Results 1 to 2 of 2

Math Help - Quadratic Inequality

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    1

    Quadratic Inequality

    Hello, I'm new to this forum but I thought it may be helpful to come here for responses when a teacher/classmate is unavailable. I'm currently studying AS level mathematics... Ok so for the homework I have been solving quadratic inequalities by graphing them but this one question out of 10 is a problem for me. The inequality is: x^2 + 2x -1 < 0 I tried factorising it like all the others but it won't factorise so I don't know what coordinates to use to plot it on a graph and therefore can't find the solution. I also tried completing the square to get (x + 1)^2 -2 < 0 and this told me that the line of symmetry of the curve is at x = -1 and that the turning point is at -2 but this didn't help either. Can anyone explain how to do this? Any response will be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,703
    Thanks
    1470
    The simplest way to solve an inequality like this, x^2+ 2x- 1< 0, or one more complicated, is to solve the corresponding equation, x^2+ 2x- 1= 0. That can be solved by completing the square or by using the quadratic formula.
    x= \frac{-3\pm\sqrt{4+4}}{2}= \frac{-3\pm 2\sqrt{2}}{2}

    The point is that a continuous function, such as this, can change from "> 0" to "< 0" only where it is equal to 0.

    Those two points, x= \frac{-3- 2\sqrt{2}}{2} and x= \frac{-3+ 2\sqrt{2}}{2} divides the number line into three intervals:
    x< \frac{-3- 2\sqrt{2}+}{2}
    \frac{-3- 2\sqrt{2}+}{2}< x< \frac{-3+ 2\sqrt{2}+}{2}
    x> \frac{-3+ 2\sqrt{2}+}{2}

    Choose one point in each interval: since [tex]2\sqrt{2}[/itex] is between 2 and 3, \frac{-3- 2\sqrt{2}+}{2} is between -5/2 and 1/2. -3 is in the first interval and 0 is in the second.
    (-3)^2+ 2(-3)-1= 9- 6- 1= 2> 0 so every x< \frac{-1- 2\sqrt{2}}{2} does not satisify the inequality.

    0^2+ 2(0)-1= -1< 0 so every \frac{-3- 2\sqrt{2}+}{2}< x< \frac{-3+ 2\sqrt{2}+}{2} satifies the inequality.

    x= 2 is in the last interval and 2^2+ 2(2)- 1= 8- 1= 7> 0 so any x> \frac{-3+ 2\sqrt{2}+}{2} does not satisfy the inequality.

    The inequality is satisfied for \frac{-3- 2\sqrt{2}+}{2}< x< \frac{-3+ 2\sqrt{2}+}{2}.


    Of course, for this simple inequality, that is exactly the same as saying that the graph of [tex]y= x^2+ 2x- 1[/itex] is a parabola opening upward and so y< 0 for x between the two x-intercepts.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic inequality
    Posted in the Algebra Forum
    Replies: 9
    Last Post: April 11th 2011, 08:47 PM
  2. [SOLVED] Quadratic Inequality
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 8th 2011, 09:27 AM
  3. Quadratic Inequality
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 12th 2009, 10:51 PM
  4. quadratic inequality
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 27th 2008, 11:56 PM
  5. quadratic inequality
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 19th 2005, 12:16 AM

Search Tags


/mathhelpforum @mathhelpforum