Question

What is the solution set of the quadratic inequality 4(x + 2)2 mc009-1.jpg 0?
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Answer 1

I can’t read the attached images — could you either re-upload them or type the inequality exactly?

Meanwhile, if the expression is 4(x + 2)^2 with one of the usual signs to 0, here are the possibilities:

- 4(x + 2)^2 > 0: holds for every real x except x = −2 (since the square is positive unless it is zero). Solution: x ∈ R \ {−2}.
- 4(x + 2)^2 ≥ 0: holds for every real x (always nonnegative). Solution: x ∈ R.
- 4(x + 2)^2 < 0: impossible (a square times 4 cannot be negative). Solution: no real x.
- 4(x + 2)^2 ≤ 0: holds only when 4(x + 2)^2 = 0, i.e. x = −2. Solution: {−2}.

If your inequality is different (for example contains additional terms), please paste it exactly and I’ll solve it.