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avatar+386 

Find value of K in the following:

 

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

 

Knowing that P=7  (P=c/a)

 

Thanks!

 Dec 18, 2016

Best Answer 

 #3
avatar+118667 
+5

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

 Dec 18, 2016
 #1
avatar+118667 
+5

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.

 Dec 18, 2016
 #2
avatar+129849 
+5

(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

 

 

cool cool cool

 Dec 18, 2016
 #3
avatar+118667 
+5
Best Answer

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

Melody  Dec 18, 2016
 #4
avatar+118667 
0

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!

 Dec 18, 2016

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