Inequality Problem

**dd86** · February 19th 2011, 03:21 AM

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.

**Prove It** · February 19th 2011, 03:28 AM

Try completing the square on $\displaystyle y^2 - 2(k - 1)y - 4k \geq 0$ .

Then you will be able to solve for $\displaystyle k$ .

**dd86** · February 19th 2011, 04:15 AM

I did attempt that. But I guess I messed up somewhere.

I'll have another go at it.

**dd86** · February 19th 2011, 04:55 AM

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.... :@

**Prove It** · February 19th 2011, 05:06 AM

Assuming you have completed the square properly you should have

$\displaystyle y^2 - 2(k - 1)y - 4k \geq 0$

$\displaystyle y^2 - 2(k - 1)y + [-(k-1)]^2 - [-(k-1)]^2 - 4k \geq 0$

$\displaystyle [y - (k-1)]^2 - (k - 1)^2 - 4k \geq 0$

$\displaystyle (y - k +1)^2 \geq (k - 1)^2 + 4k$

$\displaystyle (y - k + 1)^2 \geq k^2 - 2k + 1 + 4k$

$\displaystyle (y - k + 1)^2 \geq k^2 + 2k + 1$

$\displaystyle (y - k + 1)^2 \geq (k + 1)^2$

$\displaystyle |y - k + 1| \geq |k + 1|$ .

This should give you four inequalities to solve for $\displaystyle k$ .

**dd86** · February 19th 2011, 05:46 AM

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.

**dd86** · February 22nd 2011, 02:09 AM

Hi

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

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