KETIDAKSAMAAN Inequality

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.

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 (-∞, ∞)

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:

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

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

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

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∣∣

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

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

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

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 }

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

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

2 ≤ x2 – x < 6 2 ≤ x2 – x dan x2 – x < 6 x2 – x ≥ 2 x2 – x – 6 < 0 x2 – x – 2 ≥ 0 (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 }

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

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

Problems ∣3x+4∣< 8 2x2- 5x – 4 ≤ 0 .