1. KETIDAKSAMAAN
Inequality

2. Inequality
In solving inequality problems, we will looking for a set of every real number that make the inequality solved.
That solving set is written in line number interval form.

3. Interval Hp = { x | a < x < b } ( a, b ) Hp = { x | a ≤ x ≤ b }
Set
Interval
Grafik
Hp = { x | a < x < b }
( a, b )
Hp = { x | a ≤ x ≤ b }
[ a, b ]
Hp = { x | a ≤ x < b }
[ a, b )
Hp = { x | a < x ≤ b }
( a, b ]
Hp = { x | x ≤ b }
(-∞,b ]
Hp = { x | x < b }
(-∞,b )
Hp = { x | x ≥ a }
[ a, ∞)
Hp = { x | x > a }
( a, ∞ )
Hp = R
(-∞, ∞)

4. Solving inequality Add same number at both side of inequality
Multiply at both side with a positive number
Multiply at both side with a negative number, notice that simbol will change direction.
General form of inequality:

5. Examples (1) 2x – 7 < 4x – 2 …(grouped with same variable)
-5 ≤ 2x + 6 < 4 …(divided into 2 group)
≤ 2x < 4 – 6
-11 ≤ 2x < -2
-11/2 ≤ x < -1

6. Examples (2) x2 – x < 6 x2 –x – 6 < 0 (x+2) (x-3) <0
Notice that factorization result will divide 3 area, so we should find area which < 0

7. Fraction inequality Condition : division from a and b  where b ≠ 0
example :
x – 1 = 0 and x – 2 = 0
x1 = x2 = 2
Condition : x – 2 ≠ 0
x ≠ 0
Result : Hp = { x | x ≤ 1 x > 2 }

8. Absolute value Absolute = ∣x∣ Absolute Characteristic : ∣x∣ = x if x≥0
∣ab∣ = ∣a∣.∣b∣
, b ≠ 0
3. ∣a+b∣ ≤ ∣a∣+∣b∣
∣a-b∣ ≥ ∣∣a∣-∣b∣∣

9. Absolute inequality General statement : ∣x∣ < a ⇔ -a<x<a
∣x∣ > a ⇔ x<-a atau x>a

10. examples ∣x-4∣<1 -1<x-4<1 -1+4<x<1+4 3<x<5
3x-5≤-1 or 3x-5≥ 1
3x≤-1+5 or 3x ≥ 1+5
3x≤ 4 or 3x ≥ 6
x≤4/3 or x ≥ 2

11. Square Every positif number will have two square root.
For a ≥ 0 notation √a defined as main square root
√x2 = ∣x∣
∣ x∣2 = x2
∣x∣ < ∣y∣ ⇔ x2<y2
If x  ℝ and n > 0, then = n√xn =∣x∣
If x  ℝ and m, n > 0, then = n√xm =∣x∣m/n

12. Example √(2x-4) < 2 (√(2x-4))2 < 22 2x-4<4 2x<8 x<4
Condition : (2x-4)≥0
2x ≥ 4
x ≥ 2
Result :
Hp = { x | 2 ≤ x < 4 }

13. Quadratic formula The solution of ax2+bx+c=0, a ≠0, is given by:
ax2 + bx + c = 0 can be factorization as :
( x + x1 )( x + x2 )
D =b2 – 4ac
D > 0 → have 2 different root square
D = 0 → have same root square
D < 0 → have no real root square

14. Example x2-2x-3≤0 X1 and X2 = 3 = -1 Result :
x2-2x-3=(x -x1)(x-x2) = (x-3)(x+1)

15. 2 ≤ x2 – x < 6
2 ≤ x2 – x dan x2 – x < 6
x2 – x ≥ x2 – x – 6 < 0
x2 – x – 2 ≥ (x+2)(x-3)<0
( x + 1)( x – 2 ) ≥ 0 x3 = -2 , x4 = 3
x1 = - 1, x2 = 2
Sehingga : Hp = { x|-2<x≤-1 ∪ 2≤x<3 }

16. Example ∣3x+1∣<2∣x-6∣ ∣3x+1∣<2∣x-6∣ ⇔∣3x+1∣ < ∣2x-12∣

17. Problems (23 Sept 2015) 3x2 – x -2 > 0 4x – 7 < 3x + 5

18. Problems
∣3x+4∣< 8
2x2- 5x – 4 ≤ 0 .