0.1 Wortelvorme

Siyavula textbooks: wiskunde Page 1 / 1

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: $- 3 \sqrt{6} \times - 2 \sqrt{24} = - \sqrt{18} \times . . . . . . . . . . .$

Questions & Answers

what is defense mechanism

Chinaza Reply

what is defense mechanisms

Chinaza

I'm interested in biological psychology and cognitive psychology

Tanya Reply

what does preconceived mean

sammie Reply

physiological Psychology

Nwosu Reply

How can I develope my cognitive domain

Amanyire Reply

why is communication effective

Dakolo Reply

Communication is effective because it allows individuals to share ideas, thoughts, and information with others.

effective communication can lead to improved outcomes in various settings, including personal relationships, business environments, and educational settings. By communicating effectively, individuals can negotiate effectively, solve problems collaboratively, and work towards common goals.

it starts up serve and return practice/assessments.it helps find voice talking therapy also assessments through relaxed conversation.

miss

Every time someone flushes a toilet in the apartment building, the person begins to jumb back automatically after hearing the flush, before the water temperature changes. Identify the types of learning, if it is classical conditioning identify the NS, UCS, CS and CR. If it is operant conditioning, identify the type of consequence positive reinforcement, negative reinforcement or punishment

Wekolamo Reply

please i need answer

Wekolamo

because it helps many people around the world to understand how to interact with other people and understand them well, for example at work (job).

Manix Reply

Agreed 👍 There are many parts of our brains and behaviors, we really need to get to know. Blessings for everyone and happy Sunday!

ARC

A child is a member of community not society elucidate ?

JESSY Reply

Isn't practices worldwide, be it psychology, be it science. isn't much just a false belief of control over something the mind cannot truly comprehend?

Simon Reply

compare and contrast skinner's perspective on personality development on freud

namakula Reply

Skinner skipped the whole unconscious phenomenon and rather emphasized on classical conditioning

war

explain how nature and nurture affect the development and later the productivity of an individual.

Amesalu Reply

nature is an hereditary factor while nurture is an environmental factor which constitute an individual personality. so if an individual's parent has a deviant behavior and was also brought up in an deviant environment, observation of the behavior and the inborn trait we make the individual deviant.

Samuel

I am taking this course because I am hoping that I could somehow learn more about my chosen field of interest and due to the fact that being a PsyD really ignites my passion as an individual the more I hope to learn about developing and literally explore the complexity of my critical thinking skills

Zyryn Reply

good👍

Jonathan

and having a good philosophy of the world is like a sandwich and a peanut butter 👍

Jonathan

generally amnesi how long yrs memory loss

Kelu Reply

interpersonal relationships

Abdulfatai Reply

Got questions? Join the online conversation and get instant answers!

Jobilize.com Reply

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Siyavula textbooks: wiskunde (graad 11). OpenStax CNX. Sep 20, 2011 Download for free at http://cnx.org/content/col11339/1.4

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Siyavula textbooks: wiskunde (graad 11)' conversation and receive update notifications?

Ask

	6 Biology 06 Metabolism MCQ By OpenStax Start Quiz
©flickr: Justin	Music Appreciation Final Practice By Madison Christian Start Exam
	11 AP Key Terms 11 The Muscular System By OpenStax Start Key Terms
	24 Biology 24 Fungi MCQ By OpenStax Start Quiz
©flickr: U.S.	Biology Chapter 10 By Michael Sag Start Exam
	2 AP 02 Chemical Level of Organization MCQ By OpenStax Start Quiz
©flickr:	Math for Economists MCQ By Tony Pizur Start Quiz
©flickr: Gareth	Professional Etiquette MCQ By Abby Sharp Start Quiz
	2 AP Key Terms 02 Chemical Level of Organization By OpenStax Start Key Terms
	Concepts of biology By OpenStax Read Online Course