Daffodil International University
Educational => Mathematics => Math Helps => Topic started by: msu_math on November 21, 2012, 10:52:19 AM

Problem: For what integer value n is n^{2} + 6x + 10 also a perfect square?
Solution: Let us first express the given polynomial n^{2} + 6x + 10 in the form "a perfect square + constant". We have, n^{2} + 6x + 10 = (n+3)^{2} +1. If n^{2} + 6x + 10 = m^{2} for some integer m, then 1 = m^{2}  (n+3)^{2}. The left hand side gives the difference of two perfect squares which is 1 as in the right side. The only perfect squares that differ by 1 are 0 and 1. Hence, (n+3)^{2} = 0 , having the solution n = 3 which gives the required value of n.

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