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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- June 23rd 2011, 06:57 AMcastleQuadratic Proof
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. - June 23rd 2011, 07:17 AMProve ItRe: Quadratic Proof

For this inequation to be true, then the discriminant of must be negative.

So

- June 23rd 2011, 07:57 AMArchie MeadeRe: Quadratic Proof

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.