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10.1: Tatua Ulinganisho wa Quadratic Kutumia Mali ya Mizizi ya Mraba

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    Malengo ya kujifunza

    Mwishoni mwa sehemu hii, utaweza:

    • Tatua usawa wa quadratic wa fomu kwa\(ax^2=k\) kutumia Mali ya Mizizi ya Mraba
    • Tatua usawa wa quadratic wa fomu kwa\(a(x−h)^2=k\) kutumia Mali ya Mizizi ya Mraba
    Kabla ya kuanza, fanya jaribio hili la utayari.
    1. Kurahisisha:\(\sqrt{75}\).
    2. Kurahisisha:\(\sqrt{\dfrac{64}{3}}\)
    3. Sababu:\(4x^{2} − 12x + 9\).

    Equations Quadratic ni equations ya fomu\(ax^{2} + bx + c = 0\), wapi\(a \neq 0\). Wanatofautiana na equations linear kwa kujumuisha neno na variable iliyofufuliwa kwa nguvu ya pili. Sisi kutumia mbinu mbalimbali ya kutatua quadratic equation s kuliko equations linear, kwa sababu tu kuongeza, kutoa, kuzidisha, na kugawa maneno si kujitenga kutofautiana.

    Tumeona kwamba baadhi ya equations quadratic inaweza kutatuliwa kwa factoring. Katika sura hii, tutatumia mbinu nyingine tatu kutatua equations quadratic.

    Tatua Ulinganisho wa Quadratic wa Fomu\(ax^2=k\) Kutumia Mizizi ya Mizizi ya Mraba

    Tayari tumetatua equations za quadratic kwa kuzingatia. Hebu tathmini jinsi tulivyotumia factoring kutatua equation quadratic\(x^{2} = 9\).

    \[\begin{array}{ll} {}&{x^2=9}\\ {\text{Put the equation in standard form.}}&{x^2−9=0}\\ {\text{Factor the left side.}}&{(x - 3)(x + 3) = 0}\\ {\text{Use the Zero Product Property.}}&{(x - 3) = 0, (x + 3) = 0}\\ {\text{Solve each equation.}}&{x = 3, x = -3}\\ {\text{Combine the two solutions into} \pm \text{form}}&{x=\pm 3}\\ \nonumber \end{array}\]

    (Suluhisho linasomewa '\(x\)ni sawa na chanya au hasi\(3\).')

    Tunaweza kutumia kwa urahisi factoring kupata ufumbuzi wa equations sawa, kama\(x^{2}=16\) na\(x^{2} = 25\), kwa sababu\(16\) na\(25\) ni mraba kamilifu. Lakini nini kinatokea wakati tuna equation kama\(x^{2}=7\)? Kwa kuwa\(7\) si mraba kamili, hatuwezi kutatua equation kwa factoring.

    Hizi equations ni wote wa fomu\(x^{2}=k\).
    Tulifafanua mizizi ya mraba ya namba kwa njia hii:

    Ikiwa\(n^{2} = m\), basi\(n\) ni mizizi ya mraba ya\(m\).

    Hii inasababisha Mali ya Mizizi ya Mraba.

    Ufafanuzi: SQUARE ROOT PRO

    Ikiwa\(x^{2}=k\), na\(k \geq 0\), basi\(x = \sqrt{k}\) au\(x = -\sqrt{k}\).

    Angalia kwamba Mizizi ya Mizizi ya Mraba inatoa ufumbuzi mbili kwa usawa wa fomu\(x^2=k\): mizizi kuu ya mraba ya k na kinyume chake. Tunaweza pia kuandika ufumbuzi kama\(x=\pm \sqrt{k}\)

    Sasa, sisi kutatua equation\(x^{2} = 9\) tena, wakati huu kwa kutumia Mizizi Square Mali.

    \[\begin{array}{ll} {}&{x^{2} = 9}\\ {\text{Use the Square Root Property.}}&{x = \pm\sqrt{9}}\\ {\text{Simplify the radical.}}&{x = \pm 3}\\ {\text{Rewrite to show the two solutions.}}&{x = 3, x = −3}\\ \nonumber \end{array}\]

    Nini kinatokea wakati mara kwa mara si mraba kamilifu? Hebu kutumia Mizizi ya Mizizi ya Mraba ili kutatua equation\(x^2=7\).

    \[\begin{array} {ll} {\text{Use the Square Root Property. }}&{x = \pm\sqrt{7}}\\ {\text{Rewrite to show two solutions.}}&{x = \sqrt{7}, x = −\sqrt{7}}\\ {\text{We cannot simplify} \sqrt{7} \text{ so we leave the answer as a radical.}}&{}\\ \nonumber \end{array}\]

