# 6th degree equation [duplicate]

The number of real roots of $$\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19} = x^2 -11x -4$$ How to solve without actually finding the roots or is it the only way? I know that if the equation changes it's sign between two no.s it will have at least one root contained in them.

## marked as duplicate by TheSimpliFire, Maria Mazur, Number, Lord Shark the Unknown, Lee David Chung LinApr 20 at 1:57

• What you say isn't quite true. When $x$ is just a little less than $3,$ then first term is very large in absolute value, but negative. When $x$ is just a little bigger than $3$ the first time is very large and positive. There is no root between these two values, because the function isn't defined at $x=3$. – saulspatz Apr 19 at 20:06
• Wouldn't $3$,$5$,$17$,$19$ be roots? – Fareed AF Apr 19 at 20:07