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8.3: Kuzingatia na Kutafuta Solutions Polynomial (Zeroes)

  • Page ID
    164599
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    Kuna njia kadhaa za kupata ufumbuzi wa polynomials ambayo ni trinomials ya fomu\(ax^2 + bx + c = 0\). Ufumbuzi huu pia huitwa zero halisi za polynomials.

    1. Jaribio na Check Factoring Method: Kwa njia hii, lengo ni kujenga binomials mbili kwamba wakati kuzidisha pamoja, kusababisha trinomial kupewa. Njia hii inaweza kuwa vigumu sana wakati trinomial iliyotolewa ina maadili makubwa ya\(a\) na\(c\). Mara baada ya factoring ni kamili kupata zeroes wote halisi kwa kutumia zero sababu mali na kuweka kila sababu sawa\(0\) na na kutatua kwa\(x\).
    2. Factor kwa Grouping Factor Method: Kwa njia hii, lengo ni kujenga maneno manne kwa kugawanya muda wa kati katika maneno mawili, ambao coefficients wana bidhaa\(a ∗ c\) na kuwa na jumla ya\(b\). Utaratibu wa maneno ya kituo haijalishi. Mara baada ya maneno manne kuundwa, jozi masharti mawili ya kwanza na mabano, jozi maneno mawili ya pili na mabano, na uangalie GCF kutoka kwa jozi zote mbili. Binomial ya mara kwa mara ni sababu moja, na mambo ya GCF huchanganya ili kufanya binomial ya pili. Hii ni njia rahisi ya kutumia kwenye trinomial yoyote factorable ya fomu\(ax^2 + bx + c\), lakini inaweza kuwa na kidogo ya Curve kujifunza. Mara baada ya factoring ni kamili kupata zeroes wote halisi kwa kutumia zero sababu mali na kuweka kila sababu sawa\(0\) na na kutatua kwa\(x\).
    3. Mfumo wa Quadratic: Mfumo wa Quadratic unaweza kutumika kupata zero halisi za trinomial factorable. Tafadhali angalia Jedwali la Yaliyomo ili kupata sehemu inayoelezea jinsi ya kutumia Mfumo wa Quadratic.

    Fanya maneno kwa kutumia njia yoyote iliyojadiliwa katika sehemu hii (matatizo haya ya mfano yataonyesha Njia ya Kikundi):

