Answer: x∈(-∞,-2√6-2)U(2√6-2,+∞)
Step-by-step explanation:
[tex]\displaystyle\\y=\frac{1}{4} (x+2)^2-6\\y > 0\\Hence,\\\frac{1}{4}(x+2)^2-6 > 0 \\\\\frac{1}{4}(x+2)^2-6+6 > 0+6\\\\\frac{1}{4}(x+2)^2 > 6\\[/tex]
Multiply both parts of the equation by 4:
[tex](x+2)^2 > 24[/tex]
Extract the square root of both parts of the inequality:
[tex]|x+2| > \sqrt{24} \\|x+2| > \sqrt{4*6} \\|x+2| > 2\sqrt{6}[/tex]
Expand the modulus - we get a set of inequalities:
[tex]\displaystyle\\\left [{{x+2 > 2\sqrt{6}\ \ \ \ (1) } \atop {-(x+2) > 2\sqrt{6} \ \ \ (2)}} \right.\\[/tex]
Multiply both parts of inequality (2) by -1 and reverse the sign of the inequality:
[tex]\displaystyle\\\left [ {{x+2-2 > 2\sqrt{6}-2 } \atop {x+2 < -2\sqrt{6} }} \right. \ \ \ \ \ \left [ {{x > 2\sqrt{6}-2 } \atop {x+2-2 < -2\sqrt{6} -2}} \right. \ \ \ \ \ \left [ {{x > 2\sqrt{6}-2 } \atop {x < -2\sqrt{6}-2 }} \right.[/tex]
Thus,
[tex]x\in(-\infty,-2\sqrt{6}-2)U(2\sqrt{6} -2,+\infty)[/tex]
