# Sum of roots of a functional equation

Let $$f(x)$$ be a function which satisfies $$f(29+x)=f(29-x)$$ for all $$x \in \mathbb{R} .$$ Suppose $$f(x)$$ has (exactly) three real roots $$a, b, c$$, determine the value of $$a+ b+c$$.

My work:

From $$f(29 + x) = f(29 - x) \tag{1}$$ , it is observed that it is symmetric along the line $$x = 29$$.

Now, let $$g(x) = f(29 + x).$$ On substituting $$x \mapsto (-x)$$ we get $$g(-x) = f(29 -x)$$ But from (1), we conclude that $$g(x) = g(-x)$$ and hence $$g(x)$$ is an even function.

$$\implies$$ One of the roots of $$g(x)$$ must be $$0.$$

My questions:

1. $$\star$$ How do I decide the other $$2$$ roots ? Due to the symmetric nature of $$g(x)$$ around origin, can I conclude that the other two roots, say, $$x_0 , -x_0$$ for some $$x_0 \in \{\alpha, \beta, \gamma\}$$
2. By the conclusion that $$g(x) = 0$$ for some $$x \in \{\alpha, \beta, \gamma\}$$, is it true that $$f(29 + x)$$ has one of the roots $$= 29$$ for some $$x \in \{\alpha, \beta, \gamma\}$$ $$?$$
• You don't have to determine the other two roots, only their sum. And you know that $g$ is even ... Dec 20, 2021 at 8:25

If $$x$$ is a root then so is $$58-x$$ because $$f(58-x)=f(x)$$. Hence, the three roots are of the form $$x, 58-x,29$$ with $$x\neq 29$$. The sum is $$87$$.
• I understood the rest, but can you explain the $x\neq 29 ?$ I am sure I am overlooking something here. Dec 20, 2021 at 8:40
• Since there are three roots one of the roots has to be different from $29$. @noobman Dec 20, 2021 at 8:42