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

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mirai123 mirai123 · Answer 1 · 2020-05-03T15:23:40+02:00

Answer:

[tex]x<-4\quad \mathrm{or}\quad \:x>4[/tex]

[tex]\{x\in \mathbb{R}| \:\left(-\infty \:,\:-4\right)\cup \left(4,\:\infty \:\right) \}[/tex]

Step-by-step explanation:

[tex]x^2>16[/tex]

[tex]x<-\sqrt{16}\quad \mathrm{or}\quad \:x>\sqrt{16}[/tex]

[tex]x<-4\quad \mathrm{or}\quad \:x>4[/tex]

The interval notation will be:

[tex]\{x\in \mathbb{R}| \:\left(-\infty \:,\:-4\right)\cup \left(4,\:\infty \:\right) \}[/tex]

The contrary,

[tex]x^2<16[/tex]

is [tex]-4<x<4[/tex]

markoskontoulis markoskontoulis · Answer 2 · 2020-05-03T15:43:14+02:00

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! :)

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