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Thread: Solve the inequality

  1. #1
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    Solve the inequality

    Solve the inequation (x^2+3x+1)(x^2+3x-3)\ge 5



    on opening the brackets i got

    x^2(x^2+3x-3)+3x(x^2+3x-3)+1(x^2+3x-3)\ge 5
    x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5\ge 0
    x^4+6x^3+7x^2-6x-8\ge 0

    am i right??
    after that what should i do??
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Solve the inequality

    It looks fine, notice you can write x^4+6x^3+7x^2-6x-8=(x-1)\left(ax^3+bx^2+cx+d)
    Determine a,b,c,d by Horner's scheme or the euclidian division, afterwards do the same for ax^3+bx^2+cx+d so you can write finally the function as:
    x^4+6x^3+7x^2-6x-8=(x-1)(x-a)(x-b)(x-c)
    Where you've determined a,b,c and which is useful to make the sign-table.

    Is this clear? ...
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Solve the inequality

    Quote Originally Posted by sumedh View Post
    am i right?? after that what should i do??
    Right, now express x^4+6x^3+7x^2-6x-8=(x-1)(x+1)(x+2)(x+4) .

    Edited: Sorry, I didn't see Siron's post.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Solve the inequality

    I doens't matter . But I thought the only difficulty of this exercice was the 4th degree inequality so sumedh could solve it by himself but offcourse he can decide what he thinks which is the best for himself.
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  5. #5
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    Re: Solve the inequality

    thank you i got the answer
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