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Math Help - I have solved most equations but I do need some help please :)

  1. #1
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    I have solved most equations but I do need some help please :)

    We are doing a review of algebra 2 and I really need some help. I have a new teacher and I never learned it the way he taught us. He doesn't let us use calculators. So I was wondering if some of you could please help me with some equations! Solve the following polynomial inequalities.

    1. (x+2)^2<25

    2. 5z^2+26z+3<-2

    3. X^4(x-3)<0

    4. -2x^2-4x+2x^3 <0

    5. 13x+4+5x^2>-2

    If anyone could help me with these it would be great. And yes I have tried and been sitting here for an hour stuck on them so don't think I didn't even try. Thank you!
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  2. #2
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    Re: I have solved most equations but I do need some help please :)

    (x+2)^2 < 25

    (x+2)^2 - 25 < 0

    [(x+2) - 5][(x+2) + 5] < 0

    (x-3)(x+7) < 0

    critical values (where the left side equals 0) are x = 3 and x = -7

    solution set is -7 < x < 3

    for further info, go to the link ...

    Pauls Online Notes : Algebra - Polynomial Inequalities
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  3. #3
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    Re: I have solved most equations but I do need some help please :)

    Quote Originally Posted by Allibvarsity View Post
    We are doing a review of algebra 2 and I really need some help. I have a new teacher and I never learned it the way he taught us. He doesn't let us use calculators. So I was wondering if some of you could please help me with some equations! Solve the following polynomial inequalities.

    1. (x+2)^2<25

    2. 5z^2+26z+3<-2

    3. X^4(x-3)<0

    4. -2x^2-4x+2x^3 <0

    5. 13x+4+5x^2>-2

    If anyone could help me with these it would be great. And yes I have tried and been sitting here for an hour stuck on them so don't think I didn't even try. Thank you!
    Isolate the squares, take the negative and positive square roots, and deal with the inequalities accordingly.

    Here are a couple examples to get you started:

    (1.) (x+2)^2<25
    -5<x+2<5 (assuming x is real)
    -7<x<3

    You sure you wrote #2 down correctly? It's gonna get awfully messy when you complete the square.

    (3.) x^4(x-3)<0
    (x-3)<0 because x^4 has to be nonnegative
    x<3
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  4. #4
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    Re: I have solved most equations but I do need some help please :)

    Hello, Allibvarsity!

    2.\;5z^2+26z+3\:<\:-2

    We have: . 5z^2 + 26z + 5 \:<\:0

    When is . y \:=\:5z^2 + 25z + 5 .negative?

    The graph is an up-opening parabola: \cup
    It is negative between its x-intercepts.

    5z^2 + 25z + 5 \:=\:0 \quad\Rightarrow\quad (z + 5)(5z+1) \:=\:0 \quad\Rightarrow\quad z \:=\:\text{-}5,\,\text{-}\tfrac{1}{5}

    Therefore: . \text{-}5\;<\;x\;<\;\text{-}\tfrac{1}{5}


    This technique can be applied to #5.
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  5. #5
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    Re: I have solved most equations but I do need some help please :)

    For (2), write the inequality as 5x^2+ 26x+ 5< 0. You can use the quadratic formula to find the roots of the equation, [tex]5x^2+ 26x+ 5= 0, a and b, and then factor as 5(x- a)(x- b)< 0. Of course, such a product will be negative if and only x- a and x- b have opposite signs. And that means x is larger than one of a and b and less than the other.

    (4), 2x^3- 2x^2- 4x< 0 can be factored as 2x(x^2- x- 2)< 0. And it is easy to factor x^2- x- 2. The product of three factors, (x- a)(x- b)(x- c) is negative if all three factors are negative or if two are positive and one negative.

    (5),  13x+4+5x^2>-2 or 5x^2+ 13x+ 6> 0, can be solved by using the quadratic formula to solve the equation 5x^2+ 13x+ 6= 0 for the roots a and b. And then the problem can be written as 5(x- a)(x- b)> 0. That will be true if and only if x-a and x- b have the same sign- in other word if x is larger than both a and b or x is less than both a and b.
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