KETIDAKSAMAAN Inequality Inequality In solving inequality problems we
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.
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) � 2 x – 7 < 4 x – 2 …(grouped with same variable) 2 x < 4 x + 5 2 x - 4 x < 5 -2 x < 5 x > - 5/2 � -5 ≤ 2 x + 6 < 4 …(divided into 2 group) -5 - 6 ≤ 2 x < 4 – 6 -11 ≤ 2 x < -2 -11/2 ≤ x < -1
Examples (2) � x 2 –x<6 x 2 –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 x 1 = 1 x 2 = 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 ∣x∣ = -x if x≤ 0 1. ∣ab∣ = ∣a∣. ∣b∣ 2. , b≠ 0 3. ∣a+b∣ ≤ ∣a∣+∣b∣ 4. ∣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 � ∣ 3 x-5∣ ≥ 1 3 x-5≤-1 3 x≤-1+5 3 x≤ 4 or x≤ 4/3 or or 3 x-5≥ 1 or 3 x ≥ 1+5 3 x ≥ 6 x≥ 2
Square � Every positif number will have two square root. � For a ≥ 0 notation √a defined as main square root � √x 2 = ∣x∣ � ∣ x∣ 2 = x 2 � ∣x∣ < ∣y∣ ⇔ x 2<y 2 � If x � ℝ and n > 0, then = n√xn =∣x∣ � If x � ℝ and m, n > 0, then = n√xm =∣x∣m/n
Example � √(2 x-4) <2 (√(2 x-4))2 < 22 2 x-4<4 2 x<8 x<4 � Condition : (2 x-4)≥ 0 2 x ≥ 4 x≥ 2 � Result : Hp = { x | 2 ≤ x < 4 }
Quadratic formula � The � ax 2 solution of ax 2+bx+c=0, a ≠ 0, is given by: + bx + c = 0 can be factorization as : ( x + x 1 )( x + x 2 ) � D =b 2 – 4 ac D > 0 → have 2 different root square D = 0 → have same root square D < 0 → have no real root square
Example � x 2 -2 x-3≤ 0 X 1 and X 2 =3 = -1 � Result : x 2 -2 x-3=(x -x 1)(x-x 2) = (x-3)(x+1)
Example � ∣ 3 x+1∣<2∣x-6∣ ⇔∣ 3 x+1∣ < ∣ 2 x-12∣ ⇔ (3 x+1)2 < (2 x-12)2 ⇔ 9 x 2+6 x+1<4 x 2 -48 x+144 ⇔ 5 x 2+54 x-143 < 0 ⇔(5 x-11)(x+13)<0
Problems (23 Sept 2015) 1. 2. 3. 4. 5. 3 x 2 – x -2 > 0 4 x – 7 < 3 x + 5 7 x – 1 ≤ 10 x + 4 2 x – 4 ≤ 6 – 7 x ≤ 3 x + 6 3 x 2 – 11 x – 4 ≤ 0
Problems ∣ 3 x+4∣< 8 2 x 2 - 5 x – 4 ≤ 0 6) 7) 8) . 9) 10) .
- Slides: 18