How do I solve this quadratic equation with imaginary numbers using the pq formula? Reddit algebra

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  • Опубліковано 29 вер 2024
  • We will solve the quadratic equation with complex coefficients, x^2-2ix+(2-4i)=0 by using the pq formula. We don't really teach or use the pq formula to solve quadratic equations here in the US. However, I did derive the formula in my cubic formula video here: • How to prove the cubic...
    In the end, we also have to simplify the square root of a complex number. This question is from Reddit r/homeworkhelp: / p6pxwd4oam
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КОМЕНТАРІ • 26

  • @bprpmathbasics
    @bprpmathbasics  6 місяців тому +1

    Proof of the pq formula:
    ua-cam.com/video/NlX_VR-e8qo/v-deo.html

  • @Ninja20704
    @Ninja20704 6 місяців тому +5

    Another way would be to just substitue x=a+bi directly where a and b are real and try to solve for a and b.
    (a+bi)^2 - 2i(a+bi) + 2 - 4i=0
    a^2+2abi-b^2-2ai+2b+2-4i=0
    [a^2-b^2+2b+2]+[2ab-2a-4]i=0+0i
    So a^2-b^2+2b+2=0 and 2ab-2a-4 =0
    2ab-2a-4=0
    ab-a-2=0
    ab=a+2
    b=1+2/a
    Substitute into the other equation.
    a^2-(1+2/a)^2+2(1+2/a)+2=0
    a^2-(1+4/a+4/a^2)+2+4/a+2=0
    a^2-1-4/a-4/a^2+2+4/a+2=0
    a^2-4/a^2+3=0
    a^4+3a^2-4=0
    (a^2-1)(a^2+4)=0
    a^2 = 1 or a^2 = -4(impossible since a is real)
    So a^2=1
    a=1 or a=-1
    b=1+2/1 b=1+2/(-1)
    =3 =-1
    Thus the two answers are x=1+3i or x=-1-i
    This was actually the way i was taught to solve a quadratic equation if you get a complex number under the sqrt. It isn’t much easier but at least we get the answers immediately instead of needing to work out the sqrt first.

  • @muktakanodia1495
    @muktakanodia1495 6 місяців тому +2

    There r only 2 comments?

    • @keescanalfp5143
      @keescanalfp5143 6 місяців тому

      could mean that you have had the luck to not be the first .

  • @88kgs
    @88kgs 6 місяців тому +2

    Thank you so much for sharing this.
    Regards 🙏

  • @alibasim2537
    @alibasim2537 6 місяців тому +1

    How do I solve this to find x
    Sinx=1+Cosx

  • @ianfowler9340
    @ianfowler9340 6 місяців тому +3

    Another cool thing about squaring complex numbers. For the complex number: z = a + bi where you restrict a and b to be integers. z^2 in standard form will have real and imag. parts that will ALWAYS be the 2 smaller values (ignoring any - signs) of a Pythagorean Triplet! (2 - 3i)^2 = -5 - 12i . Now I know, strictly speaking, Pyth. Triplets a,b,c are always positive (for triangles) but you can extend them to the integers as a^2,b^2,c^2 are always positive i.e. (-5)^2 + (-12)^2 = (13)^2. It works every time. What this means for this video: In reverse, if you find the 2 square roots of a complex number that already has the 2 smaller values (ignoring the - sign) of a Pyth. Triplet for real and imag. parts you will ALWAYS get INTEGERS for the real and imag. parts of the 2 square roots. sqrt(-3+4i) = 1+3i or -1-i. Now, how cool is that!
    This was a "fudged" question as the quadratic equation was designed to have -3 and 4 in the discriminant.
    So who can be the first to offer a proof of this? It's not difficult - high school algebra is all that is needed.

  • @elmer6123
    @elmer6123 6 місяців тому

    At 3:16 I would say -3+4i=5e^[i*atan2(4,-3)] and √(-3+4i=)=±√5e^[i*atan2(4,-3)/2]=±√5e(0.4472...+i*0.8944...)=±(1+2i)

  • @lizhang3073
    @lizhang3073 6 місяців тому +1

    8:18 ah yes exclude imaginary numbers where the entire video is an imaginary quadratic video

    • @bprpmathbasics
      @bprpmathbasics  6 місяців тому +1

      a+bi, a and b have to be real.

    • @lizhang3073
      @lizhang3073 6 місяців тому

      @@bprpmathbasics thanks for the clarification im not that good at math

  • @abusayemas5158
    @abusayemas5158 6 місяців тому

    X²-x³=12, find value of x

  • @Satnam7275
    @Satnam7275 6 місяців тому +1

    Engrossing. Thanks.

  • @SalutLunar
    @SalutLunar 6 місяців тому +1

    Great video as always! Just remember to say, "Real and Imaginary parts", as opposed to, "Real and Complex parts."

    • @david4649
      @david4649 6 місяців тому

      what's the difference?

    • @SalutLunar
      @SalutLunar 6 місяців тому

      @@david4649 Complex numbers are encompassing of real and imaginary numbers. So all real numbers are complex numbers but not all complex numbers are real numbers. Likewise for imaginary numbers: all imaginary numbers are complex numbers but not all complex numbers are imaginary numbers.
      Have a look for blackpenredpen's video when he accidently labelled the axes on complex plane as Real and Complex. There were at least two videos apologising for/correcting the error.

  • @tobybartels8426
    @tobybartels8426 6 місяців тому

    You can always use the formula √(a+bi) = ½√(2√(a²+b²)+2a) + ½b̂i√(2√(a²+b²)−2a), where b̂ means 1 for b≥0 and −1 for b

  • @thejaegerbomber99
    @thejaegerbomber99 6 місяців тому

    There is an easier way to simplify sqrt(-3 + 4i). I realized -3 + 4i is a perfect square, so I did this:
    sqrt(-3 + 4i) = sqrt(1 + 4i - 4) = sqrt(1 + 2*2i + (2i)^2) = sqrt(1 + 2i)^2 = 1 + 2i (using the principal root mentioned in the video)

  • @Rithida.Math2024
    @Rithida.Math2024 6 місяців тому

    Compute the Integral x^2024/(x^2+1)^2023 dx

  • @konradnowak159
    @konradnowak159 6 місяців тому

    What a sick time travel! That montage!

  • @ZipplyZane
    @ZipplyZane 6 місяців тому

    I feel like I've seen you do the "all over" mistake before. Is this a reupload, or did you almost make that same mistake twice when dealing with the pq formula?

    • @foogod4237
      @foogod4237 6 місяців тому +1

      It wouldn't be surprising to make that mistake more than once. The pq formula is actually just a simplified form of the quadratic formula, so when you're writing it out, if's not unreasonable for your brain to fall into that trap of trying to continue writing "the rest of it"..

  • @muktakanodia1495
    @muktakanodia1495 6 місяців тому +3

    I don't have any idea how to solve these equations but I can understand him solving it😮

  • @Sk8YouInDaGround
    @Sk8YouInDaGround 6 місяців тому +8

    I never learned about komplex numbers but i was curious… not anymore, i leave it to the mathematicians from this point 🫡🥲

    • @david4649
      @david4649 6 місяців тому

      Complex numbers really aren't that hard. I find it sad that they are not taught in school. A skill you need to have is knowing how to simplify, because without that skill, you won't be able to work with i. Schools make problems for calculators.