1.4: Radicals na Maneno ya busara
- Page ID
- 180806
- Tathmini mizizi ya mraba.
- Tumia utawala wa bidhaa ili kurahisisha mizizi ya mraba.
- Tumia utawala wa quotient ili kurahisisha mizizi ya mraba.
- Ongeza na uondoe mizizi ya mraba.
- Rationalize madhehebu.
- Tumia mizizi ya busara.
Duka la vifaa anauza\(16\) ngazi\(24\) -ft na ngazi -ft. Dirisha iko\(12\) miguu juu ya ardhi. Ngazi inahitaji kununuliwa ambayo itafikia dirisha kutoka hatua kwenye\(5\) miguu ya chini kutoka jengo hilo. Ili kujua urefu wa ngazi inahitajika, tunaweza kuteka pembetatu sahihi kama inavyoonekana kwenye Mchoro\(\PageIndex{1}\), na tumia Theorem ya Pythagorean.
Kielelezo \(\PageIndex{1}\): Pembetatu ya kulia
\[ \begin{align*} a^2+b^2&=c^2 \label{1.4.1} \\[4pt] 5^2+12^2&=c^2 \label{1.4.2} \\[4pt] 169 &=c^2 \label{1.4.3} \end{align*}\]
Now, we need to find out the length that, when squared, is \(169\), to determine which ladder to choose. In other words, we need to find a square root. In this section, we will investigate methods of finding solutions to problems such as this one.
Evaluating Square Roots
When the square root of a number is squared, the result is the original number. Since \(4^2=16\), the square root of \(16\) is \(4\).The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root.
In general terms, if \(a\) is a positive real number, then the square root of \(a\) is a number that, when multiplied by itself, gives \(a\).The square root could be positive or negative because multiplying two negative numbers gives a positive number. The principal square root is the nonnegative number that when multiplied by itself equals \(a\). The square root obtained using a calculator is the principal square root.
The principal square root of \(a\) is written as \(\sqrt{a}\). The symbol is called a radical, the term under the symbol is called the radicand, and the entire expression is called a radical expression.
Je,\(\sqrt{25} = \pm 5\)?
Suluhisho
Hapana. Ingawa wote wawili\(5^2\) na\((−5)^2\) ni\(25\), ishara kubwa ina maana tu mizizi isiyo na negative, mizizi kuu ya mraba. kuu mraba mizizi ya\(25\) ni\(\sqrt{25}=5\).
Mizizi kuu ya mraba\(a\) ni nambari isiyo na nambari ambayo, wakati imeongezeka kwa yenyewe, inalingana\(a\). Imeandikwa kama usemi mkali, na ishara inayoitwa radical juu ya neno inayoitwa radicand:\(\sqrt{a}\).
Tathmini kila kujieleza.
- \(\sqrt{\sqrt{16}}\)
- \(\sqrt{49}\)-\(\sqrt{81}\)
Suluhisho
- \(\sqrt{\sqrt{16}}= \sqrt{4} =2\)kwa sababu\(4^2=16\) na\(2^2=4\)
- \(\sqrt{49} -\sqrt{81} =7−9 =−2\)kwa sababu\(7^2=49\) na\(9^2=81\)
Kwa\(\sqrt{25+144}\), tunaweza kupata mizizi ya mraba kabla ya kuongeza?
Suluhisho
Hapana. \(\sqrt{25} + \sqrt{144} =5+12=17\). Hii si sawa na\(\sqrt{25+144}=13\). Utaratibu wa shughuli unatuhitaji kuongeza maneno katika radicand kabla ya kupata mizizi ya mraba.
Tathmini kila kujieleza.
- \(\sqrt{\sqrt{81}}\)
- \(\sqrt{36} + \sqrt{121}\)
- Jibu
-
\(3\)
- Jibu b
-
\(17\)
Kutumia Utawala wa Bidhaa ili kurahisisha mizizi ya Mraba
Ili kurahisisha mizizi ya mraba, tunaandika tena kama kwamba hakuna mraba kamili katika radicand. Kuna mali kadhaa za mizizi ya mraba ambayo inatuwezesha kurahisisha maneno magumu ya ngumu. Utawala wa kwanza tutaangalia ni utawala wa bidhaa kwa kurahisisha mizizi ya mraba, ambayo inatuwezesha kutenganisha mizizi ya mraba ya bidhaa ya namba mbili katika bidhaa za maneno mawili tofauti ya busara. Kwa mfano, tunaweza kuandika upya\(\sqrt{15}\) kama\(\sqrt{3}\times\sqrt{5}\). Tunaweza pia kutumia utawala wa bidhaa kueleza bidhaa ya maneno mbalimbali radical kama kujieleza moja radical.
