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Math Help - Math 111 Help

  1. #1
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    Question Math 111 Help

    Ive tried but these are the ones that are giving me trouble....


    1.) write in vertex form. Identify the vertex
    g(x)=6+5x-10x^2


    2.)solve quadratic equation with any method
    3x^2-6x-2=0


    3.)solve inequality
    -3x > 9-12x^2
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  2. #2
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    vertex of parabola \displaystyle \frac{-b}{2a}

    2. factor \displaystyle (3x\pm g)(x\pm f)=0 where \displaystyle f,g\in\mathbb{R}

    3. solve this \displaystyle 0>9+3x-12x^2
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  3. #3
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    Quote Originally Posted by Peyton Sawyer View Post
    1.) write in vertex form. Identify the vertex
    g(x)=6+5x-10x^2
    do you know how to complete the square?



    Quote Originally Posted by Peyton Sawyer View Post
    2.)solve quadratic equation with any method
    3x^2-6x-2=0

    for ax^2+bx+c = 0 , x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}



    Quote Originally Posted by Peyton Sawyer View Post
    3.)solve inequality
    -3x > 9-12x^2
    12x^2-3x-9 > 0

    Use the same method in Q2 if you like.
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  4. #4
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    i dont understand 2 and 3.
    can you walk me through them?
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  5. #5
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    For number two, use what pickleslide suggested the quadratic since it isn't factorable.

    3 factor out a 3(4x^2-x-3)>0 divide by 3 4x^2-x-3>0 and this can be factored.
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  6. #6
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    for #1
    i got -5/12
    wat would the vertex be?
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  7. #7
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    I think it should be 1/4
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  8. #8
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    how? i though vertex form was -b/2(a)
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  9. #9
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    b= and a=-10 Thus, -b=-5 and 2a=-20. \displaystyle \frac{-5}{-20}\rightarrow\frac{1}{4}

    Also note that is only the x coordinate. You must plug it in to obtain the y value.
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  10. #10
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    i didnt rearrange the problem that is why.
    okay now i plug in the x coordinate into -10x^2-5x+6?
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  11. #11
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    Yes. You don't need to rearrange always remember it goes ax^2+bx+c. a is with the squared term and b the x.
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  12. #12
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    -10(1/4)^2-5(1/4)+6=

    5?

    is that correct?
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  13. #13
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    Nope.
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  14. #14
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    can you help me get y value?
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  15. #15
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    \displaystyle \frac{-10}{4^2}-\frac{5}{4}+6=\frac{-10}{16}-\frac{20}{16}+\frac{96}{16}=\frac{-10-20+96}{16}=\frac{33}{8}
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