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Ons het gesien dat die vergelyking in die vorm:

a 2 x 2 - b 2 a 2 x 2 - b 2
(1)

bekend is as die verskil in vierkante en kan as volg gefaktoriseer word:

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

Hierdie eenvoudige faktorisering lei na 'n ander tegniek om kwadratiese vergelykings op te los wat bekend staan ​​as kwadraatsvoltooiing.

Ons wys met 'n eenvoudige voorbeeld, deur te probeer om vir xx op te los in:

x 2 - 2 x - 1 = 0 . x 2 - 2 x - 1 = 0 .
(3)

Ons kan nie maklik faktore van hierdie term vind nie, maar die eerste twee terme lyk soortgelyk aan die eerste twee terme van die volmaakte vierkant:

( x - 1 ) 2 = x 2 - 2 x + 1 . ( x - 1 ) 2 = x 2 - 2 x + 1 .
(4)

Ons kan egter kul en 'n volmaakte vierkant skep deur 2 aan beide kante van die vergelyking te voeg.

x 2 - 2 x - 1 = 0 x 2 - 2 x - 1 + 2 = 0 + 2 x 2 - 2 x + 1 = 2 ( x - 1 ) 2 = 2 ( x - 1 ) 2 - 2 = 0 x 2 - 2 x - 1 = 0 x 2 - 2 x - 1 + 2 = 0 + 2 x 2 - 2 x + 1 = 2 ( x - 1 ) 2 = 2 ( x - 1 ) 2 - 2 = 0
(5)

Ons weet dat:

2 = ( 2 ) 2 2 = ( 2 ) 2
(6)

wat beteken dat:

( x - 1 ) 2 - 2 ( x - 1 ) 2 - 2
(7)

is 'n verskil van vierkante. Ons kan dus skryf:

( x - 1 ) 2 - 2 = [ ( x - 1 ) - 2 ] [ ( x - 1 ) + 2 ] = 0 . ( x - 1 ) 2 - 2 = [ ( x - 1 ) - 2 ] [ ( x - 1 ) + 2 ] = 0 .
(8)

Die oplossing vir x2-2x-1=0x2-2x-1=0 is dus:

( x - 1 ) - 2 = 0 ( x - 1 ) - 2 = 0
(9)

of

( x - 1 ) + 2 = 0 . ( x - 1 ) + 2 = 0 .
(10)

Dit beteken x=1+2x=1+2 of x=1-2x=1-2. Hierdie voorbeeld toon die gebruik van kwadraatsvoltooiing om 'n kwadratiese vergelyking op te los.

1. Skryf die vergelyking in die vorm ax2+bx+c=0ax2+bx+c=0. bv. x2+2x-3=0x2+2x-3=0
2. Neem die konstante oor na die regterkant van die vergelyking. Bv. x2+2x=3x2+2x=3
3. Indien nodig stel die koëffisiënt van x2x2 = 1, deur te deel deur die bestaande koëffisiënt.
4. Neem die helfte van die koëffisiënt van die xx-term, kwadreer dit en voeg dit aan beide kante van die vergelyking. Bv. in x2+2x=3x2+2x=3, die helfte van die koëffisiënt van die xx-term is 1 en 12=112=1. Daarom voeg ons 1 aan albei kante by om x2+2x+1=3+1x2+2x+1=3+1 te kry.
5. Skryf die linkerkant as 'n volkome vierkant: (x+1)2-4=0(x+1)2-4=0
6. Jy moet dan in staat wees om die vergelyking in terme van die verskil in vierkante te faktoriseer en dan vir xx op te los: (x+1-2)(x+1+2)=0(x+1-2)(x+1+2)=0

Los op:

x 2 - 10 x - 11 = 0 x 2 - 10 x - 11 = 0
(11)

##### Solution
1. Stap 1. Skryf die vergelyking in die vorm ax2+bx+c=0ax2+bx+c=0 :
x 2 - 10 x - 11 = 0 x 2 - 10 x - 11 = 0
(12)
2. Stap 2. Neem die konstante oor na die regterkant van die vergelyking:
x 2 - 10 x = 11 x 2 - 10 x = 11
(13)
3. Stap 3. Kyk dat die koëffisiënt van die term x2x2 1 is:

Die koëffisiënt van die term x2x2 is 1.

