How do you solve ${(x + 3)}^{2} \leq 0$ $(x+3)^2<=0$ ?

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Answer 1 · 2017-02-28T09:52:01

The solution is $x = - 3$ $x = -3$

There is one solution because

$AA x in RR, (x+3)^2>=0$

Therefore,

$x=-3$ , $=>$ , $(x+3)^2=0$

graph{(x+3)^2 [-10, 10, -5, 5]}

How do you solve (x+3)^2<=0(x+3)2≤0(x+3)^2<=0?