Ikiwa\(a\) na\(b\) sio hasi, mizizi ya mraba ya bidhaa\(ab\) ni sawa na bidhaa za mizizi ya mraba ya\(a\) na\(b\)
\[\sqrt{ab}=\sqrt{a}\times\sqrt{b}\]
- Fanya mraba yoyote kamili kutoka kwa radicand.
- Andika maneno makubwa kama bidhaa ya maneno makubwa.
- Kurahisisha.
Kurahisisha kujieleza radical.
- \(\sqrt{300}\)
- \(\sqrt{162a^5b^4}\)
Suluhisho
a. mraba\(\sqrt{100\times3}\) Factor kamili kutoka radicand.
\(\sqrt{100}\times\sqrt{3}\)Andika maneno makubwa kama bidhaa ya maneno makubwa.
\(10\sqrt{3}\)Kurahisisha
b\(\sqrt{81a^4b^4\times2a}\) Factor mraba kamili kutoka radicand
\(\sqrt{81a^4b^4}\times\sqrt{2a}\)Andika maneno makubwa kama bidhaa ya maneno makubwa
\(9a^2b^2\sqrt{2a}\)Kurahisisha
Kurahisisha\(\sqrt{50x^2y^3z}\)
- Jibu
-
\(5|x||y|\sqrt{2yz}\)
Angalia thamani kamili ishara karibu\(x\) na\(y\)? Hiyo ni kwa sababu thamani yao lazima iwe chanya!
- Express bidhaa ya maneno mengi radical kama kujieleza moja radical.
- Kurahisisha.
Kurahisisha kujieleza radical.
\(\sqrt{12}\times\sqrt{3}\)
Suluhisho
\[\begin{align*} &\sqrt{12\times3}\qquad \text{Express the product as a single radical expression}\\ &\sqrt{36}\qquad \text{Simplify}\\ &6 \end{align*}\]
Kurahisisha\(\sqrt{50x}\times\sqrt{2x}\) kuchukua\(x>0\).
- Jibu
-
\(10|x|\)
Kutumia Utawala wa Quotient ili kurahisisha mizizi ya Mraba
Kama vile tunaweza kuandika upya mizizi ya mraba ya bidhaa kama bidhaa za mizizi ya mraba, pia tunaweza kuandika upya mizizi ya mraba ya quotient kama quotient ya mizizi ya mraba, kwa kutumia utawala wa quotient kwa kurahisisha mizizi ya mraba. Inaweza kuwa na manufaa kutenganisha nambari na denominator ya sehemu chini ya radical ili tuweze kuchukua mizizi yao ya mraba tofauti. Tunaweza kuandika tena
\[\sqrt{\dfrac{5}{2}} = \dfrac{\sqrt{5}}{\sqrt{2}}. \nonumber \]
Mzizi wa mraba wa quotient\(\dfrac{a}{b}\) ni sawa na quotient ya mizizi ya mraba ya\(a\) na\(b\), wapi\(b≠0\).
\[\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\]
- Andika maneno makubwa kama quotient ya maneno mawili makubwa.
- Kurahisisha nambari na denominator.
Kurahisisha kujieleza radical.
\(\sqrt{\dfrac{5}{36}}\)
Suluhisho
\[\begin{align*} &\dfrac{\sqrt{5}}{\sqrt{36}}\qquad \text{Write as quotient of two radical expressions}\\ &\dfrac{\sqrt{5}}{6}\qquad \text {Simplify denominator} \end{align*}\]
Kurahisisha\(\sqrt{\dfrac{2x^2}{9y^4}}\)
- Jibu
-
\(\dfrac{x\sqrt{2}}{3y^2}\)
Hatuna haja ya ishara za thamani kamili kwa\(y^2\) sababu neno hilo litakuwa lisilo na hasi.
