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Oplossing van Kwadratiese Ongelykhede - Graad 11

Inleiding

Nou dat jy weet hoe om kwadratiese vergelykings op te los, is jy gereed om te leer hoe om kwadratiese ongelykhede op te los.

Kwadratiese Ongelykhede

'n Kwadratiese ongelykheid is ongelykheid van die vorm

a x 2 + b x + c > 0 a x 2 + b x + c 0 a x 2 + b x + c < 0 a x 2 + b x + c 0 a x 2 + b x + c > 0 a x 2 + b x + c 0 a x 2 + b x + c < 0 a x 2 + b x + c 0
(1)

Om 'n kwadratiese ongelykheid op te los is dieselfde as om uit te werk watter dele van die grafiek bo of onder die xx-as is.

Exercise 1: Kwadratiese Ongelykhede

Los die ongelykheid 4x2-4x+104x2-4x+10 op en interpreteer die oplossing grafies.

Solution

  1. Stap 1. Faktoriseer die kwadratiese funksie :

    Laat f(x)=4x2-4x+1f(x)=4x2-4x+1. Faktorisering van die funksie lewer f(x)=(2x-1)2f(x)=(2x-1)2.

  2. Stap 2. Vervang die oorspronklike ongelykheid met die faktore :
    ( 2 x - 1 ) 2 0 ( 2 x - 1 ) 2 0
    (2)
  3. Stap 3. Vind die wortels van die funksie :

    f(x)=0f(x)=0 slegs as x=12x=12.

  4. Stap 4. Skryf die finale antwoord neer :

    Dit beteken dat die grafiek van f(x)=4x2-4x+1f(x)=4x2-4x+1 die xx-as raak by x=12x=12, maar daar is geen areas waar die grafiek onder die xx-as is nie.

  5. Stap 5. Grafiese interpretasie van die oplossing :

    Figuur 1
    Figuur 1 (MG11C8_001.png)

Exercise 2: Oplossing van Kwadratiese Ongelykhede

Vind al die oplossings van die ongelykheid x2-5x+60x2-5x+60.

Solution

  1. Stap 1. Faktoriseer die kwadratiese funksie :

    Die faktore van x2-5x+6x2-5x+6 is (x-3)(x-2)(x-3)(x-2).

  2. Stap 2. Skryf die ongelykheid met die faktore :
    x 2 - 5 x + 6 0 ( x - 3 ) ( x - 2 ) 0 x 2 - 5 x + 6 0 ( x - 3 ) ( x - 2 ) 0
    (3)
  3. Stap 3. Bepaal watter intervalle ooreenstem met die ongelykheid :

    Ons moet bepaal watter waardes van xx die ongelykheid bevredig. Vanaf die faktorisering is daar vyf areas om na te kyk.

    Figuur 2
    Figuur 2 (MG11C8_002.png)

  4. Stap 4. Bepaal of die funksie positief of negatief is in elkeen van die areas :

    Laat f(x)=x2-5x+6f(x)=x2-5x+6. Kies 'n punt in elkeen van areas en evalueer die funksie.

    Tabel 1
        f ( x ) f ( x ) teken van f(x)f(x)
    Area A x < 2 x < 2 f ( 1 ) = 2 f ( 1 ) = 2 +
    Area B x = 2 x = 2 f ( 2 ) = 0 f ( 2 ) = 0 +
    Area C 2 < x < 3 2 < x < 3 f ( 2 , 5 ) = - 2 , 5 f ( 2 , 5 ) = - 2 , 5 -
    Area D x = 3 x = 3 f ( 3 ) = 0 f ( 3 ) = 0 +
    Area E x > 3 x > 3 f ( 4 ) = 2 f ( 4 ) = 2 +

    Ons kan sien dat die funksie positief is vir x2x2 en x3x3.

  5. Stap 5. Skryf die finale antwoord neer en stel dit voor op 'n getallelyn :

    Ons sien dat x2-5x+60x2-5x+60 waar is vir x2x2 en x3x3.

    Figuur 3
    Figuur 3 (MG11C8_003.png)

Exercise 3: Oplossing van Kwadratiese Ongelykhede

Los die kwadratiese ongelykheid -x2-3x+5>0-x2-3x+5>0 op.

Solution

  1. Stap 1. Bepaal hoe om die probleem te benader :

    Laat f(x)=-x2-3x+5f(x)=-x2-3x+5. f(x)f(x) kan nie deur inspeksie gefaktoriseer word nie, dus gebruik ons die formule vir kwadratiese vergelykings. Die xx-afsnitte is die oplossings van die kwadratiese funksie.

