Summing Four Roots of an Even FunctionDate: 07/27/98 at 17:50:01 From: Spencer Wu Subject: 1984 AHSME question 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: a) 0 b) 2 c) 4 d) 6 e) 8 Date: 07/27/98 at 17:59:54 From: Doctor Schwa Subject: Re: 1984 AHSME question We're looking for points where f(x) = 0. Let's say we found one of them, call it a. Somehow we have to use the property that f(2+x) = f(2-x). How? Well, let a = 2+x. Then we know 2-x is also a root (why?). What's that in terms of a? Well, x = a - 2, so: 2 - x = 2 - (a - 2) = 4 - a If you try to find another root by that same method, you just end up going in circles (try it and see why). But, lucky for us, that pair of roots that we found, a and 4-a, adds up to 4. If there are two more real roots, they have to be b and what? This means the total adds up to what? Another, maybe better way to think about this problem: f(2+x) = f(2-x) is sort of like the f(x) = f(-x), an even function, only we have that 2 there. What does the 2 do? It makes the function symmetric about x = 2 instead of about x = 0, since 2 + x and 2 - x are the two points a distance "x" away from the point 2 on the number line. How can you make use of the function being symmetric about x = 2? If there are four roots, they are also symmetric about x = 2, and therefore their average is what and hence their total is what? Feel free to write back if you'd like more of a hint on this problem. - Doctor Schwa, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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