Author Topic: Completing Perfect Squares  (Read 935 times)

Offline msu_math

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Completing Perfect Squares
« on: November 21, 2012, 10:52:19 AM »
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.
Mohammad Salah Uddin

Lecturer in Mathematics
Department of Natural Sciences
FSIT, DIU

Offline Nargis Akter

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Re: Completing Perfect Squares
« Reply #1 on: April 08, 2017, 01:58:49 PM »
Thanks