    Mfano\(\PageIndex{1}\)

    Kutatua:\(x^{2} = 169\)

    Jibu

    \[\begin{array}{ll} {}&{x^2=169}\\ {\text{Use the Square Root Property.}}&{x=\pm\sqrt{169}}\\ {\text{Simplify the radical.}}&{x = \pm13}\\{\text{Rewrite to show two solutions.}}&{x = 13, x = −13}\\ \nonumber \end{array}\]

    Mfano\(\PageIndex{2}\)

    Kutatua:\(x^2=81\)

    Jibu

    x=9, x=-9

    Mfano\(\PageIndex{3}\)

    Kutatua:\(y^{2} = 121\)

    Jibu

    y = 11, y = -11

    Jinsi ya Kutatua Equation Quadratic ya Fomu\(ax^{2} = k\) Kutumia Mizizi ya Mizizi ya Mraba

    Mfano\(\PageIndex{4}\)

    Kutatua:\(x^{2} − 48 = 0\)

    Jibu
    Picha inaonyesha equation iliyotolewa, x squared minus 48 sawa na sifuri. Hatua ya kwanza ni kutenganisha muda wa quadratic na kufanya mgawo wake moja ili kuongeza 48 kwa pande zote mbili za equation ili kupata x squared yenyewe.Hatua mbili ni kutumia Mizizi ya Mizizi ya Mraba ili kupata x sawa pamoja au kupunguza mizizi ya mraba ya 48.Hatua ya tatu, kurahisisha mizizi ya mraba ya 48 kwa kuandika 48 kama bidhaa ya 16 na tatu. Mzizi wa mraba wa 16 ni nne. Suluhisho kilichorahisishwa ni x sawa na plus au minus nne za mraba mizizi ya tatu.Hatua ya nne, angalia ufumbuzi kwa kubadili kila suluhisho katika usawa wa awali. Wakati x sawa nne za mraba mizizi ya tatu, kuchukua nafasi ya x katika equation awali na nne za mraba mizizi ya tatu kupata nne za mraba mizizi ya tatu squared minus 48 sawa na sifuri. Kurahisisha upande wa kushoto kupata mara 16 tatu minus 48 sawa na sifuri ambayo simplifies zaidi kwa sifuri sawa sifuri, taarifa ya kweli. Wakati x sawa hasi nne za mraba mizizi ya tatu, kuchukua nafasi ya x katika equation awali na hasi nne za mraba mizizi ya tatu kupata hasi nne za mraba mizizi ya tatu squared minus 48 sawa na sifuri. Kurahisisha upande wa kushoto kupata mara 16 tatu minus 48 sawa na sifuri ambayo simplifies zaidi kwa sifuri sawa sifuri, pia taarifa ya kweli.
    Mfano\(\PageIndex{5}\)

    Kutatua:\(x^{2} − 50 = 0\)

    Jibu

    \(x = 5\sqrt{2}, x = −5\sqrt{2}\)

    Mfano\(\PageIndex{6}\)

    Kutatua:\(y^{2} − 27 = 0\)

    Jibu

    \(y = 3\sqrt{3}, x = −3\sqrt{3}\)

    UFAFANUZI:KUTATUA EQUATION QUADRATIC KUTUMIA SQUARE MIZIZI MALI.
    1. Sulua muda wa quadratic na ufanye mgawo wake mmoja.
    2. Matumizi Square Mizizi Mali.
    3. Kurahisisha radical.
    4. Angalia ufumbuzi.

    Ili kutumia Mizizi ya Mizizi ya Mraba, mgawo wa muda wa kutofautiana lazima iwe sawa 1. Katika mfano unaofuata, tunapaswa kugawanya pande zote mbili za equation na 5 kabla ya kutumia Mizizi ya Mizizi ya Mraba.

    Mfano\(\PageIndex{7}\)

    Kutatua:\(5m^2=80\)

    Jibu
    Neno la quadratic ni pekee. \(5m^2=80\)
    Gawanya na 5 kufanya cofficient yake 1. \(\frac{5m^2}{5}=\frac{80}{5}\)
    Kurahisisha. \(m^2=16\)
    Tumia Mizizi ya Mizizi ya Mraba. \(m=\pm\sqrt{16}\)
    Kurahisisha radical. \(m=\pm 4\)
    Andika upya ili kuonyesha ufumbuzi mbili. m=4, m=-4

    Angalia ufumbuzi.
    .

    		  
    Mfano\(\PageIndex{8}\)

    Kutatua:\(2x^2=98\).

