5.1 Oplos van kwadratiese vergelykings (Page 2/2)

Siyavula textbooks: wiskunde Page 2 / 2

Hierdie strategieë kan op verskillende wyses gekombineer word en die sekerste manier om jou intuïsie te ontwikkel oor wat die beste ding is om te doen, is om te oefen. 'n Gekombineerde stel bewerkings kan byvoorbeeld wees:

\begin{matrix} \frac{1}{\sqrt{a x^{2} + b x}} & = & c \\ {(\frac{1}{a x^{2} + b x})}^{- 1} & = & {(c)}^{- 1} & (kry weerskante die resiprook) \\ \frac{\sqrt{a x^{2} + b x}}{1} & = & \frac{1}{c} \\ \sqrt{a x^{2} + b x} & = & \frac{1}{c} \\ {(\sqrt{a x^{2} + b x})}^{2} & = & {(\frac{1}{c})}^{2} & (kwadreer weerskante) \\ a x^{2} + b x & = & \frac{1}{c^{2}} \end{matrix}

Los op vir $x$ : $\sqrt{x + 2} = x$ .

Kwadreer beide kante van die vergelyking
Beide kante van die vergelyking behoort gekwadreer te word om die vierkantswortelteken te verwyder.

$x + 2 = x^{2}$
Skryf die vergelyking in die vorm $a x^{2} + b x + c = 0$
$\begin{matrix} x + 2 & = & x^{2} & (trek x^{2} af van beide kante) \\ x + 2 - x^{2} & = & 0 & (deel weerskante deur - 1) \\ - x - 2 + x^{2} & = & 0 \\ x^{2} - x - 2 & = & 0 \end{matrix}$
Faktoriseer die kwadratiese drieterm
$x^{2} - x - 2$

Die faktore van $x^{2} - x - 2$ is $(x - 2) (x + 1)$ .
Skryf die vergelyking met die faktore
$(x - 2) (x + 1) = 0$
Bepaal die 2 oplossings
Ons het

$x + 1 = 0$

of

$x - 2 = 0$

Dus, $x = - 1$ of $x = 2$ .
Kontroleer of die oplossings geldig is
Vervang $x = - 1$ in die oorspronklike vergelyking $\sqrt{x + 2} = x$ :

$\begin{matrix} LK & = & \sqrt{(- 1) + 2} \\ = & \sqrt{1} \\ = & 1 \\ maar \\ RK & = & (- 1) \end{matrix}$

Daarom LK $\neq$ RK. Die twee kante van 'n vergelyking moet altyd balanseer; 'n moontlike oplossing wat NIE die vergelyking bevredig nie, is nie geldig nie. In hierdie geval balanseer die twee kante van die vergelyking nie.

Dus $x \neq - 1$ .

Stel nou $x = 2$ in die oorspronklike vergelyking in $\sqrt{x + 2} = x$ :

$\begin{matrix} LK & = & \sqrt{2 + 2} \\ = & \sqrt{4} \\ = & 2 \\ en \\ RK & = & 2 \end{matrix}$

Dus, LK = RK

Dus $x = 2$ is die enigste geldige oplossing.
Skryf die finale antwoord neer
$\sqrt{x + 2} = x$ vir $x = 2$ alleenlik.

Los die vergelyking op: $x^{2} + 3 x - 4 = 0$ .

Kontroleer of die vergelyking in die vorm $a x^{2} + b x + c = 0$
Die vergelyking is in die verlangde vorm, met $a = 1$ .
Faktoriseer die kwadratiese drieterm
Jy benodig die faktore van 1 en 4 sodat die middelterm $+ 3$ is. So, die faktore is:

$(x - 1) (x + 4)$
Los die kwadratiese vergelyking op
$x^{2} + 3 x - 4 = (x - 1) (x + 4) = 0$

Dus $x = 1$ of $x = - 4$ .
Toets die oplossings
$1^{2} + 3 (1) - 4 = 0$

${(- 4)}^{2} + 3 (- 4) - 4 = 0$

Beide oplossings is geldig.
Gee die finale oplossing
Dus, die oplossing is $x = 1$ of $x = - 4$ .

Vind die wortels van die kwadratiese vergelyking $0 = - 2 x^{2} + 4 x - 2$ .

Bepaal of die vergelyking in die vorm $a x^{2} + b x + c = 0$ is met geen gemeenskaplike faktore
Daar is 'n gemeenskaplike faktor: -2. Dus, deel weerskante van die vergelyking deur -2.

$\begin{matrix} - 2 x^{2} + 4 x - 2 & = & 0 \\ x^{2} - 2 x + 1 & = & 0 \end{matrix}$
Faktoriseer $x^{2} - 2 x + 1$
Die middelterm is negatief. Dus, die faktore is $(x - 1) (x - 1)$ .

As ons uitvermenigvuldig $(x - 1) (x - 1)$ , kry ons $x^{2} - 2 x + 1$ .
Los die kwadratiese vergelyking op
$x^{2} - 2 x + 1 = (x - 1) (x - 1) = 0$

In hierdie geval is die kwadratiese uitdrukking 'n volkome vierkant, so daar is net een oplossing vir $x$ : $x = 1$ .
Toets die oplossing
$- 2 {(1)}^{2} + 4 (1) - 2 = 0$
Skryf die finale oplossing neer
Die wortel van $0 = - 2 x^{2} + 4 x - 2$ is $x = 1$ .

Oplos van kwadratiese vergelykings

Los op vir $x$ : $(3 x + 2) (3 x - 4) = 0$
Los op vir $x$ : $(5 x - 9) (x + 6) = 0$
Los op vir $x$ : $(2 x + 3) (2 x - 3) = 0$
Los op vir $x$ : $(2 x + 1) (2 x - 9) = 0$
Los op vir $x$ : $(2 x - 3) (2 x - 3) = 0$
Los op vir $x$ : $20 x + 25 x^{2} = 0$
Los op vir $x$ : $4 x^{2} - 17 x - 77 = 0$
Los op vir $x$ : $2 x^{2} - 5 x - 12 = 0$
Los op vir $x$ : $- 75 x^{2} + 290 x - 240 = 0$
Los op vir $x$ : $2 x = \frac{1}{3} x^{2} - 3 x + 14 \frac{2}{3}$
Los op vir $x$ : $x^{2} - 4 x = - 4$
Los op vir $x$ : $- x^{2} + 4 x - 6 = 4 x^{2} - 5 x + 3$
Los op vir $x$ : $x^{2} = 3 x$
Los op vir $x$ : $3 x^{2} + 10 x - 25 = 0$
Los op vir $x$ : $x^{2} - x + 3$
Los op vir $x$ : $x^{2} - 4 x + 4 = 0$
Los op vir $x$ : $x^{2} - 6 x = 7$
Los op vir $x$ : $14 x^{2} + 5 x = 6$
Los op vir $x$ : $2 x^{2} - 2 x = 12$
Los op vir $x$ : $3 x^{2} + 2 y - 6 = x^{2} - x + 2$

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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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