Kurahisisha kujieleza radical.
\(\dfrac{\sqrt{234x^{11}y}}{\sqrt{26x^7y}}\)
Suluhisho
\[\begin{align*} &\sqrt{\dfrac{234x^{11}y}{26x^7y}}\qquad \text{Combine numerator and denominator into one radical expression}\\ &\sqrt{9x^4}\qquad \text{Simplify fraction}\\ &3x^2\qquad \text{Simplify square root} \end{align*}\]
Kurahisisha\(\dfrac{\sqrt{9a^5b^{14}}}{\sqrt{3a^4b^5}}\)
- Jibu
-
\(b^4\sqrt{3ab}\)
Kuongeza na Kuondoa Mizizi ya Mraba
Tunaweza kuongeza au kuondoa maneno makubwa tu wakati wana radicand sawa na wakati wana aina sawa radical kama vile mizizi ya mraba. Kwa mfano, jumla ya\(\sqrt{2}\) na\(3\sqrt{2}\) ni\(4\sqrt{2}\). Hata hivyo, mara nyingi inawezekana kurahisisha maneno makubwa, na ambayo inaweza kubadilisha radicand. kujieleza radical\(\sqrt{18}\) inaweza kuandikwa na\(2\) katika radicand, kama\(3\sqrt{2}\), hivyo\(\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\)
- Kurahisisha kila kujieleza radical.
- Ongeza au uondoe maneno na radicands sawa.
Ongeza\(5\sqrt{12}+2\sqrt{3}\).
Suluhisho
Tunaweza kuandika upya\(5\sqrt{12}\) kama\(5\sqrt{4\times3}\). Kulingana na utawala wa bidhaa, hii inakuwa\(5\sqrt{4}\sqrt{3}\). Mzizi wa mraba wa\(\sqrt{4}\) ni\(2\), hivyo maneno inakuwa\(5\times2\sqrt{3}\), ambayo ni\(10\sqrt{3}\). Sasa tunaweza maneno kuwa radicand sawana hivyo tunaweza kuongeza.
\[10\sqrt{3}+2\sqrt{3}=12\sqrt{3} \nonumber\]
Ongeza\(\sqrt{5}+6\sqrt{20}\)
- Jibu
-
\(13\sqrt{5}\)
Ondoa\(20\sqrt{72a^3b^4c}-14\sqrt{8a^3b^4c}\)
Suluhisho
Andika upya kila neno ili wawe na radicands sawa.
\[\begin{align*} 20\sqrt{72a^3b^4c} &= 20\sqrt{9}\sqrt{4}\sqrt{2}\sqrt{a}\sqrt{a^2}\sqrt{(b^2)^2}\sqrt{c}\\ &= 20(3)(2)|a|b^2\sqrt{2ac}\\ &= 120|a|b^2\sqrt{2ac} \end{align*}\]
\[\begin{align*} 14\sqrt{8a^3b^4c} &= 14\sqrt{2}\sqrt{4}\sqrt{a}\sqrt{a^2}\sqrt{(b^2)^2}\sqrt{c}\\ &= 14(2)|a|b^2\sqrt{2ac}\\ &= 28|a|b^2\sqrt{2ac} \end{align*}\]
Sasa maneno yana radicana sawana hivyo tunaweza kuondoa.
\[120|a|b^2\sqrt{2ac}-28|a|b^2\sqrt{2ac}=92|a|b^2\sqrt{2ac}\]
Ondoa\(3\sqrt{80x}-4\sqrt{45x}\)
- Jibu
-
\(0\)
Kubainisha madhehebu
Wakati maneno yanayohusisha radicals mizizi ya mraba imeandikwa kwa fomu rahisi, haitakuwa na radical katika denominator. Tunaweza kuondoa radicals kutoka kwa denominators ya sehemu kwa kutumia mchakato unaoitwa rationalizing denominator.
Tunajua kwamba kuzidisha kwa\(1\) haina mabadiliko ya thamani ya kujieleza. Tunatumia mali hii ya kuzidisha kubadili maneno yaliyo na radicals katika denominator. Ili kuondoa radicals kutoka kwa denominators ya sehemu ndogo, kuzidisha kwa fomu ya\(1\) hiyo itaondoa radical.
