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Math Help - Chords easy question --Test tomorrow

  1. #1
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    Chords easy question --Test tomorrow

    Normally i wouldn't ask someone to reply quickly, but my test i tomorrow
    all help is gratfully appreiciated if you can answer it by midnight

    I have made a picture of my problem.
    Attached Thumbnails Attached Thumbnails Chords easy question --Test tomorrow-mat.png  
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  2. #2
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    Quote Originally Posted by shawnCanada View Post
    Normally i wouldn't ask someone to reply quickly, but my test i tomorrow
    all help is gratfully appreiciated if you can answer it by midnight

    I have made a picture of my problem.
    Read this Intersecting Chord Theorem - Math Open Reference

    Therefore in question 2: 2\times 5 = x(2x+1)

    Solve for x.
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  3. #3
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    Sorry i did not have tiime to explain my attempt:
    5x2 =x(2x+1)
    10 = 2x^2 + x
    0 = 2x^2 + x - 10
    2 / 0 = 2 / (x^2 + x - 10)
    0 = x^2 + x - 10
    0 = (x -10) (x + 1)
    (split into two equation)
    x = 10 <---
    x = -1

    In the end im left with x = 10, when in my textbook x = 2 ?
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  4. #4
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    Quote Originally Posted by shawnCanada View Post
    0 = 2x^2 + x - 10
    2 / 0 = 2 / (x^2 + x - 10)
    What is happening between these steps?



    Quote Originally Posted by shawnCanada View Post
    0 = x^2 + x - 10
    0 = (x -10) (x + 1)

    Sorry my friend, this is all types of wrong. You really should revise some factorisation.
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  5. #5
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    There's no way you can do the answer for me?

    i know i would learn better on my own but
    i'v done 23/24 questions and i just need
    you question done so i can understand what
    i did wrong here
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  6. #6
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     0 = 2x^2 + x - 10 \implies x =\frac{-1\pm\sqrt{1^2-4\times 2 \times -10}}{2\times 2}=\dots
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