0.1 Wortelvorme

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Berekening van wortelvorme

Daar is verskeie reëls wat dit makliker maak om met wortelvorme te werk. Ons sal elkeen lys en dan in diepte verduidelik waar die reël vandaan kom.

\begin{matrix} \sqrt[n]{a} \sqrt[n]{b} & = & \sqrt[n]{a b} \\ \sqrt[n]{\frac{a}{b}} & = & \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \\ \sqrt[n]{a^{m}} & = & a^{\frac{m}{n}} \end{matrix}

Wortelwet 1: $\sqrt[n]{a} \sqrt[n]{b} = \sqrt[n]{a b}$

Dit is gewoonlik nuttig om 'n wortelvorm in eksponensiaalnotasie te beskou, aangesien dit ons toelaat om die eksponentwette, wat ons in Graad 10 geleer het, te gebruik. In eksponensiaalnotasie, $\sqrt[n]{a} = a^{\frac{1}{n}}$ en $\sqrt[n]{b} = b^{\frac{1}{n}}$ . Vervolgens,

\begin{matrix} \sqrt[n]{a} \sqrt[n]{b} & = & a^{\frac{1}{n}} b^{\frac{1}{n}} \\ = & {(a b)}^{\frac{1}{n}} \\ = & \sqrt[n]{a b} \end{matrix}

'n Paar voorbeelde wat hierdie wet gebruik:

$\sqrt[3]{16} \times \sqrt[3]{4} = \sqrt[3]{64} = 4$
$\sqrt{2} \times \sqrt{32} = \sqrt{64} = 8$
$\sqrt{a^{2} b^{3}} \times \sqrt{b^{5} c^{4}} = \sqrt{a^{2} b^{8} c^{4}} = a b^{4} c^{2}$

Wortelwet 2: $\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

Indien ons na $\sqrt[n]{\frac{a}{b}}$ in eksponensieelnotasie kyk en die eksponentwette toepas, dan is

\begin{matrix} \sqrt[n]{\frac{a}{b}} & = & {(\frac{a}{b})}^{\frac{1}{n}} \\ = & \frac{a^{\frac{1}{n}}}{b^{\frac{1}{n}}} \\ = & \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \end{matrix}

'n Paar voorbeelde wat hierdie wet gebruik:

$\sqrt{12} \div \sqrt{3} = \sqrt{4} = 2$
$\sqrt[3]{24} \div \sqrt[3]{3} = \sqrt[3]{8} = 2$
$\sqrt{a^{2} b^{13}} \div \sqrt{b^{5}} = \sqrt{a^{2} b^{8}} = a b^{4}$

Wortelwet 3: $\sqrt[n]{a^{m}} = a^{\frac{m}{n}}$

Indien ons na $\sqrt[n]{a^{m}}$ in eksponensiaalnotasie kyk en die eksponentwette toepas, dan is

\begin{matrix} \sqrt[n]{a^{m}} & = & {(a^{m})}^{\frac{1}{n}} \\ = & a^{\frac{m}{n}} \end{matrix}

Byvoorbeeld

\begin{matrix} \sqrt[6]{2^{3}} & = & 2^{\frac{3}{6}} \\ = & 2^{\frac{1}{2}} \\ = & \sqrt{2} \end{matrix}

Gelyksoortige en ongelyksoortige wortelvorme

Twee wortelvorme $\sqrt[m]{a}$ en $\sqrt[n]{b}$ word gelyksoortige wortelvorme genoem indien $m = n$ , andersins word hulle ongelyksoortige wortelvorme genoem. Byvoorbeeld, $\sqrt{2}$ en $\sqrt{3}$ is gelyksoortig, terwyl $\sqrt{2}$ en $\sqrt[3]{2}$ ongelyksoortig is. Dit is belangrik om op te let dat die wortelwette wat ons pas geleer het almal gelyksoortige wortelvorme is.

Indien ons die wortelwette op ongelyksoortige wortelvorme wil toepas, moet ons hulle eers omskakel na gelyksoortige wortelvorme. Om hierdie reg te kry, gebruik ons die formule

\sqrt[n]{a^{m}} = \sqrt[b n]{a^{b m}}

om die ongelyksoortige wortelvorme oor te skryf sodat $b n$ dieselfde is vir alle wortelvorme.

