Question

When working with inequalities what do you do when its x squared > 16 and vice versa does the sign flip.

Answer 1

Answer:

$x4$

$\{x\in \mathbb{R}| \:\left(-\infty \:,\:-4\right)\cup \left(4,\:\infty \:\right) \}$

Step-by-step explanation:

$x^2>16$

$x\sqrt{16}$

$x4$

The interval notation will be:

$\{x\in \mathbb{R}| \:\left(-\infty \:,\:-4\right)\cup \left(4,\:\infty \:\right) \}$

The contrary,

$x^2$

is $-4$

Answer 2

No it doesn't flip. You split into two possible cases with two separate inequalities.

x^2 > 16

| x | > 4

And since now you are working with an absolute value, you split into 2 possible cases.

if x >= 0: then x > 4

if x < 0: then -x > 4

x < -4

FOR THE FIRST:

x ∈ (4, + ∞)

FOR THE SECOND:

x ∈ (-∞, -4)

Their UNITY is: x ∈ (-∞, -4) U (4, +∞)

And that's how you go about solving these. Hope I helped! :)