    1. \(4x^2 − 3x − 10\)
    2. \(8x^2 − 2x − 3\)
    3. \(12x − 14x^3 + 22x^2\)
    4. \(\dfrac{(x^2 + 1)^2 (−2) + (2x)2(x^2 + 1)(2x)}{(x^2 + 1)^4}\)
    5. \(\dfrac{(2x + 1)^{\frac{1}{2}} − (x + 2)(2x + 1)^{-\frac{1}{2}}}{2x + 1}\)
    Suluhisho
    1. \(\begin{array} &&4x^2 − 3x − 10 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Example problem} \\ &4x^2 − 3x − 10 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Product \(ac\)ni\(4∗(−10) = −40\), Sum ni\(b = −3\). Kutumia sababu kwa makundi,}\\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ Nakala {maneno mawili ya kati yanahitajika ili kuzidisha kwa bidhaa ya\(−40\) na kuongeza jumla ya\(−3\).}\\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ Nakala {\(8\)na \(5\)ni wagombea mzuri; Kwa kuwa bidhaa lazima kuwa hasi, moja ya maadili haya lazima kuwa hasi.}\\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ bidhaa ni\(−40\) na jumla yao ni\(−3\).}\\ &\(−8\)\(5\) ; 4x^ 2 - 8x + 5x - 10 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ Nakala {Masharti manne, jumla ya maneno mawili ya kati ni ya awali ya muda wa kati,\\(−3x\)}\ & (4x^ 2 - 8x) + (5x - 10) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ x (x - 2) + 5 (x - 2) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ Nakala {Factor nje ya GCF kutoka kila jozi- mara kwa mara sababu ya binomial daima sasa}\\ & (4x + 5) (x - 2) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ Hakikisha uangalie kwa FOIL.} \ mwisho {safu}\)
    1. \(\begin{array} &&8x^2 − 2x − 3 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Example problem} \\ &8x^2 − 2x − 3 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Product \(ac\)ni\(8∗(−3) = −24\), Sum ni\(b = −2\). Kutumia sababu kwa makundi,}\\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ Nakala {maneno mawili ya kati yanahitajika ili kuzidisha kwa bidhaa ya\(−24\) na kuongeza jumla ya\(−2\).}\\ & &\;\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ Nakala {\(6\)na\(4\) ni nzuri wagombea; Kwa kuwa bidhaa lazima kuwa hasi, moja ya maadili haya lazima kuwa hasi.}\\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ &8x ^ 2 + 4x - 6x - 3 &\;\;\(−6\)\(4\)\(−24\)\(−2\) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ Nakala {Masharti manne, Jumla ya maneno mawili ya kati ni ya awali ya muda wa kati,\(−2x\).}\\ & & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ maandishi {Utaratibu wa maneno mawili katikati haijalishi.}\\ & (8x ^ 2 + 4x) + (-6x- 3) &\;\;\ ;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ maandishi {Unda jozi ya maneno. Kumbuka kuongeza katika kati ya mabano;}\\ & &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ Nakala {tatu ya maneno manne ilikuwa hasi hapa, hivyo ishara anakaa na neno.}\\ &4x (2x + 1) + (-3) (2x + 1) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ maandishi {Factor nje GCF kutoka kila jozi- mara kwa mara sababu binomial daima sasa}\\ & (4x - 3) (2x + 1) &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ Hakikisha uangalie kwa FOIL.} \ mwisho {safu}\)