    - x 2 - 3 x + 5 = 0 x 2 + 3 x - 5 = 0 x = - 3 ± ( 3 ) 2 - 4 ( 1 ) ( - 5 ) 2 ( 1 ) = - 3 ± 29 2 x 1 = - 3 - 29 2 x 2 = - 3 + 29 2 - x 2 - 3 x + 5 = 0 x 2 + 3 x - 5 = 0 x = - 3 ± ( 3 ) 2 - 4 ( 1 ) ( - 5 ) 2 ( 1 ) = - 3 ± 29 2 x 1 = - 3 - 29 2 x 2 = - 3 + 29 2
    (4)
  2. Stap 2. Bepaal watter intervalle ooreenstem met die ongelykheid :

    Ons moet bepaal watter waardes van xx die ongelykheid bevredig. Vanaf die antwoorde het ons vyf areas om te ondersoek.

    Figuur 4
    Figuur 4 (MG11C8_004.png)

  3. Stap 3. Bepaal of die funksie negative of positief is in elkeen van die areas :

    Daar is nog 'n manier om die teken van die funksie te bepaal in verskillende areas: deur 'n rowwe skets van die grafiek van die funksie te maak. Ons weet dat die wortels van die funksie ooreenstem met die xx-afsnitte van die grafiek. Laat g(x)=-x2-3x+5g(x)=-x2-3x+5. Ons kan sien dat die die funksie 'n parabool is met 'n maksimum draaipunt en wat die xx-as by x1x1 en x2x2 sny.

    Figuur 5
    Figuur 5 (MG11C8_005.png)

    Dit is duidelik dat g(x)>0g(x)>0 vir x1<x<x2x1<x<x2

  4. Stap 4. Skryf die finale antwoord neer en stel die oplossing grafies voor :

    -x2-3x+5>0-x2-3x+5>0 vir x1<x<x2x1<x<x2

    Figuur 6
    Figuur 6 (MG11C8_006.png)

Wanneer die veranderlike van die ongelykheid in die noemer eerder as die teller is, is 'n ander benadering nodig.

Exercise 4: Nie-lineêre ongelykhede met die veranderlike in die noemer

Los op: 2x+31x-32x+31x-3

Solution

  1. Stap 1. Trek 1x-31x-3 af aan beide kante :
    2 x + 3 - 1 x - 3 0 2 x + 3 - 1 x - 3 0
    (5)
  2. Stap 2. Vereenvoudig die breuk deur die KGD te vind :
    2 ( x - 3 ) - ( x + 3 ) ( x + 3 ) ( x - 3 ) 0 x - 9 ( x + 3 ) ( x - 3 ) 0 2 ( x - 3 ) - ( x + 3 ) ( x + 3 ) ( x - 3 ) 0 x - 9 ( x + 3 ) ( x - 3 ) 0
    (6)
  3. Stap 3. Teken 'n getallelyn vir die ongelykheid :

    Figuur 7
    Figuur 7 (MG11C8_007.png)

    Ons sien dat die uitdrukking negatief is vir x<-3x<-3 en 3<x93<x9.

  4. Stap 4. Skryf die finale antwoord :
    x < - 3 o f 3 < x 9 x < - 3 o f 3 < x 9
    (7)

Hoofstuksoefeninge

Los die volgende ongelykhede op en wys jou antwoord of 'n getallelyn.

  1. Los op: x2-x<12x2-x<12.
  2. Los op: 3x2>-x+43x2>-x+4
  3. Los op: y2<-y-2y2<-y-2
  4. Los op: -t2+2t>-3-t2+2t>-3
  5. Los op: s2-4s>-6s2-4s>-6
  6. Los op: 07x2-x+807x2-x+8
  7. Los op: 0-4x2-x0-4x2-x
  8. Los op: 06x206x2
  9. Los op: 2x2+x+602x2+x+60
  10. Los op: xx-3<2xx-3<2 en x3x3.
  11. Los op: 4x-314x-31.
  12. Los op: 4(x-3)2<14(x-3)2<1.
  13. Los op: 2x-2x-3>32x-2x-3>3
  14. Los op: -3(x-3)(x+1)<0-3(x-3)(x+1)<0
  15. Los op: (2x-3)2<4(2x-3)2<4
  16. Los op: 2x15-xx2x15-xx
  17. Los op: x2+33x-20x2+33x-20
  18. Los op: x-23xx-23x
  19. Los op: x2+3x-45+x40x2+3x-45+x40
  20. Bepaal alle reële oplossings: x-23-x1x-23-x1

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