|x^2−4| ≤ |x^2−16|

Question

What is the solution for this inequality.

Answer 1

The inequality is |x ²- 4| ≤ |x ²- 16|

Case 1 :

(x ²- 4) ≤ (x ²- 16)

- 4 ≤ - 16

4 ≤ 16.

It is impossible.

Case 2 :

(x ²- 4) ≤ - (x ²- 16)

(x ²- 4) ≤ - x ²+ 16

2x ² ≤ 20

x ² ≤ 20/2

x ² ≤ 10

x  ≤ (± √10)²

x  ≤ ± √10.

Solution set of the inequality is {x | x  ≤ ± √10}.

 

Comments

The solution set is {x | -√10 ≤ x ≤ √10}.

Answer 2

The absolute value in equality is |x² - 4| ≤ |x² - 16|.

There are four cases are obtained as follows :

Case 1 : (x² - 4) > 0 and (x² - 16) > 0

x² - 4 ≤ x² - 16 ⟹ - 4 ≤ - 16 ⟹ 4 ≥ 16. This is false statement and there is no x - variable.

Case 2 : (x² - 4) < 0 and (x² - 16) < 0

- (x² - 4) ≤ - (x² - 16) ⟹ - x² + 4 ≤ - x² +16 ⟹ 4 ≤ 16. This is true statement and there is no x - variable.

Case 3 : (x² - 4) < 0 and (x² - 16) > 0

- (x² - 4) ≤ (x² - 16) ⟹ - x² + 4 ≤ x² - 16 ⟹ 20 ≤ 2x² ⟹ 10 ≤ x² ⟹ ± √10 ≤ x.

Case 4 : (x² - 4) > 0 and (x² - 16) < 0

(x² - 4) ≤ - (x² - 16) ⟹ x² - 4 ≤ - x² + 16 ⟹ 2x² ≤ 20 ⟹ x² ≤ 10 ⟹ x ≤ ± √10.

The two solutions ±√10 ≤ x and x ≤ ±√10 are can be writen as - √10 ≤ x ≤ √10.