1984 AHSME Problems/Problem 16

Problem

The function $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x$ . If the equation $f(x)=0$ has exactly four distinct real roots, then the sum of these roots is

$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$

Solution

Let one of the roots be $r_1$ . Also, define $x$ such that $2+x=r_1$ . Thus, we have $f(2+x)=f(r_1)=0$ and $f(2+x)=f(2-x)$ . Therefore, we have $f(2-x)=0$ , and $2-x$ is also a root. Let this root be $r_2$ . The sum $r_1+r_2=2+x+2-x=4$ . Similarly, we can let $r_3$ be a root and define $y$ such that $2+y=r_3$ , and we will find $2-y$ is also a root, say, $r_4$ , so $r_3+r_4=2+y+2-y=4$ . Therefore, $r_1+r_2+r_3+r_4=4+4=8, \boxed{\text{E}}$ .

Solution 2

The graph of this function is symmetric around 2. Therefore, two roots will be greater than $2$ and the other two roots will be less than $2$ . These roots are symmetric around $2$ , so the average of the four roots is $2$ . Then, the sum is $2\times4=8$ .

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1984 AHSME Problems/Problem 16

Contents

Problem

Solution

Solution 2

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