Find value of K

Ahh   I told you CPhill was a better guesser than me  :))

What havew P,c and a got to do with anything?

(k-2) x^2 -5x +7 = 0

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {5 \pm \sqrt{25-4*7*(k-2)} \over 2(k-2)}\\ x = {5 \pm \sqrt{25-28(k-2)} \over 2(k-2)}\\ x = {5 \pm \sqrt{25-28k+56} \over 2(k-2)}\\ x = {5 \pm \sqrt{81-28k} \over 2(k-2)}\\$

You had better explain better what you are on about.   k can be anything except 2

and if x is real then 

$81-28k\ge0\\-28k\ge-81\\ k\le\frac{81}{28}\\$

If x is rational then 

$81-28k \text{ is a perfect square}$

I see CPhill is answering - he may be a better guesser than me.

(k-2) x^2 -5x +7 = 0

Knowing that P = 7  (P = c/a)

c /a    =   7 / (k − 2)  =  7 / 1

This implies that .....

k − 2   = 1

k = 3

I just realized that this poorly worded question was asked just a few questions back.

It is a repeat!

DO NOT DO THIS TONY - IT IS RUDE!