Results 1 to 4 of 4

Math Help - inequalities help!

  1. #1
    Newbie
    Joined
    Jun 2007
    Posts
    18

    inequalities help!

    how do u solve this inequality?

    a) x^2 - x - 8 < 0


    b) x ^3 + 3x^2 - 18x - 22 < and equal to 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by calchelp View Post
    how do u solve this inequality?

    a) x^2 - x - 8 < 0
    ...
    Hello,

    to a):

    x^2-x+\frac{1}{4} < \frac{1}{4}+8...................complete the square
    \left(x-\frac{1}{2}\right)^2 < \frac{33}{4}...............calculate the square-root

    \left|x-\frac{1}{2}\right| < \frac{1}{2} \cdot \sqrt{33} \Longleftrightarrow x-\frac{1}{2} < \frac{1}{2} \cdot \sqrt{33}\  \wedge \  -\left(x-\frac{1}{2}\right) < \frac{1}{2} \cdot \sqrt{33}

    \Rightarrow x < \frac{1}{2}+\frac{1}{2} \cdot \sqrt{33} \ \wedge \  x > \frac{1}{2}-\frac{1}{2} \cdot \sqrt{33}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2007
    Posts
    18

    Unhappy

    im sorry could u explain what step u took...im confused how that went to the next
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,929
    Thanks
    333
    Awards
    1
    Quote Originally Posted by calchelp View Post
    a) x^2 - x - 8 < 0
    Let's try it this way.

    Solve
    x^2 - x - 8 = 0

    x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-8)}}{2 \cdot 1}

    x = \frac{1 \pm \sqrt{33}}{2}

    So
    x^2 - x - 8 = \left ( x - \frac{1 + \sqrt{33}}{2} \right ) \left ( x - \frac{1 - \sqrt{33}}{2} \right )

    Thus we need to solve:
    \left ( x - \frac{1 + \sqrt{33}}{2} \right ) \left ( x - \frac{1 - \sqrt{33}}{2} \right ) < 0

    Now, break the real numbers into three intervals and check the inequality on these intervals:
    \left ( -\infty, \frac{1 - \sqrt{33}}{2} \right ) ==> \left ( x - \frac{1 + \sqrt{33}}{2} \right ) \left ( x - \frac{1 - \sqrt{33}}{2} \right ) > 0 (No!)

    \left ( \frac{1 - \sqrt{33}}{2}, \frac{1 + \sqrt{33}}{2}\right ) ==> \left ( x - \frac{1 + \sqrt{33}}{2} \right ) \left ( x - \frac{1 - \sqrt{33}}{2} \right ) < 0 (Check!)

    \left ( \frac{1 + \sqrt{33}}{2}, \infty \right ) ==> \left ( x - \frac{1 + \sqrt{33}}{2} \right ) \left ( x - \frac{1 - \sqrt{33}}{2} \right ) > 0 (No!)

    So the inequality is only true on
    \left ( \frac{1 - \sqrt{33}}{2}, \frac{1 + \sqrt{33}}{2}\right )
    as Earboth said.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. inequalities
    Posted in the Algebra Forum
    Replies: 7
    Last Post: August 20th 2011, 04:56 AM
  2. inequalities
    Posted in the Algebra Forum
    Replies: 0
    Last Post: March 20th 2011, 02:42 AM
  3. Inequalities
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: July 1st 2009, 04:35 AM
  4. Inequalities
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 15th 2009, 11:26 AM
  5. inequalities
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 14th 2008, 09:57 PM

Search Tags


/mathhelpforum @mathhelpforum