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.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?
No, that is not the question. That is an equation, and it still has wrong parentheses.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]sorry. This the question
p2/x-a + q2/x-b + r2/x-c = k