Question

|X - 5| + 4 > 0 and |X2

| < 4. Then x can be:​

Answer 1

○☆ANSWER☆○

HOPE IT WILL HELP YOU..

Answer 2

Given:

$|x-5|+4>0$ ...(1)

$|x^2|<4$ ...(2)

The mode inequality can be solved by isolation.

$|x-5|+4>0$ ...(1)

$|x-5|>-4$

The mode functions are positive or 0, so this has solutions over reals.

$|x^2|<4$ ...(2)

$-4<x^2<4$

This gives two inequalities for x.

$\displaystyle{\left \{ {{x^2>-4} \atop {x^2<4}} \right. }$

The upper inequality has solutions over reals.

The inequality below gives $-2 <x<2$.

Solution

The intersection of three ranges is $-2<x<2$.

More information:

We usually find the ranges where the mode becomes 'negative', and '0 or positive'. When mode becomes negative, we apply a negative sign.

This fact is used in mode equations and inequalities.