    Jibu

    x=7, x=-7

    Mfano\(\PageIndex{9}\)

    Kutatua:\(3z^2=108\).

    Jibu

    z=6, z=-6

    Mali ya Mizizi ya Mraba ilianza kwa kusema, 'If\(x^2=k\), na\(k\ge 0\) '. Nini kitatokea kama\(k<0\)? Hii itakuwa kesi katika mfano unaofuata.

    Mfano\(\PageIndex{10}\)

    Kutatua:\(q^2+24=0\).

    Jibu

    \[\begin{array}{ll} {}&{q^2=24}\\ {\text{Isolate the quadratic term.}}&{q^2=−24}\\ {\text{Use the Square Root Property.}}&{q=\pm\sqrt{-24}}\\ {\text{The} \sqrt{-24} \text{is not a real number}}& {\text{There is no real solution}}\\ \nonumber \end{array}\]

    Mfano\(\PageIndex{11}\)

    Kutatua:\(c^2+12=0\).

    Jibu

    hakuna ufumbuzi halisi

    Mfano\(\PageIndex{12}\)

    Kutatua:\(d^2+81=0\).

    Jibu

    hakuna ufumbuzi halisi

    Kumbuka, sisi kwanza hutenganisha muda wa quadratic na kisha tufanye mgawo sawa na moja.
    Mfano\(\PageIndex{13}\)

    Kutatua:\(\frac{2}{3}u^2+5=17\).

    Jibu
      \(\frac{2}{3}u^2+5=17\)
    Sulua muda wa quadratic.

    \(\frac{2}{3}u^2=12\)
    \(\frac{3}{2}\)Kuzidisha na kufanya mgawo 1. \(\frac{3}{2}·\frac{2}{3}u^2=\frac{3}{2}·12\)
    Kurahisisha. \(u^2=18\)
    Tumia Mizizi ya Mizizi ya Mraba. \(u=\pm\sqrt{18}\)
    Kurahisisha radical. \(u=\pm\sqrt{9}\sqrt{2}\)
    Kurahisisha. \(u=\pm3\sqrt{2}\)
    Andika upya ili kuonyesha ufumbuzi mbili. \(u=3\sqrt{2}\),\(u=−3\sqrt{2}\)

    Angalia.
    .

    		  
    Mfano\(\PageIndex{14}\)

    Kutatua:\(\frac{1}{2}x^2+4=24\)

    Jibu

    \(x=2\sqrt{10}\),\(x=−2\sqrt{10}\)

    Mfano\(\PageIndex{15}\)

    Kutatua:\(\frac{3}{4}y^2−3=18\).

    Jibu

    \(y=2\sqrt{7}\),\(y=−2\sqrt{7}\)

    Ufumbuzi wa equations fulani inaweza kuwa na sehemu ndogo ndani ya radicals. Wakati hii itatokea, ni lazima rationalize denominator.

    Mfano\(\PageIndex{16}\)

    Kutatua:\(2c^2−4=45\).

    Jibu
      \(2c^2−4=45\)
    Sulua muda wa quadratic. \(2c^2=49\)
    Gawanya na 2 ili kufanya mgawo 1. \(\frac{2c^2}{2}=\frac{49}{2}\)
    Kurahisisha. \(c^2=\frac{49}{2}\)
    Tumia Mizizi ya Mizizi ya Mraba. \(c=\pm\frac{\sqrt{49}}{\sqrt{2}}\)
    Kurahisisha radical. \(c=\pm\frac{\sqrt{49}}{\sqrt{2}}\)
    Rationalize denominator. \(c=\pm\frac{\sqrt{49}\sqrt{2}}{\sqrt{2}\sqrt{2}}\)
    Kurahisisha. \(c=\pm\frac{7\sqrt{2}}{2}\)
    Andika upya ili kuonyesha ufumbuzi mbili. \(c=\frac{7\sqrt{2}}{2}\),\(c=−\frac{7\sqrt{2}}{2}\)
    Angalia. Tunaacha hundi kwako.  
    Mfano\(\PageIndex{17}\)

    Kutatua:\(5r^2−2=34\).

    Jibu

    \(r=\frac{6\sqrt{5}}{5}\),\(r=−\frac{6\sqrt{5}}{5}\)

    Mfano\(\PageIndex{18}\)

    Kutatua:\(3t^2+6=70\).

    Jibu

    \(t=\frac{8\sqrt{3}}{3}\),\(t=−\frac{8\sqrt{3}}{3}\)

    Tatua Ulinganisho wa Quadratic wa Fomu\(a(x-h)^2=k\) Kutumia Mizizi ya Mizizi ya Mraba

    Tunaweza kutumia Mizizi ya Mizizi ya Square kutatua equation kama\((x−3)^2=16\), pia. Tutachukua binomial nzima, (x-3), kama neno la quadratic.

