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Math Help - Basic quadratic formula issue

  1. #1
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    Basic quadratic formula issue

    What an I doing wrong? I'm trying to get the original equation into parenthesisBasic quadratic formula issue-imageuploadedbytapatalk-21370978007.916711.jpg
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  2. #2
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    Re: Basic quadratic formula issue

    (x+1)(x+\frac{1}{4})=0 is the same equation as (x+1)(4x+1)=0
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  3. #3
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    Re: Basic quadratic formula issue

    Then why doesn't this problem work?Basic quadratic formula issue-imageuploadedbytapatalk-21370979866.960367.jpg
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  4. #4
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    Re: Basic quadratic formula issue

    Shakarri multiplied by 4 on both sides of (x+ 1)(x+ \frac{1}{4})= 0, then took the "4" on the left into the second parentheses, using the distributive law.
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    Re: Basic quadratic formula issue

    But the book for this problem is getting a different answer somehow?
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  6. #6
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    Re: Basic quadratic formula issue

    Then tell us what answer the textbook is getting!
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    Re: Basic quadratic formula issue

    Quote Originally Posted by HallsofIvy View Post
    Then tell us what answer the textbook is getting!
    Please check my previous post with the image
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  8. #8
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    Re: Basic quadratic formula issue

    Multiply out your brackets and get rid of any fractions, see what you end up with.
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  9. #9
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    Re: Basic quadratic formula issue

    The original equation was 4x^2+ 5x+ 1= 0 and you used the quadratic formula to find x= -1 and x= -1/4 as solutions so that the factors are (x+1)(x+ 1/4).
    Now there are two ways to go:
    1) You can multiply those factors to get x^2+ x+ (1/4)x+ 1/4= x^2+ (5/4)x+ 1/4= 0. You can then multiply both sides of that equation by 4 to get 4x^2+ 5x+ 1= 0.

    2) As Shakarri said, you can immediately multiply both sides of (x+ 1)(x+ 1/4)= 0 by 4 to get (x+ 1)(4x+ 1)= 0. Then multiply that to get 4x^2+ 4x+ x+ 1= 4x^2+ 5x+ 1= 0.
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