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2.2 Die kwadratiese formule

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Oplossing met die kwadratiese formule

Dit is nie altyd moontlik om 'n kwadratiese vergelyking met faktorisering op te los nie en soms is dit langdradig en ingewikkeld om kwadraatsvoltooiing toe te pas. In sulke gevalle kan jy van die kwadratiese formule gebruik maak, 'n metode wat die antwoorde van enige kwadratiese vergelyking oplewer.

Beskou die algemene vorm van 'n kwadratiese funksie:

f ( x ) = a x 2 + b x + c .

Haal die a as gemene faktor uit om te kry:

f ( x ) = a ( x 2 + b a x + c a ) .

Nou moet ons 'n bietjie speurwerk doen om te bepaal hoe om [link] na 'n volmaakte vierkant met 'n paar oorblywende terme te omskep. Ons weet dat om 'n volmaakte vierkant te hê:

( m + n ) 2 = m 2 + 2 m n + n 2

en

( m - n ) 2 = m 2 - 2 m n + n 2

Die sleutel is die middelterm aan die regterkant nl. 2 × die eerste term × die tweede term aan die linkerkant. Met [link] weet ons die eerste term is x en die tweede term is b 2 a . Dit beteken that die tweede term aan die regterkant is 2 b 2 a x . Dus,

( x + b 2 a ) 2 = x 2 + 2 b 2 a x + ( b 2 a ) 2 .

Nou maak ons gebruik van die feit dat as jy dieselfde hoeveelheid bytel en dan aftrek die uitdrukking dieselfde bly. As ons dus b 2 a 2 aan die regterkant van [link] optel en aftrek kry ons:

f ( x ) = a ( x 2 + b a x + c a ) = a x 2 + b a x + b 2 a 2 - b 2 a 2 + c a = a x + b 2 a 2 - b 2 a 2 + c a = a x + b 2 a 2 + c - b 2 4 a

Ons stel f ( x ) = 0 om die wortels te vind, en verkry die volgende:

a ( x + b 2 a ) 2 = b 2 4 a - c

Deel nou deur a en neem die vierkantswortel van beide kante:

x + b 2 a = ± b 2 4 a 2 - c a

Om eindelik vir x op te los impliseer:

x = - b 2 a ± b 2 4 a 2 - c a = - b 2 a ± b 2 - 4 a c 4 a 2

wat verder vereenvoudig kan word tot:

x = - b ± b 2 - 4 a c 2 a

Hierdie is die algemene oplossings vir 'n kwadratiese vergelyking. Let daarop dat daar gewoonlik twee oplossings is, maar dat hul nie noodwendig bestaan nie, afhangende van die uitdrukking b 2 - 4 a c (onder die vierkantswortel) se teken. Hierdie oplossings word ook die wortels van 'n kwadratiese vergelyking genoem.

Vind die wortels van die funksie f ( x ) = 2 x 2 + 3 x - 7 .

  1. Die uitdrukking kan nie gefaktoriseer word nie. Die algemene kwadratiese formule sal dus gebruik moet word.

  2. Vanuit die vergelyking:

    a = 2
    b = 3
    c = - 7

  3. Skryf altyd eers die formule neer en stel daarna die waardes van a , b en c in.

    x = - b ± b 2 - 4 a c 2 a = - ( 3 ) ± ( 3 ) 2 - 4 ( 2 ) ( - 7 ) 2 ( 2 ) = - 3 ± 65 4 = - 3 ± 65 4

  4. Die twee wortels van f ( x ) = 2 x 2 + 3 x - 7 is x = - 3 + 65 4 en - 3 - 65 4 .

Vind die oplossings vir die kwadratiese vergelyking x 2 - 5 x + 8 = 0 .

