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 Lin Apr 20 at 1:57

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  • $\begingroup$ @MorganRodgers It becomes a sixth degree equation after cross multiplying. $\endgroup$ – Hans Engler Apr 19 at 19:59
  • $\begingroup$ This is a rational function, so if there is a sign change somewhere, it could also be due to a pole (vertical asymptote). There is a solution that you can find by just guessing. $\endgroup$ – Hans Engler Apr 19 at 20:04
  • $\begingroup$ 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 $x=3$. $\endgroup$ – saulspatz Apr 19 at 20:06
  • $\begingroup$ Wouldn't $3$,$5$,$17$,$19$ be roots? $\endgroup$ – Fareed AF Apr 19 at 20:07
  • 2
    $\begingroup$ @FareedAF No, the left hand side isn't defined at those values. $\endgroup$ – saulspatz Apr 19 at 20:10

Hint: use descartes rule of sign. After you make the expression in the general form of a polynomial. see:https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs.

  • $\begingroup$ Yes. But I think in this case it's an elementary concept. Any way I provided a link, thx for your comment. $\endgroup$ – yousef magableh Apr 19 at 20:26

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