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Kwadraatsvoltooiing

Ons het gesien dat die vergelyking in die vorm:

a 2 x 2 - b 2

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

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

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 x op te los in:

x 2 - 2 x - 1 = 0 .

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 .

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

Ons weet dat:

2 = ( 2 ) 2

wat beteken dat:

( x - 1 ) 2 - 2

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

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

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

( x - 1 ) - 2 = 0

of

( x - 1 ) + 2 = 0 .

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

Metode: los kwadratiese vergelykings op deur kwadraatsvoltooing

  1. Skryf die vergelyking in die vorm a x 2 + b x + c = 0 . bv. x 2 + 2 x - 3 = 0
  2. Neem die konstante oor na die regterkant van die vergelyking. Bv. x 2 + 2 x = 3
  3. Indien nodig stel die koëffisiënt van x 2 = 1, deur te deel deur die bestaande koëffisiënt.
  4. Neem die helfte van die koëffisiënt van die x -term, kwadreer dit en voeg dit aan beide kante van die vergelyking. Bv. in x 2 + 2 x = 3 , die helfte van die koëffisiënt van die x -term is 1 en 1 2 = 1 . Daarom voeg ons 1 aan albei kante by om x 2 + 2 x + 1 = 3 + 1 te kry.
  5. Skryf die linkerkant as 'n volkome vierkant: ( 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 x op te los: ( x + 1 - 2 ) ( x + 1 + 2 ) = 0

Los op:

x 2 - 10 x - 11 = 0

deur kwadraatsvoltooiing.

  1. x 2 - 10 x - 11 = 0
  2. x 2 - 10 x = 11
  3. Die koëffisiënt van die term x 2 is 1.

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

    x 2 - 10 x + 25 = 11 + 25
  5. ( x - 5 ) 2 - 36 = 0
  6. ( x - 5 ) 2 - 36 = 0
    [ ( x - 5 ) + 6 ] [ ( x - 5 ) - 6 ] = 0
  7. [ x + 1 ] [ x - 11 ] = 0 x = - 1 of x = 11

Los op:

2 x 2 - 8 x - 16 = 0

deur kwadraatsvoltooiing.

  1. 2 x 2 - 8 x - 16 = 0
  2. 2 x 2 - 8 x = 16
  3. Die koëffisiënt van die term x 2 is 2. Deel dus beide kante deur 2:

    x 2 - 4 x = 8
  4. Die koëffisiënt van die term x is -4; ( - 4 ) 2 = - 2 en ( - 2 ) 2 = 4 . Dus:

    x 2 - 4 x + 4 = 8 + 4
  5. ( x - 2 ) 2 - 12 = 0
  6. [ ( x - 2 ) + 12 ] [ ( x - 2 ) - 12 ] = 0
  7. [ x - 2 + 12 ] [ x - 2 - 12 ] = 0 x = 2 - 12 or x = 2 + 12
  8. Laat die linkerkant as 'n volkome vierkant geskryf

    ( x - 2 ) 2 = 12
  9. x - 2 = ± 12
  10. Dus x = 2 - 12    of     x = 2 + 12

    Vergelyk met antwoord in stap 7.

Khan academy video on solving quadratics - 1

Kwadraatsvoltooiing oefeninge

Los die volgende vergelykings op deur kwadraatsvoltooiing:

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

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Kristine 2*2*2=8
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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J, combine like terms 7x-4y
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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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