  1. Die uitdrukking kan nie gefaktoriseer word nie. Die algemene kwadratiese formule sal dus gebruik moet word.

  2. Vanuit die vergelyking:

    a = 1
    b = - 5
    c = 8
  3. x = - b ± b 2 - 4 a c 2 a = - ( - 5 ) ± ( - 5 ) 2 - 4 ( 1 ) ( 8 ) 2 ( 1 ) = 5 ± - 7 2

  4. Aangesien die uitdrukking onder die vierkantswortel negatief is sal hierdie oplossings nie-reëel wees ( - 7 is nie 'n reële getal nie). Daarom is daar geen reële oplossings vir die kwadratiese vergelyking x 2 - 5 x + 8 = 0 nie. Dit beteken dat die grafiek van die funksie f ( x ) = x 2 - 5 x + 8 geen x -afsnitte het nie, maar dat die hele grafiek bokant die x -as lê.

Khan academy video on quadratic equations - 2

Oplossing met die kwadratiese formule

Los op vir t deur gebruik te maak van die kwadratiese formule.

  1. 3 t 2 + t - 4 = 0
  2. t 2 - 5 t + 9 = 0
  3. 2 t 2 + 6 t + 5 = 0
  4. 4 t 2 + 2 t + 2 = 0
  5. - 3 t 2 + 5 t - 8 = 0
  6. - 5 t 2 + 3 t - 3 = 0
  7. t 2 - 4 t + 2 = 0
  8. 9 t 2 - 7 t - 9 = 0
  9. 2 t 2 + 3 t + 2 = 0
  10. t 2 + t + 1 = 0
  • In al die behandelde voorbeelde is die antwoorde in wortelvorm gelaat, alhoewel dit ook in desimale vorm geskryf kan word met behulp van 'n sakrekenaar. In 'n toets of eksamen, let op die vraag se instruksies of die antwoord in wortelvorm of 'n sekere aantal desimale syfers verlang word.
  • Kwadraatsvoltooiing word slegs as oplossingsmetode gebruik wanneer indien spesifiek daarvoor gevra word.

Gemenge oefeninge

Los die kwadratiese vergelykings op deur van faktorisering, kwadraatsvoltooiing of die kwadratiese formule gebruik te maak:

  • Probeer altyd om eers die trinomiaal te faktoriseer en, indien nie moontlik nie, gebruik die formule.
  • Los sommige probleme met behulp van kwadraatsvoltooiing op en vergelyk dan jou antwoorde met díe wat met ander metodes verkry is.
1. 24 y 2 + 61 y - 8 = 0 2. - 8 y 2 - 16 y + 42 = 0 3. - 9 y 2 + 24 y - 12 = 0
4. - 5 y 2 + 0 y + 5 = 0 5. - 3 y 2 + 15 y - 12 = 0 6. 49 y 2 + 0 y - 25 = 0
7. - 12 y 2 + 66 y - 72 = 0 8. - 40 y 2 + 58 y - 12 = 0 9. - 24 y 2 + 37 y + 72 = 0
10. 6 y 2 + 7 y - 24 = 0 11. 2 y 2 - 5 y - 3 = 0 12. - 18 y 2 - 55 y - 25 = 0
13. - 25 y 2 + 25 y - 4 = 0 14. - 32 y 2 + 24 y + 8 = 0 15. 9 y 2 - 13 y - 10 = 0
16. 35 y 2 - 8 y - 3 = 0 17. - 81 y 2 - 99 y - 18 = 0 18. 14 y 2 - 81 y + 81 = 0
19. - 4 y 2 - 41 y - 45 = 0 20. 16 y 2 + 20 y - 36 = 0 21. 42 y 2 + 104 y + 64 = 0
22. 9 y 2 - 76 y + 32 = 0 23. - 54 y 2 + 21 y + 3 = 0 24. 36 y 2 + 44 y + 8 = 0
25. 64 y 2 + 96 y + 36 = 0 26. 12 y 2 - 22 y - 14 = 0 27. 16 y 2 + 0 y - 81 = 0
28. 3 y 2 + 10 y - 48 = 0 29. - 4 y 2 + 8 y - 3 = 0 30. - 5 y 2 - 26 y + 63 = 0
31. x 2 - 70 = 11 32. 2 x 2 - 30 = 2 33. x 2 - 16 = 2 - x 2
34. 2 y 2 - 98 = 0 35. 5 y 2 - 10 = 115 36. 5 y 2 - 5 = 19 - y 2

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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 11). OpenStax CNX. Sep 20, 2011 Download for free at http://cnx.org/content/col11339/1.4
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