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Math Help - is this correct?

  1. #1
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    is this correct?

    Today in math I learned new concepts. However, I am unsure if these three questions are correct. This is what I did:

    Solve by considering all cases. Show each solution on a number line.
    x^2-2x-15<0

    Steps I did:
    x^2-2x-15<0
    = (x-5)(x+3)<0
    = (x-5<0) (x+3<0)
    = (x<5) (x<-3)

    Therefore: -3<x<5 is a solution.

    x^3-3x^2-5x+6<0
    = (x-1)(x^2-x-6)<0
    = (x-1)(x-3)(x+2)<0
    = (x-1<0) (x-3<0) (x+2<0)
    = (x<1) (x<3) (x<-2)

    Therefore: x<-2 or 1<x<3 is a solution.

    Solve using intervals.
     5x^3 - 12x^2 - 11x + 6 \leq 0
    \Rightarrow (x+1)(5x^2-17x+6) \leq 0
    Used quadratic equation to find the roots of 5x^2-17x+6
    = (x+1)(x-3)(x-0.4)

    Therefore: x≤-1 or 0.4 ≤ x ≤ 3 is a solution.

    Am I correct with the findings? Is what I did to solve for all cases correct?
    Last edited by mr fantastic; December 15th 2009 at 06:12 PM. Reason: Fixed latex
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  2. #2
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    Quote Originally Posted by Barthayn View Post
    Today in math I learned new concepts. However, I am unsure if these three questions are correct. This is what I did:

    Solve by considering all cases. Show each solution on a number line.
    x^2-2x-15<0

    Steps I did:
    x^2-2x-15<0
    = (x-5)(x+3)<0
    = (x-5<0) (x+3<0)
    = (x<5) (x<-3)

    Therefore: -3<x<5 is a solution.

    x^3-3x^2-5x+6<0
    = (x-1)(x^2-x-6)<0
    = (x-1)(x-3)(x+2)<0
    = (x-1<0) (x-3<0) (x+2<0)
    = (x<1) (x<3) (x<-2)

    Therefore: x<-2 or 1<x<3 is a solution.

    Solve using intervals.
    5x^3-12x^2-11x+6 \leq 0
    \Rightarrow (x+1)(5x^2-17x+6) \leq 0
    Used quadratic equation to find the roots of 5x^2-17x+6
    = (x+1)(x-3)(x-0.4)

    Therefore: x≤-1 or 0.4 ≤ x ≤ 3 is a solution.

    Am I correct with the findings? Is what I did to solve for all cases correct?
    The easiest method to solve quadratic inequalities is completing the square.


     x^2-2x-15<0

    x^2 - 2x + (-1)^2 - (-1)^2 - 15 < 0

    (x - 1)^2 - 16 < 0

    (x - 1)^2 < 16

    |x - 1| < 4

    -4 < x - 1 < 4

    -3 < x < 1.
    Last edited by mr fantastic; December 15th 2009 at 06:13 PM.
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  3. #3
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    Your method is faulty. You cannot change the "equals" statements for the factors to inequalities as you did.

    However, you do need to start by working with the "equals" for the inequality. By solving the related equation, you can find the zeroes of the quadratic. The quadratic is either above the x-axis (that is, positive) or below the x-axis (that is, negative) everywhere else. The zeroes split the number-line into intervals of positivity and negativity.

    To learn how to find the signs on these intervals, using what you already know about graphing quadratics, try here. In this case, since the quadratic is positive, it is above the axis on the ends, and below the axis in the middle. So your solution will be the middle interval.
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  4. #4
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    He wants us to know what numbers, when subbed in, will give us a number greater/lesser then 0.
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  5. #5
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    And those numbers will be the values within the solution interval. (There are infinitely-many real numbers within this interval, which is why you're supposed to give the solution as an interval or inequality, rather than attempting the impossible task of listing all the solution values.)

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