Evaluating a Radical Expression

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  • Опубликовано: 10 янв 2025

Комментарии • 31

  • @francis6888
    @francis6888 2 года назад +13

    I just rationalised the denominators of both fractions and compared the two. The answer is X/2

    • @sgg3808
      @sgg3808 2 года назад

      I did the same

    • @dakshgiriraj4468
      @dakshgiriraj4468 2 года назад

      traditionally everyone wud hav done this..even this approach clicked in my mind too..as that is what we all have been taught in our high school

  • @McGravyboat
    @McGravyboat 2 года назад +1

    Another method (notation: /# means square root of #, and _ _ around an expression is just bringing attention to the expression, or "underlining" it):
    (/3+/2) / (/5-/3)= multiply by (/3-/2) / (/3-/2)
    (/3+/2)(/3-/2) / [(/5-/3)(/3-/2)]= multiply out the conjugates
    (3-2) / [(/5-/3)(/3-/2)]= simplify
    1 / [(/5-/3)(/3-/2)]= multiply the denominator by (/5+/3) / (/5+/3) which creates a complex fraction for now
    1 / [(/5+/3)(/5-/3)(/3-/2) / (/5+/3)]= multiply out the conjugates
    1 / [(5-3)(/3-/2) / (/5+/3)]= simplify
    1 / [2 * (/3-/2) / (/5+/3)]= notice that the underlined expression...
    1 / [2 * _(/3-/2) / (/5+/3)_]= ...turns out to be the reciprocal of x; substitute
    1 / [(2 * _(1/x)_]= simplify
    1 / [2/x]= "reciprocate" if that's a word
    x/2

  • @Qermaq
    @Qermaq 2 года назад +1

    Solved this very similarly. I used the reciprocal of the first expression, it's equal to 1/x. Then I multiplied both sides by y, so 1/2 = y/x and y = x/2.

  • @SuperYoonHo
    @SuperYoonHo 2 года назад +1

    Thank you very much sir SyberMath! Your'e a very good teacher.

    • @SyberMath
      @SyberMath  2 года назад +1

      Np. Thank you!!! 💖

    • @SuperYoonHo
      @SuperYoonHo 2 года назад +1

      @@SyberMath 😊😊🙏🙏

  • @theemeraldprogamer731
    @theemeraldprogamer731 2 года назад

    Wow I'm shocked at how simple this is and I just didn't know!Keep enlightening

  • @stvp68
    @stvp68 2 года назад

    I 🧡 when the answer is so simple!

  • @drozfarnyline4940
    @drozfarnyline4940 2 года назад +1

    I have got x/2 by double rationalization
    Here is a problem suggestion
    {(log(2x-5))/(log(x^2-8)}=(1/2)
    solve for x

    • @Jha-s-kitchen
      @Jha-s-kitchen 2 года назад +2

      x=11/3 or x=3, but x=3 cause denominator to be 0,
      So x = 11/3

    • @drozfarnyline4940
      @drozfarnyline4940 2 года назад +1

      @@Jha-s-kitchen omg!
      Thank you so much
      I solved it some hours ago and got the same answer as yours
      But solution was given as 17/3(only numerical answer not full solution)
      So I was confused whether I had made any mistake
      Thanks for clearing my doughts!

    • @Green_Eclipse
      @Green_Eclipse 2 года назад +1

      Yeah I got 11/3 as well. 17/3 comes from having a typo of log(x^2+8) in the denominator.

    • @drozfarnyline4940
      @drozfarnyline4940 2 года назад

      @@Green_Eclipse oh thank you too for pointing that out!

  • @mariomestre7490
    @mariomestre7490 2 года назад

    Genial, Merci

  • @christianthomas9863
    @christianthomas9863 2 года назад

    just multiply and divide the equation of x by the conjugate of the numerator and do the same for the denominator gives
    x = 2.(V(3) + (V(2))/(V(5)-V(3)) therefore: (V(3) + (V(2))/(V(5)-V(3)) = x/2

  • @vladimirkaplun5774
    @vladimirkaplun5774 2 года назад

    Does not worth the electricity consumed by his tablet.

  • @yakupbuyankara5903
    @yakupbuyankara5903 2 года назад +1

    X/2.

  • @Jha-s-kitchen
    @Jha-s-kitchen 2 года назад

    Yup I also got x/2 😁

  • @morteza3268
    @morteza3268 2 года назад

    Good-nice 🙂

  • @rakeshsrivastava1122
    @rakeshsrivastava1122 2 года назад +1

    x/2.

  • @winnewFirst
    @winnewFirst 2 года назад

    Must be the easiest of yours!

  • @broytingaravsol
    @broytingaravsol 2 года назад +1

    x/2

  • @barakathaider6333
    @barakathaider6333 2 года назад

    👍

  • @marcelojabuti3619
    @marcelojabuti3619 2 года назад

    You explain like you know why?

  • @james46291
    @james46291 2 года назад +1

    x/2

  • @rohitsk6068
    @rohitsk6068 2 года назад +1

    x/2

  • @giuseppemalaguti435
    @giuseppemalaguti435 2 года назад +1

    x/2

  • @mathswan1607
    @mathswan1607 2 года назад

    x/2