Arithemetic

[vijay1965]

How do I show that p2/x−a +b2/x−b+r2/x−c=k where p, q, r, a, b, c and k are real and has no imaginary roots?

 

[Dr.Peterson]

How do I show that p2/x−a +b2/x−b+r2/x−c=k where p, q, r, a, b, c and k are real and has no imaginary roots?

Do you mean p^2/(x−a) +b^2/(x−b)+r^2/(x−c)=k, that is, `p^2/(x−a) +b^2/(x−b)+r^2/(x−c)=k`? That isn't what you wrote. I also suspect one b should be a q.

Furthermore, are you claiming that this is always true, or are you solving some equation (since you mention roots)? I think at least one word was omitted or should be.

Maybe you need to show us the actual problem as given to you. And this isn't arithmetic.

 

[vijay1965]

sorry. This the question
p2/x-a + q2/x-b + r2/x-c = k

 

[Dr.Peterson]

sorry. This the question
p2/x-a + q2/x-b + r2/x-c = k

No, that is not the question. That is an equation, and it still has wrong parentheses.

The most important thing is the words: what are you told about the equation, and what are you to do with it?

Please quote the entire problem, or show us an image of it.

 

[Steven G]

sorry. This the question
p2/x-a + q2/x-b + r2/x-c = k

p2/x-a + q2/x-b + r2/x-c = k = [math] \frac {p^2}{x} - a +\frac {q^2}{x}-b + \frac {r^2}{x} - c = k[/math]