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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.

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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
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    $\begingroup$ @FareedAF No, the left hand side isn't defined at those values. $\endgroup$ – saulspatz Apr 19 at 20:10
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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.

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  • $\begingroup$ 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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