# Algebra question

A function $f:R \to R$ satisfies $f(5−x)=f(5+x)$. If $f(x)=0$ has $5$ distinct real roots, what is the sum of all of the distinct real roots?

I am unable to know from where to start.

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$5+t$ is a root iff $5-t$ is a root. –  Siméon Mar 22 '13 at 8:16
Hint: since it has $5$ real roots, one of its roots $\lambda$ must satisfy $5 - \lambda = 5 + \lambda.$ –  Geoff Robinson Mar 22 '13 at 8:18
What has this got to do with number theory? Is this an ongoing competition there? –  Jyrki Lahtonen Mar 22 '13 at 8:19
@JyrkiLahtonen Yes , but I don't care about that as I have already done it wrong. .. I dont' cheat others. –  Kushashwa Ravi Shrimali Mar 22 '13 at 8:20
Ok. Followin Ju'x' hint you should get two pairs of roots with that symmetry, and one singleton root symmetric with itself :-) –  Jyrki Lahtonen Mar 22 '13 at 8:22

Note that the condition is the same as $f(y) = f(10 -y)$ for all $y$.

So if $y_{0}$ is a root, so is $10 - y_{0}$.

Now if $y_{0} = 10 - y_{0}$, it follows $y_{0} = 5$. Since the root are distinct, only one root $y_{0}$ can equal $10 -y_{0}$, and the other four must go in pairs $y_{1} \ne 10 - y_{1}$, $y_{2} \ne 10 - y_{2}$.

Hence the sum is $$5 + 10 + 10 = 25.$$

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Let $\phi(x) = f(x+5)$. Then $\phi$ is even. Hence if $\phi(x_0) = 0$, then $\phi(-x_0) =0$. Since we are given that there are a finite number of roots, we have $\sum_{x \in \phi^{-1} \{0\}} x = 0$.

Finally, since $f^{-1} \{0 \} = \phi^{-1} \{0\} + \{5\}$, and we know that $|f^{-1} \{0 \}| =|\phi^{-1} \{0 \}| = 5$, we see that $\sum_{x \in f^{-1} \{0\}} x = \sum_{x \in \phi^{-1} \{0\}} x + 5 |\phi^{-1} \{0\}| = 0 + 5\cdot 5 = 25$.

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