4. Stap 4. Neem die helfte van die koëffisiënt van die xx-term, kwadreer dit en voeg dit aan beide kante van die vergelyking:

Die koëffisiënt van die term xx is -10. Helfte van die koëffisiënt van die term xx sal wees (-10)2=-5(-10)2=-5 en die kwadraat sal wees (-5)2=25(-5)2=25. Dus:

x 2 - 10 x + 25 = 11 + 25 x 2 - 10 x + 25 = 11 + 25
(14)
5. Stap 5. Skryf die linkerkant as 'n volkome vierkant:
( x - 5 ) 2 - 36 = 0 ( x - 5 ) 2 - 36 = 0
(15)
6. Stap 6. Faktoriseer die vergelyking as die verskil in vierkante:
( x - 5 ) 2 - 36 = 0 ( x - 5 ) 2 - 36 = 0
(16)
[ ( x - 5 ) + 6 ] [ ( x - 5 ) - 6 ] = 0 [ ( x - 5 ) + 6 ] [ ( x - 5 ) - 6 ] = 0
(17)
7. Stap 7. Los die onbekende waarde op:
[ x + 1 ] [ x - 11 ] = 0 x = - 1 of x = 11 [ x + 1 ] [ x - 11 ] = 0 x = - 1 of x = 11
(18)

Los op:

2 x 2 - 8 x - 16 = 0 2 x 2 - 8 x - 16 = 0
(19)

##### Solution
1. Stap 1. Skryf die vergelyking in die vorm ax2+bx+c=0ax2+bx+c=0 :
2 x 2 - 8 x - 16 = 0 2 x 2 - 8 x - 16 = 0
(20)
2. Stap 2. Neem die konstante oor na die regterkant van die vergelyking:
2 x 2 - 8 x = 16 2 x 2 - 8 x = 16
(21)
3. Stap 3. Kyk dat die koëffisiënt van die term x2x2 1 is. :

Die koëffisiënt van die term x2x2 is 2. Deel dus beide kante deur 2:

x 2 - 4 x = 8 x 2 - 4 x = 8
(22)
4. Stap 4. Neem die helfte van die koëffisiënt van die xx-term, kwadreer dit en voeg dit aan beide kante van die vergelyking :

Die koëffisiënt van die term xx is -4; (-4)2=-2(-4)2=-2 en (-2)2=4(-2)2=4. Dus:

x 2 - 4 x + 4 = 8 + 4 x 2 - 4 x + 4 = 8 + 4
(23)
5. Stap 5. Skryf die linkerkant as 'n volkome vierkant:
( x - 2 ) 2 - 12 = 0 ( x - 2 ) 2 - 12 = 0
(24)
6. Stap 6. Faktoriseer die vergelyking as die verskil in vierkante:
[ ( x - 2 ) + 12 ] [ ( x - 2 ) - 12 ] = 0 [ ( x - 2 ) + 12 ] [ ( x - 2 ) - 12 ] = 0
(25)
7. Stap 7. Los die onbekende waarde op:
[ x - 2 + 12 ] [ x - 2 - 12 ] = 0 x = 2 - 12 or x = 2 + 12 [ x - 2 + 12 ] [ x - 2 - 12 ] = 0 x = 2 - 12 or x = 2 + 12
(26)
8. Stap 8. Die laaste drie stappe kan ook op 'n ander manier gedoen word:

Laat die linkerkant as 'n volkome vierkant geskryf

( x - 2 ) 2 = 12 ( x - 2 ) 2 = 12
(27)
9. Stap 9. Kry die vierkantswortel aan beide kante van die vergelyking:
x - 2 = ± 12 x - 2 = ± 12
(28)
10. Stap 10. Los op vir xx :

Dus x=2-12x=2-12    of    x=2+12x=2+12

Vergelyk met antwoord in stap 7.

Figuur 1

Los die volgende vergelykings op deur kwadraatsvoltooiing:

1. x 2 + 10 x - 2 = 0 x 2 + 10 x - 2 = 0
2. x 2 + 4 x + 3 = 0 x 2 + 4 x + 3 = 0
3. x 2 + 8 x - 5 = 0 x 2 + 8 x - 5 = 0
4. 2 x 2 + 12 x + 4 = 0 2 x 2 + 12 x + 4 = 0
5. x 2 + 5 x + 9 = 0 x 2 + 5 x + 9 = 0
6. x 2 + 16 x + 10 = 0 x 2 + 16 x + 10 = 0
7. 3 x 2 + 6 x - 2 = 0 3 x 2 + 6 x - 2 = 0
8. z 2 + 8 z - 6 = 0 z 2 + 8 z - 6 = 0
9. 2 z 2 - 11 z = 0 2 z 2 - 11 z = 0
10. 5 + 4 z - z 2 = 0 5 + 4 z - z 2 = 0

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