I've been having some trouble with this one:

Prove that if $\displaystyle x^2$ > 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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- Jun 23rd 2011, 06:57 AMcastleQuadratic Proof
I've been having some trouble with this one:

Prove that if $\displaystyle x^2$ > 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. - Jun 23rd 2011, 07:17 AMProve ItRe: Quadratic Proof
$\displaystyle \displaystyle \begin{align*} x^2 &> k(x - 2) \\ x^2 &> kx - 2k \\ x^2 - kx + 2k &> 0\end{align*}$

For this inequation to be true, then the discriminant of $\displaystyle \displaystyle x^2 - kx + 2k$ must be negative.

So

$\displaystyle \displaystyle \begin{align*} (-k)^2 - 4(1)(2k) &< 0 \\ k^2 - 8k &< 0 \\ k^2 - 8k + (-4)^2 &< (-4)^2 \\ (k - 4)^2 &< 16 \\ |k - 4| &< 4 \\ -4 < k - 4 &< 4 \\ 0 < k &< 8 \end{align*}$ - Jun 23rd 2011, 07:57 AMArchie MeadeRe: Quadratic Proof
$\displaystyle x^2-kx+2k>0$

is true for all x, if the U-shaped quadratic never touches the x-axis,

so the equation

$\displaystyle x^2-kx+2k=0$

is of the form

$\displaystyle ax^2+bx+c=0$

and must have complex roots

$\displaystyle \Rightarrow\ b^2-4ac<0\Rightarrow\ k^2-8k<0$

$\displaystyle \Rightarrow\ k(k-8)<0$

One factor is positive and the other is negative.

k>8 and k<0 cannot occur, so k>0 and k<8.