Question

Answer the following:

Solve the following for x, where |x| is modulus function, [x] is greatest interger function, {x} is a fractional part function

|x² − 9| + |x² − 4| = 5

Answer 1

|x² − 9| + |x² − 4| = 5

Case 1: x < − 3

If x < − 3, then x² > 9

∴ |x² − 9| = x² − 9 and |x² − 4| = x² − 4

∴ equation becomes,

∴ x² − 9 + x² − 4 = 5

∴ 2x² = 18

∴ x² = 9

∴ x = ± 3

But x < − 3, therefore x ≠ ± 3.

Case 2: − 3 ≤ x < − 2

If − 3 ≤ x < − 2, then 4 < x² ≤ 9

∴ |x² − 9| = 9 − x² and |x² − 4| = x² − 4

∴ equation becomes,

9 − x² + x² − 4 = 5

∴ 5 = 5, which is true

∴ − 3 ≤ x < − 2 is the solution    ...(1)

Case 3: − 2 ≤ x < 2

If − 2 ≤ x < 2, then 0 ≤ x² < 4

∴ |x² − 9| = 9 − x² and |x² − 4| = 4 − x² 

∴ equation becomes,

9 − x² + 4 − x²  = 5

∴ 2x² = 8

∴ x² = 4

∴ x = ± 2

But − 2 ≤ x < 2

∴ x ≠ 2

∴ x = − 2 is a solution    ...(2)

Case 4: 2 ≤ x < 3

If 2 ≤ x < 3, then 4 ≤ x < 9

∴ |x² − 9| = 9 − x² and |x² − 4| = x² − 4

∴ equation becomes,

9 − x² +  x² − 4 = 5

∴ 5 = 5, which is true

∴ 2 ≤ x < 3 is the solution   ...(3)

Case 5: x ≥ 3

If x ≥ 3, then x² ≥ 9

∴ |x² − 9| = x² − 9 and |x² − 4| = x² − 4

∴ equation becomes,

x² − 9 + x² − 4 = 5

∴ 2x² = 18

∴ x² = 9

∴ x = ± 3

But x ≥ 3, therefore x ≠ −3.

∴  x = 3 is a solution     ...(4)

From (1), (2), (3) and (4),

the solution set = [− 3, − 2] ∪ [2, 3].