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    This module and collection are included inLens: Siyavula: Wiskunde (Gr 10 - 12)
    By: Siyavula

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Inleiding

In graad 10 het ons gekyk na die oplos van lineêre vergelykings, kwadratiese vergelykings, eksponentieële vergelykings en lineêre ongelykhede. Hierdie hoofstuk bo op daardie werk. Ons kyk na verskillende metodes om kwadratiese vergelykings om te los.

Oplos deur Faktorisering

Die oplos van kwadratiese vergelylings deur faktorisering was behandel in Graad 10. Kom ons doen gou 'n voorbeeld om jou geheue te verfris.

Exercise 1: Oplos van Kwadratiese Vergelykings

Losop die vergelyking, 2x2-5x-12=02x2-5x-12=0.

Solution

  1. Stap 1. Bepaal of die vergelyking gemene faktore het :

    Hierdie vergelyking het geen gemene faktore nie.

  2. Stap 2. Bepaal of the vergelyking in die vorm ax2+bx+cax2+bx+c met a>0a>0 :

    Die verglyking is in die voorgeskrewe vorm, met a=2a=2, b=-5b=-5 en c=-12c=-12.

  3. Stap 3. Vaktoriseer die kwadrate :

    2x2-5x-122x2-5x-12 het faktore van die vorm:

    ( 2 x + s ) ( x + v ) ( 2 x + s ) ( x + v )
    (1)

    met ss en vv konstantes wat bepaal moet word. Dit word vermenigvuldig om

    2 x 2 + ( s + 2 v ) x + s v 2 x 2 + ( s + 2 v ) x + s v
    (2)

    te gee. Ons sien dat sv=-12sv=-12 en s+2v=-5s+2v=-5. Hierdie is 'n stel gelyktydige vergelykings in ss en vv, dit is maklik om numeries op te los. Al die opsies vir ss en vv word hieronder oorweeg.

    Tabel 1
    s s v v s + 2 v s + 2 v
    2 -6 -10
    -2 6 10
    3 -4 -5
    -3 4 5
    4 -3 -2
    -4 3 2
    6 -2 2
    -6 2 -2

    Ons sien dat die kombinasie van s=3s=3 en v=-4v=-4 gee vir ons s+2v=-5s+2v=-5.

  4. Stap 4. Skryf die vergelyking met die faktore :
    ( 2 x + 3 ) ( x - 4 ) = 0 ( 2 x + 3 ) ( x - 4 ) = 0
    (3)
  5. Stap 5. Los op die vergelyking :

    Indien twee hakkies vermenigvuldig word en 0 gee, moet een van die hakkies 0 wees, dus

    2 x + 3 = 0 2 x + 3 = 0
    (4)

    of

    x - 4 = 0 x - 4 = 0
    (5)

    Dus, x=-32x=-32 of x=4x=4

  6. Stap 6. Die finale antwoord is :

    Die oplossing van 2x2-5x-12=02x2-5x-12=0 is x=-32x=-32 of x=4x=4.

Dit is belangrik om te onthou dat 'n kwadratiese vergelyking in die vorm ax2+bx+c=0ax2+bx+c=0 moet wees voor ons dit kan op los met die metodes.

Exercise 2: Los op die kwadratiese vergelyking met faktorisering

Los op aa: a(a-3)=10a(a-3)=10

Solution

  1. Stap 1. Herskryf die vergelyking in die vorm ax2+bx+c=0ax2+bx+c=0 :

    Verwyder die hakkies en kry al die terme aan een kant van die gelykaanteken.

    a 2 - 3 a - 10 = 0 a 2 - 3 a - 10 = 0
    (6)
  2. Stap 2. Faktoriseer die drieterm :
    ( a + 2 ) ( a - 5 ) = 0 ( a + 2 ) ( a - 5 ) = 0
    (7)
  3. Stap 3. Los op die vergelyking :
    a + 2 = 0 a + 2 = 0
    (8)

    of

    a - 5 = 0 a - 5 = 0
    (9)

    Los die twee lineêre vergelykings op en kontroleer die antwoorde in die oorspronklike vergelyking.

