Solving a "Stanford" University Entrance Exam Question | b=?

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  • Опубликовано: 4 янв 2025

Комментарии •

  • @mandolinic
    @mandolinic 5 дней назад +2

    Did it with logs:
    3 log b = log b + log 3
    2 log b = log 3
    log b = (log 3)/2
    b = √3
    This gives one solution, but doesn't really give any hints about the other two. Finding those requires some further thought.

  • @tomtke7351
    @tomtke7351 7 дней назад +3

    b^3=3b
    b^2=3
    b=+/-sqrt(3)
    a 3rd solution is missing
    so
    b^3=3b
    b^3-3b=0
    b(b^2-3)=0
    b=0 sol.1
    b=+/-sqrt(3) sol.2, sol.3
    sqrt(3) = 1.732
    1.732^3=?3(1.732)
    =? 3•1.732
    1.732•1.732•1.732
    =? 3•1.732
    3•1.732 = 3•1.733✔️
    b=? -1.732
    (-1.732)^3=3(-1.732)
    (-1.732)^2•(-1.732)
    =? 3(-1.732)
    3•(-1.732) =❤ 3•(-1.732)✔️

  • @AlexanderSemashkevich
    @AlexanderSemashkevich 16 часов назад +1

    b³-3b=0
    b(b²-3)=0
    So roots are 0 and ±sqrt3

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 7 дней назад +1

    {b+b ➖ b+b➖ b+b ➖ }={b^2+b^2+b^2}=b^6 b^2^2^2 b^1^1^2 b^1^2 (b ➖ 2b+1).

  • @BobGillah
    @BobGillah 7 дней назад +2

    b=0

    • @tomtke7351
      @tomtke7351 7 дней назад

      since there's b^3 = 3b
      there's 3 solutions
      b=0 is 1 of 3 .. so you get 1/3 credit

  • @kajalbanerjee8220
    @kajalbanerjee8220 4 дня назад

    Root 3

  • @kajalbanerjee8220
    @kajalbanerjee8220 4 дня назад

    3root 3