What is the solution set of the quadratic inequality 4 times open paren x plus 2 close paren squared is less than or equal to 0 ?

A square is always nonnegative, so 4(x + 2)² is nonnegative for every real x. For it to be less than or equal to 0, it must equal 0. Thus (x + 2)² = 0, which gives x = −2, so the solution set is the set containing only −2.

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