How to solve for "a" || Oxford math admission test

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  • เผยแพร่เมื่อ 7 ม.ค. 2025

ความคิดเห็น • 17

  • @Zozo3-y2c
    @Zozo3-y2c 3 วันที่ผ่านมา +1

    Wowww How Have you done that ? You're genius

  • @SSrinivasan-go4yu
    @SSrinivasan-go4yu 3 วันที่ผ่านมา +2

    Great bro

  • @sirmonke8946
    @sirmonke8946 วันที่ผ่านมา

    It would be easier to just square both sides from the original equation.
    (sqrt(a)+sqrt(-a))^2=144
    a + 2sqrt(-a^2) - a = 144
    2sqrt(-a^2) = 144
    sqrt(-a^2) = 72
    -a^2 = 72^2
    a^2 = -72^2
    a = +/- sqrt(-72^2)
    a = +/- 72i

  • @abdallahabdoulhakim2998
    @abdallahabdoulhakim2998 2 วันที่ผ่านมา

    7 minutes for that is insane!
    √a+√(-a)=12
    √a+√(-a) = √a (1+√(-1))=√a (1+(e^iπ)^0.5 )=√a(1+e^(iπ/2)) =12
    square the two terms
    a(1+e^(iπ/2) )^2=a(1+2e^(iπ/2)+e^iπ )=a(1+2ae^(iπ/2)-1)=2ae^(iπ/2)=144
    a=144/(2e^(iπ/2) )=72e^(-iπ/2)=-72i
    ±a=72i

    • @schneisim
      @schneisim วันที่ผ่านมา

      he loves his handwrighting

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 3 วันที่ผ่านมา

    (6)+( ➖ 6) (3^3)+( ➖ 3^3) (1^1)+( ➖ 3^1) ( ➖ 3^1) (a ➖ 3a+1).

  • @krishgamingindia405
    @krishgamingindia405 2 วันที่ผ่านมา +1

    I don't think the last step is correct. (I) Shouldn't be there. You can't sperate (-72)² as (72²)(-1), it's probably (72)²(-1)². Which would make all the numbers positive. And if we take root of such numbers there is no + or -. It's only +. I think it's called taking a "modulus"

    • @sirmonke8946
      @sirmonke8946 วันที่ผ่านมา

      Checking the guy's answer with wolfram alpha, he is indeed correct.

    • @mateusbarbosakopp1387
      @mateusbarbosakopp1387 วันที่ผ่านมา

      actually it is not (-72)² but rather -72² and then you can separate

    • @krishgamingindia405
      @krishgamingindia405 วันที่ผ่านมา

      @@mateusbarbosakopp1387 so I would have been correct if it was -72² but since thier are brackets (-72)² i am wrong? Interesting property.

  • @wavyi8270
    @wavyi8270 วันที่ผ่านมา

    Can someone tell me whats wrong with:
    sqrt(a) + sqrt(-a) = 12
    sqrt(a)(1 + i) = 12
    sqrt(a) = 12/(1 + i)
    a = (12/[1 + i])^2
    a = 144/(1 + i)^2
    a = 144/2i
    a = 72/i

    • @mateusbarbosakopp1387
      @mateusbarbosakopp1387 วันที่ผ่านมา

      in the third line, instead of squaring both sides, you should simplify the term on the right by multiplying it by the conjugate, then it becomes:
      sqrt(a) = 12/(1+i) * ((1-i)/(1-i))
      sqrt(a) = 12(1-i)/(1+1)
      sqrt(a) = 6(1-i)
      a=(6(1-i))² = 36 -72i -36 = -72i

    • @wavyi8270
      @wavyi8270 วันที่ผ่านมา

      @mateusbarbosakopp1387 thanks so much. How come you have to multiply by the conjugate first though? Is it simply because i is in the fraction?

    • @rickie_coll
      @rickie_coll วันที่ผ่านมา

      72/i = (72/i)*(i/i) = 72i/(-1) = -72i
      There's nothing wrong in 72/i

    • @wavyi8270
      @wavyi8270 วันที่ผ่านมา +1

      @@rickie_coll ah that makes sense. since i is its own thing i forgot that its really the same as having a radical in the denominator lol