    Mfano\(\PageIndex{19}\)

    Kutatua:\((x−3)^2=16\).

    Jibu
      \((x−3)^2=16\)
    Tumia Mizizi ya Mizizi ya Mraba. \(x−3=\pm\sqrt{16}\)
    Kurahisisha. \(x−3=\pm 4\)
    Andika kama equations mbili. \(x−3=4\),\(x−3=−4\)
    Kutatua. x=7, x=-1

    Angalia.
    .

    		  
    Mfano\(\PageIndex{20}\)

    Kutatua:\((q+5)^2=1\).

    Jibu

    q=-6, q=-4

    Mfano\(\PageIndex{21}\)

    Kutatua:\((r−3)^2=25\).

    Jibu

    r=8, r=-1

    Mfano\(\PageIndex{22}\)

    Kutatua:\((y−7)^2=12\).

    Jibu
      \((y−7)^2=12\).
    Tumia Mizizi ya Mizizi ya Mraba. \(y−7=\pm\sqrt{12}\)
    Kurahisisha radical. \(y−7=\pm2\sqrt{3}\)
    Kutatua kwa y. \(y=7\pm2\sqrt{3}\)
    Andika upya ili kuonyesha ufumbuzi mbili. \(y=7+2\sqrt{3}\),\(y=7−2\sqrt{3}\)

    Angalia.
    .

    		  
    Mfano\(\PageIndex{23}\)

    Kutatua:\((a−3)^2=18\).

    Jibu

    \(a=3+3\sqrt{2}\),\(a=3−3\sqrt{2}\)

    Mfano\(\PageIndex{24}\)

    Kutatua:\((b+2)^2=40\).

    Jibu

    \(b=−2+2\sqrt{10}\),\(b=−2−2\sqrt{10}\)

    Kumbuka, tunapochukua mizizi ya mraba ya sehemu, tunaweza kuchukua mizizi ya mraba ya nambari na denominator tofauti.
    Mfano\(\PageIndex{25}\)

    Kutatua:\((x−\frac{1}{2})^2=\frac{5}{4}\).

    Jibu
      \((x−\frac{1}{2})^2=\frac{5}{4}\)
    Tumia Mizizi ya Mizizi ya Mraba. \((x−\frac{1}{2})=\pm\sqrt\frac{5}{4}\)
    Andika upya radical kama sehemu ya mizizi ya mraba. \((x−\frac{1}{2})=\pm\frac{\sqrt{5}}{\sqrt{4}}\)
    Kurahisisha radical. \((x−\frac{1}{2})=\pm\frac{\sqrt{5}}{2}\)
    Kutatua kwa x. \(x=\frac{1}{2}+\pm\frac{\sqrt{5}}{2}\)
    Andika upya ili kuonyesha ufumbuzi mbili. \(x=\frac{1}{2}+\frac{\sqrt{5}}{2}\),\(x=\frac{1}{2}−\frac{\sqrt{5}}{2}\)
    Angalia. Tunaacha hundi kwako  
    ​​​​​​
    Mfano\(\PageIndex{26}\)

    Kutatua:\((x−\frac{1}{3})^2=\frac{5}{9}\).

    Jibu

    \(x=\frac{1}{3}+\frac{\sqrt{5}}{3}\),\(x=\frac{1}{3}−\frac{\sqrt{5}}{3}\)

    Mfano\(\PageIndex{27}\)

    Kutatua:\((y−\frac{3}{4})^2=\frac{7}{16}\).

    Jibu

    \(y=\frac{3}{4}+\frac{\sqrt{7}}{4}\),\(y=\frac{3}{4}−\frac{\sqrt{7}}{4}\),

    Tutaanza suluhisho kwa mfano unaofuata kwa kutenganisha binomial.

    Mfano\(\PageIndex{28}\)

    Kutatua:\((x−2)^2+3=30\).

    Jibu
      \((x−2)^2+3=30\)
    Sulua neno la binomial. \((x−2)^2=27\)
    Tumia Mizizi ya Mizizi ya Mraba. \(x−2=\pm\sqrt{27}\)
    Kurahisisha radical. \(x−2=\pm3\sqrt{3}\)
    Kutatua kwa x. \(x=2+\pm3\sqrt{3}\)
    \(x−2=\pm3\sqrt{3}\) \(x=2+3\sqrt{3}\),\(x=2−3\sqrt{3}\)
    Angalia. Tunaacha hundi kwako  
    Mfano\(\PageIndex{29}\)

    Kutatua:\((a−5)^2+4=24\).

