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Homework Help: Simple quadratics question

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Find k such that the graph of y=9x^2 + 3kx + k
    A) intersects the x-axis in one point only
    B) intersects the x-axis in two points
    C) does not intersect the x-axis

    2. Relevant equations
    D= b^2 − 4ac
    y= ax^2+bx+c
    quadratic equation

    3. The attempt at a solution
    I got A but using the discriminent b^2-4ac= 0 meaning there is one root. but I have no idea how to get B and C. The answer is

    B) k < 0 or k >4
    C) 0<k<4
  2. jcsd
  3. Mar 5, 2008 #2
    [tex]y=9x^{2}+3kx+k[/tex], so a=9, b=3k, c=k, now

    A) [tex]D=b^{2}-4ac=0[/tex] so we will have two identical solutions, and the graph will touch the x-axis. Probbably you meant here to touch , because it does not really intersect.
    B) D>0

    SInce a=9>0 it means that the parabola will be opened upward. So when D<0, it means that the parabola does not intersect the x-axis at all, since the quadratic equation [tex]9x^{2}+3kx+k=0[/tex] does not have real roots, while when D>0, it means that there are two distinct roots of the quadratic equation.

    Do u know how to go about it now?
  4. Mar 5, 2008 #3
    Still don't understand.

    For A, by using b^2-4ac= 0 I got the answer that k=0 (which is correct)
    For B, I still don't understand how to achieve this answer k < 0 or k >4. How can I work with my equations to conclude that Teh values of k must be k< 0 or k >4 for y=9x^2 + 3kx + k to intersect two places.
  5. Mar 5, 2008 #4
    well B)[tex]D=b^{2}-4ac>0 =>(3k)^2-4*(9)*(k)>0=>9k^{2}-36k>0[/tex] , do you know how to solve this quadratic inequality?????

    One method for doint it is like this

    [tex]9k^{2}-36k>0=>k^{2}-4k>0=> k(k-4)>0[/tex] then this is greater then zero if:1) k>0, and k-4>0, or 2) k<0 and k-4<0

    another method is to first graph the function [tex]y=9k^{2}-36k[/tex] and see at what intervals the function is positive, that is above x-axis.
    Last edited: Mar 5, 2008
  6. Mar 5, 2008 #5


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    b^2-4ac = 9k^2 - 36k.

    9k^2 - 36k = 0 is a quadratic equation, and it has two solutions.

    You only found one solution (k = 0) - what is the other? :smile:
  7. Mar 5, 2008 #6
    ^ Right, 0 and 4, lol

    I'm getting the numbers but I'm not getting the greater than and less than signs.

    Referring to [tex]9k^{2}-36k>0=>k^{2}-4k>0=> k(k-4)>0[/tex] I got that but then what I concluded from that is k>0 and K>4. HOw can I end up with K<0 ?
  8. Mar 5, 2008 #7
    Try graphing the quadratic y = k2 - 4k. We are looking for when y > 0. What values of k give that?
  9. Mar 5, 2008 #8


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    Hurrah! :smile:

    Stop! … think! … k(k-4)>0 … try a few example for k …

    Got it? :smile:
  10. Mar 5, 2008 #9
    i perfectly well explained it in my post #4, that

    [tex]k(k-4)>0[/tex] is possible in two cases

    k>0 \ \ and \ \ k-4>0 \ or \ \ k<0 \ and \ \ k-4<0
    [/tex] think about this!!!
    Last edited: Mar 5, 2008
  11. Mar 6, 2008 #10

    Gib Z

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    It may be easier to do k(k-4) > 0 graphically - the find the intervals for which the curve is above the k axis. Sketching that should be easy.
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