Thursday, October 15, 2009

Period 4: Solving Inequalities Algebraically

Inequality deals with a range of values, rather than an exact number, so it is necessary to replace the equal sign in an equation with its supposed inequality (either >, <, >, or <).

For example, Solve for the following inequality:


(x + 10) (2x – 9) > 0

  1. Break up both of the brackets into two separate “sections”, x +12>0 and 2x - 9 >0.
  2. So far, we know that each individual section must exceed the value of 0, as this is the inequality. From this, we can isolate the x value to determine the roots.

    x+ 12 > 0
    x > -10

    2x-9 >0
    = 2x> 9
    x > 9/2


    Note that for section 2, the value of x can also be 0.


    So now we know that the roots of the equation when (x + 10) (2x – 9) = 0
    are x = -10, x = 9/2, and x = 0

  1. We can then break up number line into three intervals in relation to the roots that we have found.

    Because the lowest root is -10, we recognize in order for the graph to continue negatively, x must therefore be less than -10, so x <
    -10.

    Likewise, the highest root value is 9/2 so in order for the graph to continue positively, x must be bigger than 9/2, so 9/2 > x. We must also notice that the middle interval must (graphically) be larger than -10, yet smaller than 9/2, so, respectively, the inequality is represented by -10 <>



  2. You can test out each interval by plugging it into the equation and determining which one is the solution.

    For x<-10, test x = -11

    (-11 + 10) (2 (-11) -9) = 31
    because the value “31” is greater than “0”, x< -10 is a solution. For -10 <9/2,>x = 1
    (1+10) (2(1)-9) = -77
    because the value “-77” is less than “0”, -10 <>
    solution. For x > 9/2, test x = 5
    (5+10) (2(5)-9) = 135
    because the value “135” is greater than “0”, x>9/2 is a solution.

  3. Summarize the information found onto a table:

    Interval Factor

    x < -10

    -10

    x > 9/2

    (x +10)

    ( - )

    ( + )

    ( + )

    (2x – 9)

    ( - )

    ( - )

    ( + )

    f(x)

    ( + )

    ( - )

    ( + )


  4. The solution, shown on the number line, is x < -10 and x > 9/2

    Note: Page 8 of our textbooks summarizes all possible intervals

    Therefore, the solution is XE(-∞, -10] U[9/2, ∞)

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