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Algebra problem

  • 27-08-2014 5:29am
    #1
    Registered Users, Registered Users 2 Posts: 5,857 ✭✭✭


    I had to look at the solution for a geometry problem and one of the steps is beyond me at the moment, I hope someone can help:

    1. SQRT((X/2)^2)+(X/2)^2))

    becomes

    2. SQRT(2*(X^2/2^2) I understand this step but...

    becomes

    3. x/SQRT(2)

    When I try to simplify step 2 I always get SQRT(2)*X/2 - I don't see how to get to step 3? Thanks in advance and I hope my equations make sense?


Comments

  • Registered Users, Registered Users 2 Posts: 939 ✭✭✭Ciaran


    Valmont wrote: »
    I had to look at the solution for a geometry problem and one of the steps is beyond me at the moment, I hope someone can help:

    1. SQRT((X/2)^2)+(X/2)^2))

    becomes

    2. SQRT(2*(X^2/2^2) I understand this step but...

    becomes

    3. x/SQRT(2)

    When I try to simplify step 2 I always get SQRT(2)*X/2 - I don't see how to get to step 3? Thanks in advance and I hope my equations make sense?

    sqrt(2)/2 = 1/sqrt(2) because 2 is sqrt(2) squared.


  • Registered Users, Registered Users 2 Posts: 5,633 ✭✭✭TheBody


    You have:

    [latex]\sqrt{\frac{2x^2}{2^2}}=\frac{\sqrt{2}x}{2} [/latex].

    If we multiply top and bottom by [latex]\sqrt{2}[/latex] we get:


    [latex]\frac{\sqrt{2}x}{2}\frac{\sqrt{2}}{\sqrt{2}}=\frac{2x}{2\sqrt{2}}=\frac{x}{\sqrt{2}} [/latex].

    Simples! :D


  • Registered Users, Registered Users 2 Posts: 5,857 ✭✭✭Valmont


    Thank you both. TheBody, how did you get the graphics in your post?


  • Registered Users, Registered Users 2 Posts: 2,345 ✭✭✭Kavrocks


    Valmont wrote: »
    Thank you both. TheBody, how did you get the graphics in your post?
    He used LaTeX inside [LaTeX] tags. If you try and quote his response you will see what he used.


  • Registered Users, Registered Users 2 Posts: 13,075 ✭✭✭✭bnt


    I tried doing it from the inside out:
    SQRT [ (x/2)² +(x/2)² ]
    = SQRT [ x²/4 + x²/4 ]
    = SQRT [ x²/2 ]
    = x / √2

    You are the type of what the age is searching for, and what it is afraid it has found. I am so glad that you have never done anything, never carved a statue, or painted a picture, or produced anything outside of yourself! Life has been your art. You have set yourself to music. Your days are your sonnets.

    ―Oscar Wilde predicting Social Media, in The Picture of Dorian Gray



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  • Registered Users, Registered Users 2 Posts: 5,857 ✭✭✭Valmont


    I have another seemingly insoluble algebra problem. I have to solve the following equation for y in order to graph the line:
    [latex]12+5\frac{-y+7}{2}=3y[/latex]
    My answer was repeatedly, after many checks and rechecks, [latex]y=\frac{59}{11}[/latex]. After being told it was incorrect, I entered the equation into this online calculator which again gives the answer as [latex]y=\frac{59}{11}[/latex].

    Naturally I marked the question as wrong I looked up the solution which has an equally logical but completely different solution. They take [latex]12+5\frac{-y+7}{2}=3y[/latex] and turn it into [latex]24-5y-35=6y[/latex] and I understand how this is done. The only problem is that if we now solve for y we get [latex]y=-1[/latex]! One has to be wrong which means either Khanacademy are wrong or the mathpapa.com online calculator is! Please help!


  • Registered Users, Registered Users 2 Posts: 939 ✭✭✭Ciaran


    Valmont wrote: »
    I have another seemingly insoluble algebra problem. I have to solve the following equation for y in order to graph the line:
    [latex]12+5\frac{-y+7}{2}=3y[/latex]
    My answer was repeatedly, after many checks and rechecks, [latex]y=\frac{59}{11}[/latex]. After being told it was incorrect, I entered the equation into this online calculator which again gives the answer as [latex]y=\frac{59}{11}[/latex].

    Naturally I marked the question as wrong I looked up the solution which has an equally logical but completely different solution. They take [latex]12+5\frac{-y+7}{2}=3y[/latex] and turn it into [latex]24-5y-35=6y[/latex] and I understand how this is done. The only problem is that if we now solve for y we get [latex]y=-1[/latex]! One has to be wrong which means either Khanacademy are wrong or the mathpapa.com online calculator is! Please help!
    24-5y-35=6y should be 24-5y+35=6y which gives your original answer.


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