    Jibu

    \(a=5+2\sqrt{5}\),\(a=5−2\sqrt{5}\)

    Mfano\(\PageIndex{30}\)

    Kutatua:\((b−3)^2−8=24\).

    Jibu

    \(b=3+4\sqrt{2}\),\(b=3−4\sqrt{2}\)

    Mfano\(\PageIndex{31}\)

    Kutatua:\((3v−7)^2=−12\).

    Jibu
      \((3v−7)^2=−12\)
    Tumia Mizizi ya Mizizi ya Mraba. \(3v−7=\pm\sqrt{−12}\)
    The\(\sqrt{−12}\) si idadi halisi. Hakuna suluhisho halisi.
    Mfano\(\PageIndex{32}\)

    Kutatua:\((3r+4)^2=−8\).

    Jibu

    hakuna ufumbuzi halisi

    Pande za kushoto za equations katika mifano miwili ijayo hazionekani kuwa za fomu\(a(x−h)^2\). Lakini ni trinomials kamili ya mraba, kwa hiyo tutazingatia kuiweka katika fomu tunayohitaji.

    Mfano\(\PageIndex{33}\)

    Kutatua:\(p^2−10p+25=18\).

    Jibu

    Upande wa kushoto wa equation ni trinomial kamili ya mraba. Tutaifanya kwanza.

      \(p^2−10p+25=18\)
    Factor kamili mraba trinomial. \((p−5)^2=18\)
    Tumia Mizizi ya Mizizi ya Mraba. \(p−5=\pm\sqrt{18}\)
    Kurahisisha radical. \(p−5=\pm3\sqrt{2}\)
    Tatua kwa p. \(p=5\pm3\sqrt{2}\)
    Andika upya ili kuonyesha ufumbuzi mbili. \(p=5+3\sqrt{2}\),\(p=5−3\sqrt{2}\)
    Angalia. Tunaacha hundi kwako.  
    Mfano\(\PageIndex{34}\)

    Kutatua:\(x^2−6x+9=12\).

    Jibu

    \(x=3+2\sqrt{3}\),\(x=3−2\sqrt{3}\)

    Mfano\(\PageIndex{35}\)

    Kutatua:\(y^2+12y+36=32\).

    Jibu

    \(y=−6+4\sqrt{2}\),\(y=−6−4\sqrt{2}\)

    Mfano\(\PageIndex{36}\)

    Kutatua:\(4n^2+4n+1=16\).

    Jibu

    Tena, tunaona upande wa kushoto wa equation ni kamili mraba trinomial. Tutaifanya kwanza.

      \(4n^2+4n+1=16\)
    Factor kamili mraba trinomial. \((2n+1)^2=16\)
    Tumia Mizizi ya Mizizi ya Mraba. \((2n+1)=\pm\sqrt{16}\)
    Kurahisisha radical. \((2n+1)=\pm4\)
    Kutatua kwa n. \(2n=−1\pm4\)
    Gawanya kila upande kwa 2.

    \(\frac{2n}{2}=\frac{−1\pm4}{2}\)

    \(n=\frac{−1\pm4}{2}\)
    Andika upya ili kuonyesha ufumbuzi mbili. \(n=\frac{−1+4}{2}\),\(n=\frac{−1−4}{2}\)
    Kurahisisha kila equation. \(n=\frac{3}{2}\),\(n=−\frac{5}{2}\)

    Angalia.
    .

    		  
    Mfano\(\PageIndex{37}\)

    Kutatua:\(9m^2−12m+4=25\).

    Jibu

    \(m=\frac{7}{3}\),\(m=−1\)

    Mfano\(\PageIndex{38}\)

    Kutatua:\(16n^2+40n+25=4\).

    Jibu

    \(n=−\frac{3}{4}\), \(n=−\frac{7}{4}\)

    Fikia rasilimali hizi za mtandaoni kwa maelekezo ya ziada na mazoezi na kutatua usawa wa quadratic:

    Dhana muhimu

    • Mizizi ya Mizizi ya mraba
      Ikiwa\(x^2=k\)\(k\ge 0\), na, basi\(x=\sqrt{k}\) au\(x=−\sqrt{k}\).

    faharasa

    quadratic equation
    equation quadratic ni equation ya fomu\(ax^2+bx+c=0\) ambapo\(a \ne 0\).
    Mizizi ya mraba Mali
    Mali ya Mizizi ya Mraba inasema kwamba\(x^2=k\), ikiwa\(k\ge 0\), na, basi\(x=\sqrt{k}\) au\(x=−\sqrt{k}\).