**Generalized Method of
Intervals**

**Generalized Method of Intervals for Solving
Inequalities:**

Let \(F(x)={{(x-{{a}_{1}})}^{{{k}_{1}}}}{{(x-{{a}_{2}})}^{{{k}_{2}}}}…{{(x-{{a}_{n-1}})}^{{{k}_{(n-1)}}}}{{(x-{{a}_{n}})}^{{{k}_{n}}}}\).

Where k₁, k₂,
… k_{n} ϵ Z and a₁, a₂, …., a_{n} are fixed real numbers
satisfying the condition.

a₁ < a₂
< a₃ < … < a _{n}₋₁ < a_{n}

For solving F(x) > 0 (or) F(x) < 0, consider the following algorithm:

- We
mark the number a₁, a₂, …, a
_{n}on the number axis and put a plus sign in the interval on the right of the largest of these numbers, i.e., on the right of a_{n} - Then
we put the plus sign in the interval on the left of a
_{n}. if k_{n}is an even number and minus sign if k_{n }is an odd number. In the next interval, we put a sign according to the following rule. - When
passing through the point a
_{n}₋ ₁, the polynomial F(x) changes sign if k_{ n}₋₁ is an odd number. Then we consider the next interval and put a sign in it using the same rule - Thus, we consider all the intervals. The solution of the inequality F(x) > 0 is the union of all intervals in which we have put the plus sign, and the solution of the inequality F(x) < 0 is the union of all intervals in which we have put the minus sign.

**Key points:**

(i) (x – a) (x – b) < 0 ⇒ x ϵ (a, b), where a < b

(ii) (x – a) (x – b) > 0 ⇒ x ϵ (-∞, a) ∪ (b, ∞), where a<b

(iii) x² ≤ a² ⇒ x ϵ [-a, a]

(iv) x² ≥ a² ⇒ x ϵ (-∞, a] ∪ [a, ∞)

(v) if ax² + bx + c =0, (a > 0) x ϵ (α, β), where α, β (α < β) are roots of the equation ax² + bx + c = 0.

**Example**:
find x² – x – 2 > 0.

**Solution:
**Given that x² – x – 2 > 0

(x – 2) (x + 1) > 0

x² – x – 2 > 0 ⇒ X = -1, 2

now on the number line (x – axis) mark x = -1 and x = 2

now x > 2, x – 2 > 0

(x + 1) (x – 2) > 0

When -1 < x < 2

(x + 1) (x – 2) < 0

When x < -1, but x – 2 > 0

(x + 1) (x – 2) > 0

Hence sign number x² – x – 2 > 0.

From the figure x² – x – 2, x ϵ (-∞, -1) ∪ (2, ∞),