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$How many integers satisfy the inequality $(x+3)^{2}\leq1$?$

Guest Oct 26, 2014

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Best Answer

(x+3)² will always be positive.

The only integers less than or equal to 1 are 0 and 1.

How can (x+3)² = 0 ? ---> x = -3

How can (x+3)² = 1 ? ---> (x + 3) = 1 or (x + 3) = -1 ---> x = -4 or x = -2

Answer: 3 integers: -2, -3, and -4

geno3141 Oct 26, 2014

geno3141 · Answer 1 · 2014-10-26T07:51:35

(x+3)² will always be positive.

The only integers less than or equal to 1 are 0 and 1.

How can (x+3)² = 0 ? ---> x = -3

How can (x+3)² = 1 ? ---> (x + 3) = 1 or (x + 3) = -1 ---> x = -4 or x = -2

Answer: 3 integers: -2, -3, and -4