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 $x \in \{\alpha, \beta, \gamma\}$ $?$
  • 2
    $\begingroup$ You don't have to determine the other two roots, only their sum. And you know that $g$ is even ... $\endgroup$
    – Martin R
    Dec 20, 2021 at 8:25

1 Answer 1


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 $87$.

  • $\begingroup$ I understood the rest, but can you explain the $x\neq 29 ?$ I am sure I am overlooking something here. $\endgroup$
    – noobman
    Dec 20, 2021 at 8:40
  • 1
    $\begingroup$ Since there are three roots one of the roots has to be different from $29$. @noobman $\endgroup$ Dec 20, 2021 at 8:42

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