Find value of K in the following:
(k-2) x^2 -5x +7 = 0
Knowing that P=7 (P=c/a)
Thanks!
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.