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Oplos van kwadratiese vergelykings

'n Kwadratiese vergelyking, is 'n vergelyking waar die mag van die veranderlike hoogstens 2 is. Die volgende is voorbeelde van kwadratiese vergelykings.

2 x 2 + 2 x = 1 2 - x 3 x + 1 = 2 x 4 3 x - 6 = 7 x 2 + 2

Kwadratiese vergelykings verskil van lineêre vergelykings daarin dat 'n lineêre vergelyking slegs een oplossing het, terwyl ‘n kwadratiese vergelyking hoogstens 2 oplossings het. Daar is spesiale gevalle waar 'n kwadratiese vergelyking slegs een oplossing het.

Om 'n kwadratiese vergelyking op te los, herskryf ons dit met 'n 0 aan die een kant van die gelykaanteken en die produk van twee lineêre uitdrukkings, in hakies, aan die anderkant. Ons weet byvoorbeeld dat:

( x + 1 ) ( 2 x - 3 ) = 2 x 2 - x - 3

Om op te los:

2 x 2 - x - 3 = 0

moet ons in staat wees om 2 x 2 - x - 3 te herskryf as ( x + 1 ) ( 2 x - 3 ) , en ons weet reeds hoe om dit te doen.

Ondersoek: faktorisering van 'n kwadratiese uitdrukking

Faktoriseer die volgende kwadratiese uitdrukkings:

  1. x + x 2
  2. x 2 + 1 + 2 x
  3. x 2 - 4 x + 5
  4. 16 x 2 - 9
  5. 4 x 2 + 4 x + 1

As jy 'n kwadratiese uitdrukking kan faktoriseer, is jy een stap weg daarvan om 'n kwadratiese vergelyking op te los. Byvoorbeeld, x 2 - 3 x + 2 = 0 kan geskryf word as ( x - 1 ) ( x - 2 ) = 0 . Dit beteken dat x - 1 = 0 of x - 2 = 0 , wat x = 1 en x = 2 gee as die 2 oplossings van die kwadratiese vergelyking x 2 - 3 x + 2 = 0 .

Metode: oplos van kwadratiese vergelykings

  1. Deel heel eerste die hele vergelyking deur enige gemene faktore van die koëffisiënte, ten einde 'n vergelyking te kry van die vorm a x 2 + b x + c = 0 waar a , b en c geen gemeenskaplike faktore het nie. Byvoorbeeld, 2 x 2 + 4 x + 2 = 0 kan geskryf word as x 2 + 2 x + 1 = 0 deur te deel met 2.
  2. Skryf a x 2 + b x + c in terme van sy faktore ( r x + s ) ( u x + v ) . Dit beteken ( r x + s ) ( u x + v ) = 0 .
  3. Wanneer ons die vergelyking geskryf het in die vorm ( r x + s ) ( u x + v ) = 0 , volg dit dat die oplossing sal wees x = - s r of x = - v u .
  4. Vervang elke moontlike waarde van die oplossing in die oorspronklike vergelyking in om te toets of dit 'n geldige oplossing is.

Oplossing (wortels) van kwadratiese vergelykings

'n Kwadratiese vergelyking het 2 wortels omdat enige een van die 2 waardes die vergelyking kan bevredig.

Khan akademie video oor vergelykings - 3

Los op vir x : 3 x 2 + 2 x - 1 = 0 .

  1. Ons het gesien die faktore van 3 x 2 + 2 x - 1 is ( x + 1 ) and ( 3 x - 1 ) .

  2. ( x + 1 ) ( 3 x - 1 ) = 0
  3. Ons het

    x + 1 = 0

    of

    3 x - 1 = 0

    Dus, x = - 1 of x = 1 3 .

  4. As ons die antwoorde instel in die oorspronklike vergelyking in, vind ons die vergelyking is waar vir beide antwoorde.
  5. 3 x 2 + 2 x - 1 = 0 vir x = - 1 of x = 1 3 .

Dit mag gebeur dat die vergelyking met die eerste oogopslag nie soos 'n kwadratiese vergelyking lyk nie, maar deur 'n paar bewerkings in een verander kan word. Onthou dat dieselfde bewerking aan elke kant gedoen moet word om die vergelyking geldig (waar) te hou.

Dit mag nodig wees om een of meer van die volgende te doen:

  • Byvoorbeeld,
    a x + b = c x x ( a x + b ) = x ( c x ) a x 2 + b x = c
  • Dit beteken om beide kante te verhef tot die mag - 1 . Byvoorbeeld,
    1 a x 2 + b x = c ( 1 a x 2 + b x ) - 1 = ( c ) - 1 a x 2 + b x 1 = 1 c a x 2 + b x = 1 c
  • Dit beteken om weerskante te verhef tot die mag 2. Byvoorbeeld,
    a x 2 + b x = c ( a x 2 + b x ) 2 = c 2 a x 2 + b x = c 2

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
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The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
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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
hmm
Abhi
is it a question of log
Abhi
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Commplementary angles
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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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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 10) [caps]. OpenStax CNX. Aug 04, 2011 Download for free at http://cnx.org/content/col11328/1.4
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