Kwa denominator iliyo na muda mmoja, kuzidisha kwa radical katika denominator juu ya yenyewe. Kwa maneno mengine, kama denominator ni\(b\sqrt{c}\), kuzidisha na\(\dfrac{\sqrt{c}}{\sqrt{c}}\).
Kwa denominator iliyo na jumla au tofauti ya neno la busara na lisilo na maana, kuzidisha namba na denominator kwa conjugate ya denominator, ambayo hupatikana kwa kubadilisha ishara ya sehemu kubwa ya denominator. Ikiwa denominator ni\(a+b\sqrt{c}\), basi conjugate ni\(a-b\sqrt{c}\).
- Panua nambari na denominator kwa radical katika denominator.
- Kurahisisha.
Andika\(\dfrac{2\sqrt{3}}{3\sqrt{10}}\) kwa fomu rahisi.
Suluhisho
Radical katika denominator ni\(\sqrt{10}\). Hivyo kuzidisha sehemu na\(\dfrac{\sqrt{10}}{\sqrt{10}}\). Kisha kurahisisha.
\[\begin{align*} &\dfrac{2\sqrt{3}}{3\sqrt{10}}\times\dfrac{\sqrt{10}}{\sqrt{10}}\\ &\dfrac{2\sqrt{30}}{30}\\ &\dfrac{\sqrt{30}}{15} \end{align*}\]
Andika\(\dfrac{12\sqrt{3}}{\sqrt{2}}\) kwa fomu rahisi.
- Jibu
-
\(6\sqrt{6}\)
- Pata conjugate ya denominator.
- Panua nambari na denominator kwa conjugate.
- Tumia mali ya usambazaji.
- Kurahisisha.
Andika\(\dfrac{4}{1+\sqrt{5}}\) kwa fomu rahisi.
Suluhisho
Anza kwa kutafuta conjugate ya denominator kwa kuandika denominator na kubadilisha ishara. Hivyo conjugate ya\(1+\sqrt{5}\) ni\(1-\sqrt{5}\). Kisha kuzidisha sehemu na\(\dfrac{1-\sqrt{5}}{1-\sqrt{5}}\).
\[\begin{align*} &\dfrac{4}{1+\sqrt{5}}\times\dfrac{1-\sqrt{5}}{1-\sqrt{5}}\\ &\dfrac{4-4\sqrt{5}}{-4}\qquad \text{Use the distributive property}\\ &\sqrt{5}-1\qquad \text{Simplify} \end{align*}\]
Andika\(\dfrac{7}{2+\sqrt{3}}\) kwa fomu rahisi.
- Jibu
-
\(14-7\sqrt{3}\)
Kutumia Mizizi ya busara
Ingawa mizizi ya mraba ni mizizi ya kawaida ya busara, tunaweza pia kupata mizizi ya mchemraba,\(4^{th}\)\(5^{th}\) mizizi, na zaidi. Kama vile kazi ya mizizi ya mraba ni inverse ya kazi ya mraba, mizizi hii ni inverse ya kazi zao za nguvu. Kazi hizi zinaweza kuwa na manufaa wakati tunahitaji kuamua namba ambayo, wakati imefufuliwa kwa nguvu fulani, inatoa idadi fulani.
Kuelewa \(n^{th}\)Mizizi
Tuseme tunajua kwamba\(a^3=8\). Tunataka kupata nambari gani iliyoinuliwa kwa\(3^{rd}\) nguvu ni sawa na\(8\). Tangu\(2^3=8\), tunasema kwamba\(2\) ni mchemraba mzizi wa\(8\).
\(n^{th}\)mzizi wa\(a\) ni idadi ambayo, wakati kukulia kwa\(n^{th}\) nguvu, anatoa. kwa mfano,\(−3\) ni\(5^{th}\) mzizi wa\(−243\) sababu\({(-3)}^5=-243\). Kama\(a\) ni idadi halisi na angalau\(n^{th}\) mizizi moja, basi\(n^{th}\) mizizi kuu ya\(a\) ni idadi na ishara sawa na\(a\) kwamba, wakati kukulia kwa\(n^{th}\) nguvu, sawa\(a\).