    1. \(\begin{array} && 12x − 14x^3 + 22x^2 &\text{Example problem} \\ &−14x^3 + 22x^2 + 12x &\text{Reorder the terms in decreasing order of variable degree.} \\ &2x(−7x^2 + 11x + 6) &\text{Factor out the GCF so a trinomial results that can be factored using factor by grouping.} \\ & &\text{The GCF of \(2x\)itakuwa ni pamoja na katika jibu la mwisho, hivyo usisahau kuhusu hilo.}\\ &18-7x^2 + 11x + 6 &\ maandishi {Bidhaa\(ac\) ni\(−7 ∗ 6 = −42\), Sum ni\(b = 11\). Kutumia sababu kwa kikundi,}\\ & &\ Nakala {maneno mawili ya kati yanahitajika ili kuzidisha kwa bidhaa\(−42\) na kuongeza kwa jumla ya\(11\).}\\ & &\ maandishi {Hakuna namba zinazotimiza mahitaji haya yote,}\\ & &\ maandishi {ambayo ina maana kwamba trinomial haipatikani sababu integer.}\\ &18-7x ^ 2 + 11x + 6 &\ maandishi {Ili kupata mambo na sifuri za polynomial, tumia Mfumo wa Quadratic.}\\ & &\ maandishi {Hebu\(a = −7\)\(b = 11\),\(c = 6\)}\\ &x =\ dfrac {-11 ±\ sqrt {11^2 ‡ 4 (-7) (6)}} {2 (‡ 7)}} &\ maandishi {Mfumo wa Quadratic}\\ &x =\ dfrac {-11 ±\ sqrt {121 + 168}} {-14} &\ maandishi {Kurahisisha}\\ &x =\ dfrac {11 ±\ sqrt {289}} {14} &\ maandishi {Gawanya\(−1\) kutoka maneno yote}\\ &x =\ dfrac {11 ±\ sqrt {289}} {14} = 2,\;\; x =\ dfrac {11 ±\ sqrt {289}} {14} = -\ dfrac {3} {7} &\ maandishi {Majibu halisi kwa sifuri katika fomu radical, ikifuatiwa na idadi halisi aina.}\\ & (x - 2),\;\; (x + -\ dfrac {3} {7}) &\ maandishi {mambo. Jihadharini kuingiza sahihi\(±\) katika mambo.}\\ & &\ Nakala {Pata ufumbuzi, halafu mhandisi wa reverse kutambua sababu ambayo itazalisha suluhisho hilo.}\\ & &\ maandishi {Suluhisho la kwanza kutoka kwa Mfumo wa Quadratic lilikuwa\(x = 2\).}\\ & &\ maandishi {Sababu ya \((x − 2)\)wakati kuweka sawa na\(0\) itazalisha ufumbuzi wa\(x = 2\).}\\ & &\ Nakala {Suluhisho la pili kutoka Mfumo wa Quadratic lilikuwa\(x = −\dfrac{3}{7}\).}\\ & &\ maandishi {Sababu ya\((x + −\dfrac{3}{7})\) kuzalisha ufumbuzi wa\(x = −\dfrac{3}{7}\).}\\ &2x (x- 2) (x +\\ dfrac {3} {7}) &\ maandishi { Mambo ya polynomial, ikiwa ni pamoja na GCF ya awali ambayo ilibadilishwa mwanzoni mwa tatizo hili.} \ mwisho {safu}\)
    1. \(\begin{array} && \dfrac{(x^2 + 1)^2 (−2) + (2x)2(x^2 + 1)(2x)}{(x^2 + 1)^4} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Example problem} \\ &\dfrac{2(x^2 + 1)[(x^2 + 1)(−1) + (x)2(2x)]}{(x^2 + 1)(x^2 + 1)^3} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Factor out the GCF from the numerator.} \\ &\dfrac{2\cancel{(x^2 + 1)}[(x^2 + 1)(−1) + (x)2(2x)]}{\cancel{(x^2 + 1)}(x^2 + 1)^3} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Remove common factors.} \\ &\dfrac{2[(x^2 + 1)(−1) + (x)2(2x)]}{(x^2 + 1)^3} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Remove common factors.} \\ &\dfrac{2[−x^2 − 1 + 4x^2]}{(x^2 + 1)^3} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Simplify} \\ &\dfrac{2(3x^2 − 1)}{(x^2 + 1)^3} &\;\;\;\;\;\;\;\;\;\;\;\;\text{Final answer.} \end{array}\)
    1. \(\begin{array} &&\dfrac{(2x + 1)^{\frac{1}{2}} − (x + 2)(2x + 1)^{-\frac{1}{2}}}{2x + 1} &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Example problem} \\ &\dfrac{(2x + 1)^{\frac{1}{2}} −\dfrac{(x + 2)}{(2x + 1)^{\frac{1}{2}}}}{2x + 1} &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Write the expression with a positive exponent (move it to the denominator).} \\ &\dfrac{(2x + 1)^{\frac{1}{2}}(2x + 1)^{\frac{1}{2}} − (x + 2)}{\dfrac{(2x + 1)^{\frac{1}{2}}}{2x + 1}} &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Write the numerator with a common denominator.} \\ &\dfrac{2x + 1 − x − 2}{(2x + 1)^{\frac{1}{2}} (2x + 1)} &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Simplified.} \\ &\dfrac{2x + 1 − x − 2}{(2x + 1)^{\frac{3}{2}}} &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{Final answer.} \end{array}\)

