I've been having some trouble with this one: Prove that if > k(x - 2) for all real x, then 0 < k < 8. I started with taking the k(x-2) to the other side, but I don't think I'm going anywhere with this. Thanks in advance.
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For this inequation to be true, then the discriminant of must be negative. So
Last edited by Prove It; June 23rd 2011 at 07:45 AM.
is true for all x, if the U-shaped quadratic never touches the x-axis, so the equation is of the form and must have complex roots One factor is positive and the other is negative. k>8 and k<0 cannot occur, so k>0 and k<8.