Question

solve the inequality by any method.
$(x + 6)^2 \leq 0$
choose the correct answer below
\\(\bigcirc\\) a. \\(-6\\) and \\(6\\)
\\(\bigcirc\\) b. \\((-\infty, -6)\\) and \\((-6, \infty)\\)
\\(\bigcirc\\) c. \\((-\infty, \infty)\\)
\\(\bigcirc\\) d. \\((-6, 6)\\)
\\(\bigcirc\\) e. \\((-\infty, 6)\\) and \\((6, \infty)\\)
\\(\bigcirc\\) f. \\(-6\\)
\\(\bigcirc\\) g. \\(6\\)
\\(\bigcirc\\) h. no real solutions

Explanation:

Step 1: Recall the property of squares

A square of any real number is non - negative, i.e., for any real number \(a\), \(a^{2}\geq0\). In the inequality \((x + 6)^{2}\leq0\), since \((x + 6)^{2}\geq0\) for all real \(x\), the only way for \((x + 6)^{2}\leq0\) to hold is when \((x + 6)^{2}=0\).

Step 2: Solve for \(x\) when \((x + 6)^{2}=0\)

If \((x + 6)^{2}=0\), then by taking the square root of both sides, we get \(x+6 = 0\). Solving for \(x\), we subtract 6 from both sides of the equation: \(x=-6\).

Answer:

F. \(-6\)