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Inequality problem

  1. Mar 18, 2006 #1
    for f(x)=x^2+kx+k, determine all values of k such that f(x)>0

    could someone please help me get started here? I am reviewing for a highschool math contest and it's been quite some time since I've learned about inequalities. I've managed to solve some rudimentary problems, but this one really has me stumped.
  2. jcsd
  3. Mar 18, 2006 #2
    simple enough. this equation represents a curve called parabula.
    try drawing this curve on XY surface.
    try with different k values such as: k=0, k=1, k=-1... and see how this changes the curve.

    *please note that k=0 doesn't match your question because if does get the value of 0 and you're required to find f(x)>0.

    if you need more help, let me know how it goes...
  4. Mar 18, 2006 #3


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    Try completing the square: [tex]x^2+kx+k=\left( x+\frac{k}{2}\right) ^2+k-\frac{k^2}{4}>0[/tex] for what values of k?
  5. Mar 18, 2006 #4
    since this is a parabula, you can find its roots (the points where f(x)=0). there are always 2 roots.
    roots equations: (-B +- sqrt(B^2 - 4AC)) / 2A
    A is the coefficient of X^2
    B is the coefficient of X^1
    C is the coefficient of x^0

    the parabula has 3 different possible conditions:
    #1: intersecting the X axis twice.
    #2: intersecting the X axis once.
    #3: not intersecting the X axis at all.

    think which one you want it to be, and use the roots equation to calculate accordingly
    Last edited: Mar 18, 2006
  6. Mar 18, 2006 #5


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    For a fixed value of k, f(x) is an upwards opening parabola whose vertex occurs at [tex]x=-\frac{k}{2},[/tex] we require that all points on the parabola be above the x-axis and hence require that the lowest point on the parabola be above the x-axis, that is the y-coordinate of the vertex must be >0, i.e. we require [tex]f\left( -\frac{k}{2}\right) >0 .[/tex]
  7. Mar 18, 2006 #6
    @benorin: i'm not sure your suggestion leads to the full solution. i might be wrong though...
  8. Mar 18, 2006 #7
    Benorin's solution works completely because it's the same as yours (using the discriminant).
  9. Mar 19, 2006 #8
    Thanks a lot, benorin and greytomato!

    Visualising how the parabola whose vertex is above y=0 was the connection that I had needed to make on my own. :P From there, I just completed the square, and solved for the values of k that would keep the vertex above y=0. I probably wouldn't have thought to use discriminants that way though, so it's good that I saw an alternative way of solving the problem.

    Another problem came up for me today, and I think it's probably better to post it in this thread instead of clogging the forums up with a new thread:

    Find the minimum value of f(x)=3^(x^2+4x)

    For this question, I figured f(x)>0, since a number raised to a power cannot yield a negative number. so, I have


    written down, but once again, I find myself stuck =\
  10. Mar 19, 2006 #9


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    One big hint for you is that the function f(x) := 3x is a strictly increasing function, that is for every x1 < x2, you'll always have:
    f(x1) < f(x2).
    Now to find the minimum value of [tex]3 ^ {x ^ 2 + 4x}[/tex], you must find a minimum value of: x2 + 4x, which can be done by completing the squares.
    Can you go from here? :)
  11. Mar 19, 2006 #10
    yes! thank you very much!
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