\(n^{th}\)mizizi kuu ya\(a\) imeandikwa kama\(\sqrt[n]{a}\), ambapo\(n\) ni integer chanya kubwa kuliko au sawa na\(2\). Katika kujieleza radical,\(n\) inaitwa index ya radical.
Kama\(a\) ni idadi halisi na mizizi angalau moja, basi\(n^{th}\) \(n^{th}\)mizizi kuu ya\(a\), imeandikwa kama\(\sqrt[n]{a}\), ni idadi na ishara sawa na\(a\) kwamba, wakati kukulia kwa\(n^{th}\) nguvu, sawa\(a\). Ripoti ya radical ni\(n\).
Kurahisisha kila moja ya yafuatayo:
- \(\sqrt[5]{-32}\)
- \(\sqrt[4]{4}\times\sqrt[4]{10234}\)
- \(-\sqrt[3]{\dfrac{8x^6}{125}}\)
- \(8\sqrt[4]{3}-\sqrt[4]{48}\)
Suluhisho
a.\(\sqrt[5]{-32}=-2\) kwa sababu\((-2)^5=-32\)
b Kwanza, kueleza bidhaa kama kujieleza moja radical. \(\sqrt[4]{4096}=8\)kwa sababu\(8^4=4096\)
c.\[\begin{align*} &\dfrac{-\sqrt[3]{8x^6}}{\sqrt[3]{125}}\qquad \text{Write as quotient of two radical expressions}\\ &\dfrac{-2x^2}{5}\qquad \text{Simplify} \end{align*}\]
d.\[\begin{align*} &8\sqrt[4]{3}-2\sqrt[4]{3}\qquad \text{Simplify to get equal radicands}\\ &6\sqrt[4]{3}\qquad \text{Add} \end{align*}\]
Rahisisha
- \(\sqrt[3]{-216}\)
- \(\dfrac{3\sqrt[4]{80}}{\sqrt[4]{5}}\)
- \(6\sqrt[3]{9000}+7\sqrt[3]{576}\)
- Jibu
-
\(-6\)
- Jibu b
-
\(6\)
- Jibu c
-
\(88\sqrt[3]{9}\)
Kutumia Watazamaji wa busara
Maneno makubwa yanaweza pia kuandikwa bila kutumia ishara kali. Tunaweza kutumia vielelezo vya busara (fractional). Ripoti lazima iwe integer nzuri. Ikiwa index\(n\) ni hata, basi haiwezi kuwa hasi.
Pia tunaweza kuwa na exponents busara na nambari zaidi ya\(1\). Katika kesi hizi, exponent lazima kuwa sehemu katika suala la chini kabisa. Sisi kuongeza msingi kwa nguvu na kuchukua mizizi nth. Nambari inatuambia nguvu na denominator inatuambia mizizi.
Yote ya mali ya exponents kwamba sisi kujifunza kwa exponents integer pia kushikilia kwa exponents busara.
Watazamaji wa busara ni njia nyingine ya kueleza\(n^{th}\) mizizi kuu. Fomu ya jumla ya kugeuza kati ya kujieleza kwa radical na ishara radical na moja na exponent busara ni
\[a^{\tfrac{m}{n}}=(\sqrt[n]{a})^m=\sqrt[n]{a^m}\]
- Kuamua nguvu kwa kuangalia namba ya exponent.
- Kuamua mizizi kwa kuangalia denominator ya exponent.
- Kutumia msingi kama radicand, ongeza radicand kwa nguvu na kutumia mizizi kama index.
Andika\(343^{\tfrac{2}{3}}\) kama radical. Kurahisisha.
Suluhisho
The\(2\) inatuambia nguvu na\(3\) inatuambia mizizi.
\(343^{\tfrac{2}{3}}={(\sqrt[3]{343})}^2=\sqrt[3]{{343}^2}\)
Tunajua kwamba\(\sqrt[3]{343}=7\) kwa sababu\(7^3 =343\). Kwa sababu mizizi ya mchemraba ni rahisi kupata, ni rahisi kupata mizizi ya mchemraba kabla ya kukata tatizo hili. Kwa ujumla, ni rahisi kupata mizizi kwanza na kisha kuinua kwa nguvu.
\[343^{\tfrac{2}{3}}={(\sqrt[3]{343})}^2=7^2=49\]
Andika\(9^{\tfrac{5}{2}}\) kama radical. Kurahisisha.
