I can't solve this question:
Solve for k where y^2-4k>=2(k-1)y where y=[(x^2-2x+6)/(x-3)]. (Ans. k=5)
Your tips/ hints are sufficient. Please don't give the specific steps to solve this question.
Thanks in advance.
Assuming you have completed the square properly you should have
$\displaystyle \displaystyle y^2 - 2(k - 1)y - 4k \geq 0$
$\displaystyle \displaystyle y^2 - 2(k - 1)y + [-(k-1)]^2 - [-(k-1)]^2 - 4k \geq 0 $
$\displaystyle \displaystyle [y - (k-1)]^2 - (k - 1)^2 - 4k \geq 0$
$\displaystyle \displaystyle (y - k +1)^2 \geq (k - 1)^2 + 4k$
$\displaystyle \displaystyle (y - k + 1)^2 \geq k^2 - 2k + 1 + 4k$
$\displaystyle \displaystyle (y - k + 1)^2 \geq k^2 + 2k + 1$
$\displaystyle \displaystyle (y - k + 1)^2 \geq (k + 1)^2$
$\displaystyle \displaystyle |y - k + 1| \geq |k + 1|$.
This should give you four inequalities to solve for $\displaystyle \displaystyle k$.
Four inequalities? Hmm...
OK, I've only come across one way on how to solve inequalities with absolute values on both sides. And that's by squaring both sides.
However, it's now the other way.
And I'm boggled by the fact that I have two unknowns on the LHS. So, I need more guidance, please.