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.

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- Feb 19th 2011, 03:21 AMdd86Inequality Problem
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. - Feb 19th 2011, 03:28 AMProve It
Try completing the square on $\displaystyle \displaystyle y^2 - 2(k - 1)y - 4k \geq 0$.

Then you will be able to solve for $\displaystyle \displaystyle k$. - Feb 19th 2011, 04:15 AMdd86
I did attempt that. But I guess I messed up somewhere.

I'll have another go at it. - Feb 19th 2011, 04:55 AMdd86
I've completed the square correctly. But I think I'm still missing out on something very obvious but I can't seem to get it.

I substituted y for the equation it stood for but I still came to a dead end.... :@ - Feb 19th 2011, 05:06 AMProve It
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 $\displaystyle \displaystyle k$. - Feb 19th 2011, 05:46 AMdd86
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. - Feb 22nd 2011, 02:09 AMdd86
Hi

The problem still can't be solved yet. Any one else has got an idea how to?