMediocre 8 часов назад

    Initially, I thought "why don't you just do the ±sqrt twice or just the fourth root but with the 1, i, -1, -i attatched too"

    • @trueriver1950
      @trueriver1950 8 часов назад

      I think that's a valid destination of the same answer

  • @-Yousof-
    @-Yousof- 9 часов назад +1

    noice! 2nd comment btw

  • @fernandocardenaspiepereit4097
    @fernandocardenaspiepereit4097 7 часов назад +1

    It has no solutions, Imaginary numbers don't exist.