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, … kn ϵ Z and a₁, a₂, …., an are fixed real numbers satisfying the condition.
a₁ < a₂ < a₃ < … < a n₋₁ < an
For solving F(x) > 0 (or) F(x) < 0, consider the following algorithm:
- We mark the number a₁, a₂, …, an 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 an
- Then we put the plus sign in the interval on the left of an. if kn is an even number and minus sign if kn is an odd number. In the next interval, we put a sign according to the following rule.
- When passing through the point an ₋ 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, ∞),