Question

Solve `(x^2 - 4)/(x^2 - 2x - 15) ≤ 0`

Answer 1

The given inequality is f(x) = `(x^2 - 4)/(x^2 + 4x - 15)`

= `((x + 2)(x - 2))/(x^2 - 5x + 3x - 15)`

= `((x + 2)(x - 2))/(x(x - 5) + 3(x - 5))`

f(x) = `((x + 2)(x - 2))/((x + 3)(x - 5)) ≤ 0`

[The critical numbers of f(x) are those values of x for which f(x) = 0, and those values of x for which f(x) is not defined. When x = – 3, 5. f(x) = ∞ ⇒ f(x) is not defined.]

The critical numbers are x = – 2 , 2, – 3, 5

Divide the number line into five intervals

(- ∞, – 3), (- 3, – 2), (- 2, 2), (2, 5), (5, ∞)

(i) (– ∞, – 3)

When x < – 3 say x = – 4

The factor x + 2 = – 4 + 2 = – 2 < 0

The factor x – 2 = – 4 – 2 = 6 < 0

The factor x + 3 = – 4 + 3 = – 1 < 0

The factor x – 5 = – 4 – 5 = – 9 < 0

Thus x + 2 < 0, x + 3 < 0, x – 2 < 0, x – 5 < 0

∴ `(x^2 - 4)/(x^2 + 4x - 15) > 0`

Thus `(x^2 - 4)/(x^2 + 4x - 15) ≤  0` is not true in the interval (– ∞, – 3).

∴ It has no solution in (– ∞, – 3)

(ii) (– 3, – 2)

When – 3 < x ≤ – 2 say x = – 2.5

The factor x + 2 = – 2.5 + 2 = – 0.5 < 0

The factor x – 2 = – 2.5 – 2 = – 4.5 < 0

The factor x + 3 = – 2.5 + 3 = 0.5 > 0

The factor x – 5 = – 2.5 – 5 = – 7.5 < 0

Thus x + 2 < 0, x + 3 > 0

and

x – 2 < 0

x – 5 < 0

∴ `(x^2 - 4)/(x^2 + 4x - 15) < 0`

Thus `(x^2 - 4)/(x^2 + 4x - 15) ≤  0` is not true in the interval (– 3, – 2).

∴ It has no solution in (– 3, – 2)

(iii) (– 2, 2)

When – 2 ≤ x ≤ 2 say x = 0

The factor x + 2 = 0 + 2 = 2 > 0

The factor x – 2 = 0 – 2 = – 2 < 0

The factor x + 3 = 0 + 3 = 3 > 0

The factor x – 5 = 0 – 5 = – 5 < 0

Interval	Sign of x + 2	Sign of x – 2	Sign of x + 3	Sign of x – 5	Sign of  `((x + 2)(x - 2))/((x + 3)(x - 5))`
(– ∞, – 3)	–	–	–	–	+
(– 3, – 2)	–	–	+	–	–
(– 2, 2)	+	–	+	–	+
(2, 5)	+	+	+	–	–
(5, ∞)	+	+	+	+	+

Thus x + 2 > 0 

x + 3 > 0

and

x – 2 < 0

x – 5 < 0

∴ `(x^2 - 4)/(x^2 + 4x - 15) > 0`

Thus `(x^2 - 4)/(x^2 + 4x - 15) ≤  0` is not true in the interval (– 2, – 2).

∴ It has no solution in (– 2, – 2)

(iv) (2, 5)

When 2 ≤ x < 5 say x = 3

The factor x + 2 = 3 = 3 + 2 = 5 > 0

The factor x – 2 = 3 – 2 = 1 > 0

The factor x + 3 = 3 + 3 = 6 > 0

The factor x – 5 = 3 – 5 = – 2 < 0

Thus x + 2 > 0,

x + 3 > 0

and

x – 2 > 0

x – 5 < 0

∴ `(x^2 - 4)/(x^2 + 4x - 15) < 0`

Thus `(x^2 - 4)/(x^2 + 4x - 15) ≤  0` is not true in the interval (2, 5).

∴ It has no solution in (2, 5)

(v) (5, ∞)

When 5 < x < ∞ say x = 6

The factor x + 2 = 6 + 2 = 8 > 0

The factor x – 2 = 6 – 2 = 4 > 0

The factor x + 3 = 6 + 3 = 9 > 0

The factor x – 5 = 6 – 5 = 1 > 0

Thus

x + 2 > 0,

x + 3 > 0

and

x – 2 > 0,

x – 5 > 0

∴ `(x^2 - 4)/(x^2 + 4x - 15) > 0`

Thus `(x^2 - 4)/(x^2 + 4x - 15) ≤  0` is not true in the interval (5, ∞).

∴ It has no solution in (5, ∞)

The given inequality f(x) = `(x^2 - 4)/(x^2 + 4x - 15) ≤  0` has solution in the intervals (– 3, – 2]

∴ The solution set is (– 3, 2] ∪ [2, 5)