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Math Help - inequality

  1. #1
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    inequality

    Solve the inequality below by using algebra. You must show your work (for the test) on this problem to receive full credit. Type in your answer using interval notation (using inf for infinity and u for union).
    6x2+3x<4-2x


    for an answer i got, (-inf,-4/3)u(1/2,inf).

    it says I'm in correct, I'm at a complete loss. I'm almost positive the algebra is right, but I think I'm screwing up writing it out in interval notation.

    please help me out, very much appreciated.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Twan View Post
    Solve the inequality below by using algebra. You must show your work (for the test) on this problem to receive full credit. Type in your answer using interval notation (using inf for infinity and u for union).
    6x2+3x<4-2x


    for an answer i got, (-inf,-4/3)u(1/2,inf).

    it says I'm in correct, I'm at a complete loss. I'm almost positive the algebra is right, but I think I'm screwing up writing it out in interval notation.

    please help me out, very much appreciated.
    Show us what you did.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    Show us what you did.

    6x^2+5x-4<0

    (6x^2+8x)(-3x-4)

    2x(3x+4)-1(3x+4)

    2x-1<0 3x+4<0
    x<1/2 x<-4/3
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  4. #4
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    Hello, Twan!

    Your answer is on the "wrong side".


    Solve the inequality: . 6x^2 + 3x \:<\:4 - 2x

    We have: . 6x^2 + 5x - 4 \:<\:0

    The graph of: . y \:=\:6x^2 + 5x - 4 is an up-opening parabola: \cup
    . . It is negative between its x-intercepts.

    To find the x-intercepts: . 6x^2 + 5x - 4 \:=\:0\quad\Rightarrow\quad (3x+4)(2x-1) \:=\:0 \quad\Rightarrow\quad x\:=\:\text{-}\tfrac{4}{3},\:\tfrac{1}{2}

    Therefore, the interval is: . \left(\text{-}\frac{4}{3},\:\frac{1}{2}\right)

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  5. #5
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    oh ic, thank you very much, i thought i'd have to do a union. If that problem was altered slightly what should i look for if there's a possible union or just a simple point like (-4/3,1/2)
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Twan View Post
    oh ic, thank you very much, i thought i'd have to do a union. If that problem was altered slightly what should i look for if there's a possible union or just a simple point like (-4/3,1/2)
    This is not a point! This is the interval \frac{-4}{3}<x<\frac{1}{2}!
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  7. #7
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    6x^2+5x-4<0
    =>(x-1/2)(x+4/3)<0
    =>either (x-1/2)<0 and (x+4/3)>0
    or (x-1/2)>0 and (x+4/3)<0 (not possible)
    => -4/3 <x<1/2
    =>x belongs to (-4/3,1/2) which is the required interval.
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