Faktorisering

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Inleiding

In graad 10 het ons gekyk na die oplos van lineêre vergelykings, kwadratiese vergelykings, eksponentieële vergelykings en lineêre ongelykhede. Hierdie hoofstuk bo op daardie werk. Ons kyk na verskillende metodes om kwadratiese vergelykings om te los.

Oplos deur faktorisering

Die oplos van kwadratiese vergelylings deur faktorisering was behandel in Graad 10. Kom ons doen gou 'n voorbeeld om jou geheue te verfris.

Losop die vergelyking, $2 x^{2} - 5 x - 12 = 0$ .

Bepaal of die vergelyking gemene faktore het
Hierdie vergelyking het geen gemene faktore nie.
Bepaal of the vergelyking in die vorm $a x^{2} + b x + c$ met $a > 0$
Die verglyking is in die voorgeskrewe vorm, met $a = 2$ , $b = - 5$ en $c = - 12$ .
Vaktoriseer die kwadrate
$2 x^{2} - 5 x - 12$ het faktore van die vorm:

$(2 x + s) (x + v)$

met $s$ en $v$ konstantes wat bepaal moet word. Dit word vermenigvuldig om

$2 x^{2} + (s + 2 v) x + s v$

te gee. Ons sien dat $s v = - 12$ en $s + 2 v = - 5$ . Hierdie is 'n stel gelyktydige vergelykings in $s$ en $v$ , dit is maklik om numeries op te los. Al die opsies vir $s$ en $v$ word hieronder oorweeg.

$s$ $v$ $s + 2 v$

2 -6 -10

-2 6 10

3 -4 -5

-3 4 5

4 -3 -2

-4 3 2

6 -2 2

-6 2 -2

Ons sien dat die kombinasie van $s = 3$ en $v = - 4$ gee vir ons $s + 2 v = - 5$ .
Skryf die vergelyking met die faktore
$(2 x + 3) (x - 4) = 0$
Los op die vergelyking
Indien twee hakkies vermenigvuldig word en 0 gee, moet een van die hakkies 0 wees, dus

$2 x + 3 = 0$

of

$x - 4 = 0$

Dus, $x = - \frac{3}{2}$ of $x = 4$
Die finale antwoord is
Die oplossing van $2 x^{2} - 5 x - 12 = 0$ is $x = - \frac{3}{2}$ of $x = 4$ .

Dit is belangrik om te onthou dat 'n kwadratiese vergelyking in die vorm $a x^{2} + b x + c = 0$ moet wees voor ons dit kan op los met die metodes.

Los op $a$ : $a (a - 3) = 10$

Herskryf die vergelyking in die vorm $a x^{2} + b x + c = 0$
Verwyder die hakkies en kry al die terme aan een kant van die gelykaanteken.

$a^{2} - 3 a - 10 = 0$
Faktoriseer die drieterm
$(a + 2) (a - 5) = 0$
Los op die vergelyking
$a + 2 = 0$

of

$a - 5 = 0$

Los die twee lineêre vergelykings op en kontroleer die antwoorde in die oorspronklike vergelyking.
Skryf die finale antwoord neer
Dus, $a = - 2$ of $a = 5$

Los op $b$ : $\frac{3 b}{b + 2} + 1 = \frac{4}{b + 1}$

Deel beide kante met die KGV
$\frac{3 b (b + 1) + (b + 2) (b + 1)}{(b + 2) (b + 1)} = \frac{4 (b + 2)}{(b + 2) (b + 1)}$
Bepaal die beperkings
Die noemers is dieselfde, daarom moet die tellers ook dieselfde wees.

Maar $b \neq - 2$ en $b \neq - 1$
Vereenvoudig die vergelyking na die standaard vorm
$\begin{matrix} 3 b^{2} + 3 b + b^{2} + 3 b + 2 & = & 4 b + 8 \\ 4 b^{2} + 2 b - 6 & = & 0 \\ 2 b^{2} + b - 3 & = & 0 \end{matrix}$
Vaktoriseer die drieterm en los op die vergelyking
$\begin{matrix} (2 b + 3) (b - 1) & = & 0 \\ 2 b + 3 = 0 & o f & b - 1 = 0 \\ b = \frac{- 3}{2} & o f & b = 1 \end{matrix}$
Kontroleer die oplossing in die oorspronklike vergelyking
Albei oplossing is geldig

Dus, $b = \frac{- 3}{2}$ of $b = 1$

Oplossing deur faktorisering

Los op die volgende kwadratiese vergelykings op deur faktorisering. Sommige van die antwoorde kan gelos word in die wortel vorm.

$2 y^{2} - 61 = 101$
$2 y^{2} - 10 = 0$
$y^{2} - 4 = 10$
$2 y^{2} - 8 = 28$
$7 y^{2} = 28$
$y^{2} + 28 = 100$
$7 y^{2} + 14 y = 0$
$12 y^{2} + 24 y + 12 = 0$
$16 y^{2} - 400 = 0$
$y^{2} - 5 y + 6 = 0$
$y^{2} + 5 y - 36 = 0$
$y^{2} + 2 y = 8$
$- y^{2} - 11 y - 24 = 0$
$13 y - 42 = y^{2}$
$y^{2} + 9 y + 14 = 0$
$y^{2} - 5 k y + 4 k^{2} = 0$
$y (2 y + 1) = 15$
$\frac{5 y}{y - 2} + \frac{3}{y} + 2 = \frac{- 6}{y^{2} - 2 y}$
$\frac{y - 2}{y + 1} = \frac{2 y + 1}{y - 7}$

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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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$s$	$v$	$s + 2 v$
2	-6	-10
-2	6	10
3	-4	-5
-3	4	5
4	-3	-2
-4	3	2
6	-2	2
-6	2	-2