- Jibu
-
\({(\sqrt{9})}^5=3^5=243\)
Andika\(\dfrac{4}{\sqrt[7]{a^2}}\) kwa kutumia kielelezo cha busara.
Suluhisho
Nguvu ni\(2\) na mizizi ni\(7\), hivyo mtangazaji wa busara atakuwa\(\dfrac{2}{7}\). Tunapata\(\dfrac{4}{a^{\tfrac{2}{7}}}\). Kutumia mali ya exponents, sisi kupata\(\dfrac{4}{\sqrt[7]{a^2}}=4a^{\tfrac{-2}{7}}\)
Andika\(x\sqrt{{(5y)}^9}\) kwa kutumia kielelezo cha busara.
- Jibu
-
\(x(5y)^{\dfrac{9}{2}}\)
Kurahisisha:
a.\(5(2x^{\tfrac{3}{4}})(3x^{\tfrac{1}{5}})\)
b.\(\left(\dfrac{16}{9}\right)^{-\tfrac{1}{2}}\)
Suluhisho
a.
\[\begin{align*} &30x^{\tfrac{3}{4}}\: x^{\tfrac{1}{5}}\qquad \text{Multiply the coefficients}\\ &30x^{\tfrac{3}{4}+\tfrac{1}{5}}\qquad \text{Use properties of exponents}\\ &30x^{\tfrac{19}{20}}\qquad \text{Simplify} \end{align*}\]
b.
\[\begin{align*} &{\left(\dfrac{9}{16}\right)}^{\tfrac{1}{2}}\qquad \text{Use definition of negative exponents}\\ &\sqrt{\dfrac{9}{16}}\qquad \text{Rewrite as a radical}\\ &\dfrac{\sqrt{9}}{\sqrt{16}}\qquad \text{Use the quotient rule}\\ &\dfrac{3}{4}\qquad \text{Simplify} \end{align*}\]
Kurahisisha\({(8x)}^{\tfrac{1}{3}}\left(14x^{\tfrac{6}{5}}\right)\)
- Jibu
-
\(28x^{\tfrac{23}{15}}\)
Kupata rasilimali hizi online kwa maelekezo ya ziada na mazoezi na radicals na exponents busara.
Dhana muhimu
- Mzizi mkuu wa mraba wa namba \(a\)ni nambari isiyo ya nambari ambayo inapoongezeka kwa yenyewe ni sawa\(a\) . Angalia Mfano.
- Ikiwa \(a\)na\(b\) sio hasi, mizizi ya mraba ya bidhaa\(ab\) ni sawa na bidhaa za mizizi ya mraba ya\(a\) na\(b\) Angalia Mfano na Mfano.
- Ikiwa \(a\)na\(b\) sio hasi, mizizi ya mraba ya quotient\(\dfrac{a}{b}\) ni sawa na quotient ya mizizi ya mraba ya\(a\) na\(b\) Angalia Mfano na Mfano.
- Tunaweza kuongeza na kuondoa maneno makubwa ikiwa wana radicand sawa na index sawa. Angalia Mfano na Mfano.
- Maneno makubwa yaliyoandikwa kwa fomu rahisi hayana radical katika denominator. Ili kuondokana na mizizi ya mraba kutoka kwa denominator, kuzidisha namba zote na denominator kwa conjugate ya denominator. Angalia Mfano na Mfano.
- \(n^{th}\)mizizi kuu ya\(a\) ni idadi na ishara sawa na\(a\) kwamba wakati kukulia kwa\(n^{th}\) nguvu sawa\(a\). Mizizi hii ina mali sawa na mizizi ya mraba. Angalia Mfano.
- Radicals inaweza kuandikwa upya kama exponents busara na exponents busara inaweza kuandikwa upya kama radicals. Angalia Mfano na Mfano.
- Mali ya watazamaji hutumika kwa watazamaji wa busara. Angalia Mfano.