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Completing Perfect Squares

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msu_math:
Problem: For what integer value n is n2 + 6x + 10 also a perfect square?

Solution: Let us first express the given polynomial n2 + 6x + 10 in the form "a perfect square + constant". We have, n2 + 6x + 10 = (n+3)2 +1. If n2 + 6x + 10 = m2 for some integer m, then 1 = m2 - (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.

Nargis Akter:
Thanks

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