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Algebra Level 5

Find the number of real solutions to the equation $\sum_{r=1}^{100} \dfrac{r}{x-r}=x$

The answer is 101.

2 solutions

Deepanshu Gupta
Nov 13, 2014

Draw approximate graph of this question :

Graph Graph

hence no: of solution is 101

q.e.d

Nice one! If someone is unable to find the solution in the third quadrant, you could also reason it this way -

There are a total of $101$ roots to the whole polynomial as the $100$ degree denominator on the left will get multiplied by $x$ on the right. And $100$ real roots are clearly visible due to the asymptotes. So, the $101$ th root also has to be real as complex roots to polynomials with real coefficients occur as conjugates.

Pratik Shastri - 6 years, 7 months ago

Nice graph, Deepanshu. Is this Geogebra? WolframAlpha wasn't able to duplicate it.

Brian Charlesworth - 6 years, 7 months ago

Thanks Sir ! Actually I used Paint to draw this graph :)

Deepanshu Gupta - 6 years, 7 months ago

Did the same way !

Keshav Tiwari - 6 years, 7 months ago

Did it same way! :)

Pranjal Jain - 6 years, 6 months ago

Nice solution..

Satya Howladar - 6 years, 3 months ago

Gandhar Joshi
Nov 14, 2014

There are a total of 101 roots to the whole polynomial as the100 degree denominator on the left will get multiplied by x on the right. And 100 real roots are clearly visible due to the asymptotes. So, the101th root also has to be real as complex roots to polynomials with real coefficients occur as conjugates.

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The answer is 101.

2 solutions

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