  4. Stap 4. Skryf die finale antwoord neer :

    Dus, a=-2a=-2 of a=5a=5

Exercise 3: Op los van breuke wat lei na 'n kwadratiese vergelyking

Los op bb: 3bb+2+1=4b+13bb+2+1=4b+1

Solution

  1. Stap 1. Deel beide kante met die KGV :
    3 b ( b + 1 ) + ( b + 2 ) ( b + 1 ) ( b + 2 ) ( b + 1 ) = 4 ( b + 2 ) ( b + 2 ) ( b + 1 ) 3 b ( b + 1 ) + ( b + 2 ) ( b + 1 ) ( b + 2 ) ( b + 1 ) = 4 ( b + 2 ) ( b + 2 ) ( b + 1 )
    (10)
  2. Stap 2. Bepaal die beperkings :

    Die noemers is dieselfde, daarom moet die tellers ook dieselfde wees.

    Maar b-2b-2 en b-1b-1

  3. Stap 3. Vereenvoudig die vergelyking na die standaard vorm :
    3 b 2 + 3 b + b 2 + 3 b + 2 = 4 b + 8 4 b 2 + 2 b - 6 = 0 2 b 2 + b - 3 = 0 3 b 2 + 3 b + b 2 + 3 b + 2 = 4 b + 8 4 b 2 + 2 b - 6 = 0 2 b 2 + b - 3 = 0
    (11)
  4. Stap 4. Vaktoriseer die drieterm en los op die vergelyking :
    ( 2 b + 3 ) ( b - 1 ) = 0 2 b + 3 = 0 o f b - 1 = 0 b = - 3 2 o f b = 1 ( 2 b + 3 ) ( b - 1 ) = 0 2 b + 3 = 0 o f b - 1 = 0 b = - 3 2 o f b = 1
    (12)
  5. Stap 5. Kontroleer die oplossing in die oorspronklike vergelyking :

    Albei oplossing is geldig

    Dus, b=-32b=-32 of b=1b=1

Oplossing deur Faktorisering

Los op die volgende kwadratiese vergelykings op deur faktorisering. Sommige van die antwoorde kan gelos word in die wortel vorm.

  1. 2 y 2 - 61 = 101 2 y 2 - 61 = 101
  2. 2 y 2 - 10 = 0 2 y 2 - 10 = 0
  3. y 2 - 4 = 10 y 2 - 4 = 10
  4. 2 y 2 - 8 = 28 2 y 2 - 8 = 28
  5. 7 y 2 = 28 7 y 2 = 28
  6. y 2 + 28 = 100 y 2 + 28 = 100
  7. 7 y 2 + 14 y = 0 7 y 2 + 14 y = 0
  8. 12 y 2 + 24 y + 12 = 0 12 y 2 + 24 y + 12 = 0
  9. 16 y 2 - 400 = 0 16 y 2 - 400 = 0
  10. y 2 - 5 y + 6 = 0 y 2 - 5 y + 6 = 0
  11. y 2 + 5 y - 36 = 0 y 2 + 5 y - 36 = 0
  12. y 2 + 2 y = 8 y 2 + 2 y = 8
  13. - y 2 - 11 y - 24 = 0 - y 2 - 11 y - 24 = 0
  14. 13 y - 42 = y 2 13 y - 42 = y 2
  15. y 2 + 9 y + 14 = 0 y 2 + 9 y + 14 = 0
  16. y 2 - 5 k y + 4 k 2 = 0 y 2 - 5 k y + 4 k 2 = 0
  17. y ( 2 y + 1 ) = 15 y ( 2 y + 1 ) = 15
  18. 5 y y - 2 + 3 y + 2 = - 6 y 2 - 2 y 5 y y - 2 + 3 y + 2 = - 6 y 2 - 2 y
  19. y - 2 y + 1 = 2 y + 1 y - 7 y - 2 y + 1 = 2 y + 1 y - 7

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