Vereenvoudig die gelyksoortige wortelvorme so ver moontlik, en wys alle stappe: $\sqrt[3]{3} \times \sqrt[5]{5}$

Vind die gemene wortel
$\begin{matrix} = & \sqrt[15]{3^{5}} \times \sqrt[15]{5^{3}} \end{matrix}$
Gebruik wortelwet 1
$\begin{matrix} = & \sqrt[15]{3^{5} . 5^{3}} \\ = & \sqrt[15]{243 \times 125} \\ = & \sqrt[15]{30375} \end{matrix}$

Eenvoudigste wortelvorm

Wanneer daar gewerk word met wortelvorme, word antwoorde meestal in die eenvoudigste wortelvorm gegee. Byvoorbeeld

\begin{matrix} \sqrt{50} & = & \sqrt{25 \times 2} \\ = & \sqrt{25} \times \sqrt{2} \\ = & 5 \sqrt{2} \end{matrix}

$5 \sqrt{2}$ is die eenvoudigste wortelvorm van $\sqrt{50}$ .

Herskryf $\sqrt{18}$ in die eenvoudigste wortelvorm:

Breek die getal 18 op in sy kleinste faktore
$\begin{matrix} \sqrt{18} & = & \sqrt{2 \times 9} \\ = & \sqrt{2 \times 3 \times 3} \\ = & \sqrt{2} \times \sqrt{3 \times 3} \\ = & \sqrt{2} \times \sqrt{3^{2}} \\ = & 3 \sqrt{2} \end{matrix}$

Vereenvoudig: $\sqrt{147} + \sqrt{108}$

Vereenvoudig elke vierkantswortel afsonderlik
$\begin{matrix} \sqrt{147} + \sqrt{108} & = & \sqrt{49 \times 3} + \sqrt{36 \times 3} \\ = & \sqrt{7^{2} \times 3} + \sqrt{6^{2} \times 3} \end{matrix}$
Neem die waardes wat $^{2}$ het onder die wortelvorm na die buitekant van die vierkantswortelteken
$\begin{matrix} = & 7 \sqrt{3} + 6 \sqrt{3} \end{matrix}$
Die wortelvorme wat presies dieselfde is kan as "gelyksoortige terme" geneem word en bymekaar getel word
$\begin{matrix} = & 13 \sqrt{3} \end{matrix}$

Rasionaliserende delers

Dit is nuttig om met breuke te werk wat rasionale delers het eerder as wortelvormige delers. Dit is moontlik om enige breuk wat 'n wortelvorm as deler het oor te skryf as 'n breuk met rasionale deleter. Ons sal nou sien hoe dit gedoen kan word.

Enige uitdrukking van die vorm $\sqrt{a} + \sqrt{b}$ (waar $a$ and $b$ rasionaal is) kan verander word in 'n rasionale getal deur dit te vermenigvuldig met $\sqrt{a} - \sqrt{b}$ (soortgelyk kan $\sqrt{a} - \sqrt{b}$ gerasionaliseer word deur dit te vermenigvuldig met $\sqrt{a} + \sqrt{b}$ ). Dit is omdat

(\sqrt{a} + \sqrt{b}) (\sqrt{a} - \sqrt{b}) = a - b

wat rasionaal is (aangesien $a$ en $b$ rasionaal is).

Indien ons 'n breuk het met 'n deler wat soos $\sqrt{a} + \sqrt{b}$ lyk, kan ons bloot bo en onder vermenigvuldig met $\sqrt{a} - \sqrt{b}$ en kry sodoende 'n rasionale deler.

\begin{matrix} \frac{c}{\sqrt{a} + \sqrt{b}} & = & \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} - \sqrt{b}} \times \frac{c}{\sqrt{a} + \sqrt{b}} \\ = & \frac{c \sqrt{a} - c \sqrt{b}}{a - b} \end{matrix}

of soortgelyk

\begin{matrix} \frac{c}{\sqrt{a} - \sqrt{b}} & = & \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} \times \frac{c}{\sqrt{a} - \sqrt{b}} \\ = & \frac{c \sqrt{a} + c \sqrt{b}}{a - b} \end{matrix}

Rasionaliseer die deler: $\frac{5 x - 16}{\sqrt{x}}$

Raak ontslae van die vierkantswortel onder die deler
Om ontslae te raak van die $\sqrt{x}$ in die deler, kan 'n mens dit uitvermenigvuldig met nog 'n $\sqrt{x}$ . Dit "rasionaliseer" die wortelvorm in die deler. Let op dat $\frac{\sqrt{x}}{\sqrt{x}}$ = 1, en dus word die vergelyking gerasionaliseer deur met 1 te vermenigvuldig en bly dit dieselfde ding.

$\begin{matrix} \frac{5 x - 16}{\sqrt{x}} \times \frac{\sqrt{x}}{\sqrt{x}} \end{matrix}$
Daar is nie meer 'n wortelvorm in die deler nie.
Die wortelvorm is uitgedruk in die noemer, wat die verkose manier is om wortelvorme te skryf. (Dit is hoekom noemers nie gerasionaliseer word nie.)

