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By: Siyavula

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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)

Los op:

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

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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Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

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