A2853. Find the product of the real solutions to (x2 −7x+ 11)(x2−13x+42) = 1.
Solution by the Missouri State University Problem Solving Group Department of Mathematics
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Springfield, MO 65897
Letf(x) =x2−7x+ 11 and g(x) =x2−13x+ 42.
We have f(x) = 1 if and only if (x−2)(x−5) = 0, and f(2)g(2) = 120 = 1, f(5)g(5) = 12 = 1.
So we have two solutions x= 2 andx= 5.
Now note that f(x) = −1 if and only if (x−3)(x−4) = 0, and f(3)g(3) = (−1)12 = 1, f(4)g(4) = (−1)6 = 1, so we have two more solutions, x= 3 andx= 4.
Finally g(x) = 0 if and only if (x−6)(x−7) = 0 and f(6)g(6) = 50 = 1, f(7)g(7) = 110 = 1, so we have solutions x= 6 andx= 7.
The product of the real solutions is then 2·3·4·5·6·7 = 5040.
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