$\begin{matrix} \frac{5 x \sqrt{x} - 16 \sqrt{x}}{x} \\ = & \frac{(\sqrt{x}) (5 x - 16)}{x} \end{matrix}$

Rasionaliseer die volgende: $\frac{5 x - 16}{\sqrt{y} - 10}$

Rasionaliseer hierdie deler deur gebruik te maak van 'n slim "1"
$\begin{matrix} \frac{5 x - 16}{\sqrt{y} - 10} \times \frac{\sqrt{y} + 10}{\sqrt{y} + 10} \end{matrix}$
Vermenigvuldig die noemers en delers
$\begin{matrix} \frac{5 x \sqrt{y} - 16 \sqrt{y} + 50 x - 160}{y - 100} \end{matrix}$
In hierdie geval is daar geen volgende stap nie
Al die terme in die noemer is verskillend en kan nie verder vereenvoudig word nie en daar is geen wortelvorme meer in die deler nie.

Vereenvoudig die volgende: $\frac{y - 25}{\sqrt{y} + 5}$

Vermenigvuldig hierdie vergelykings met 'n slim vorm van "1" wat dit sal rasionaliseer
$\frac{y - 25}{\sqrt{y} + 5} \times \frac{\sqrt{y} - 5}{\sqrt{y} - 5}$
Vermenigvuldig die noemers en delers
$\begin{matrix} \frac{y \sqrt{y} - 25 \sqrt{y} - 5 y + 125}{y - 25} & = & \frac{\sqrt{y} (y - 25) - 5 (y - 25)}{(y - 25)} \\ = & \frac{(y - 25) (\sqrt{y} - 25)}{(y - 25)} \\ = & \sqrt{y} - 25 \end{matrix}$

Oefeninge

Brei uit:
$(\sqrt{x} - \sqrt{2}) (\sqrt{x} + \sqrt{2})$
Rasionaliseer die deler:
$\frac{10}{\sqrt{x} - \frac{1}{x}}$
Skryf as 'n enkele breuk:
$\frac{3}{2 \sqrt{x}} + \sqrt{x}$

Skryf in die eenvoudigste wortelvorm:

(a) $\sqrt{72}$	(b) $\sqrt{45} + \sqrt{80}$
(c) $\frac{\sqrt{48}}{\sqrt{12}}$	(d) $\frac{\sqrt{18} \div \sqrt{72}}{\sqrt{8}}$
(e) $\frac{4}{(\sqrt{8} \div \sqrt{2})}$	(f) $\frac{16}{(\sqrt{20} \div \sqrt{12})}$

Brei uit en vereenvoudig:
${(2 + \sqrt{2})}^{2}$
Brei uit en vereenvoudig:
$(2 + \sqrt{2}) (1 + \sqrt{8})$
Brei uit en vereenvoudig:
$(1 + \sqrt{3}) (1 + \sqrt{8} + \sqrt{3})$
Rasionaliseer die deler:
$\frac{y - 4}{\sqrt{y} - 2}$
Rasionaliseer die deler:
$\frac{2 x - 20}{\sqrt{y} - \sqrt{10}}$
Bewys, sonder die gebruik van 'n sakrekenaar, dat:
$\sqrt{\frac{8}{3}} + 5 \sqrt{\frac{5}{3}} - \sqrt{\frac{1}{6}} = \frac{13}{2} \sqrt{\frac{2}{3}}$
Vereenvoudig, sonder die gebruik van 'n sakrekenaar:
$\frac{\sqrt{98} - \sqrt{8}}{\sqrt{50}}$
Vereenvoudig, sonder die gebruik van 'n sakrekenaar:
$\sqrt{5} (\sqrt{45} + 2 \sqrt{80})$
Skryf die volgende met 'n rasionale deler:
$\frac{\sqrt{5} + 2}{\sqrt{5}}$
Vereenvoudig:
$\sqrt{98 x^{6}} + \sqrt{128 x^{6}}$
Evalueer, sonder die gebruik van 'n sakrekenaar: ${(2 - \frac{\sqrt{7}}{2})}^{\frac{1}{2}} . {(2 + \frac{\sqrt{7}}{2})}^{\frac{1}{2}}$
Die gebruik van 'n sakrekenaar word toegelaat in hierdie vraag. Vereenvoudig volledig en wys alle stappe: $3^{- \frac{1}{2}} [\sqrt{12} + \sqrt[3]{(3 \sqrt{3})}]$
Vul in die ontbrekende wortelvormgetal wat die volgende vergelyking waar sal maak: $- 3 \sqrt{6} \times - 2 \sqrt{24} = - \sqrt{18} \times . . . . . . . . . . .$

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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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