    Sababu kwa kutumia njia yoyote iliyojadiliwa katika sehemu hii:

    1. \(5x^2 − 23x − 10\)
    2. \(8x^2 + 2x − 3\)
    3. \(3x^2 − 7x − 6\)
    4. \(10x^2 + 13x − 5\)
    5. \(12x^5 − 17x^4 + 6x^3\)
    6. \(\dfrac{(2x^2 − 1)^2 (−2) + (2x)2(2x^2 − 1)(2x)}{(2x^2 − 1)^4}\)
    7. \(\dfrac{2(2x − 3)^{\frac{1}{3}} − (x − 1)(2x − 3)^{-\frac{2}{3}}}{2x − 3^{\frac{2}{3}}}\)

    Mfumo wa Quadratic

    Ufafanuzi: Mfumo wa Quadratic

    Mfumo wa Quadratic hutumiwa kutatua (au kupata zero) ya Polynomial (Quadratic Equation) ya shahada\(2\) iliyo katika fomu\(ax^2 + bx + c = 0\). Mfumo wa Quadratic ni:

    \[x = \dfrac{−b ± \sqrt{b^2 − 4ac}}{2a} \nonumber \]

    ambapo\(a\),\(b\), na\(c\) ni coefficients ya fomu ya kawaida ya Equation Quadratic,\(ax^2 + bx + c = 0\).

    Kwa kazi zifuatazo, tafuta zero zote za\(f\) kutumia Mfumo wa Quadratic. Eleza jibu la mwisho kama majibu halisi (kwa fomu kali) na pia kama decimals, iliyozunguka mahali pa elfu.

    1. \(f(x) = −2x^2 + 4x − 1\)
    2. \(f(x) = x^3 − 3x^2 − 4x\)
    Suluhisho
    1. Weka\(f(x) = 0: −2x^2 + 4x − 1 = 0\). Kazi hii imeandikwa kwa fomu\(ax^2 + bx + c = 0\), na\(a = −2\),\(b = 4\) na\(c = −1\).

    Kubadilisha\(a\),\(b\) na\(c\) katika Mfumo wa Quadratic na maadili haya:

    \(\begin{array} &&x = \dfrac{−4 ± \sqrt{4^2 − 4(−2)(−1)}}{2(−2)} &\;\;\;\;\;\text{Quadratic Formula} \\ &x = \dfrac{−4 ± \sqrt{(16 − 8)}}{−4} &\;\;\;\;\;\text{Simplify} \\ &x = \dfrac{−4 ± \sqrt{8}}{−4} &\;\;\;\;\;\text{Simplify} \\ &x = \dfrac{−4 ± 2 \sqrt{2}}{−4} &\;\;\;\;\;\text{Simplify the radical} \\ &x = \dfrac{2 ± \sqrt{2}}{2} &\;\;\;\;\;\text{Exact answers in radical form} \\ &x = \dfrac{2 − \sqrt{2}}{2} ,\;\; x = \dfrac{2 + \sqrt{2}}{2} &\;\;\;\;\;\text{Exact answers written as two roots} \\ &x = 0.293 \text{ and } x = 1.707 &\;\;\;\;\;\text{Approximation answers rounded to the thousandths place} \end{array}\)

    1. Kazi\(f(x) = x^3 − 3x^2 − 4x\) ni kazi ya ujazo. Factor\(x\) kutoka maneno yote matatu kabla ya kutumia Quadratic Equation juu ya sababu trinomial:\(x(x^2 − 3x − 4) = 0\), na\(a = 1\),\(b = −3\) na\(c = −4\).

    Usisahau kwamba\(x\) kwamba alikuwa factored nje ni mzizi, yaani\(x = 0\).

    Kubadilisha\(a\),\(b\) na\(c\) katika Mfumo wa Quadratic na maadili haya:

    \(\begin{array} &&x = \dfrac{3 ± \sqrt{(−3)2 − 4(1)(−4)}}{2(1)} &\text{Quadratic Formula} \\ &x = \dfrac{3 ± \sqrt{(16 + 9)}}{2} &\text{Simplify} \\ &x = \dfrac{3 ± \sqrt{25}}{2} &\text{Simplify} \\ &x = \dfrac{3 ± 5}{2} &\text{Simplify further} \\ &x = \dfrac{3 − 5}{2} ,\;\;x = \dfrac{−2}{2} ,\;\; x = −1 &\text{Second root (first root is \(x = 0\))}\\ &x =\ dfrac {3 + 5} {2},\;\; x =\ dfrac {8} {2},\;\; x = 4 &\ maandishi {mizizi ya tatu}\ mwisho {safu}\)

    Kuna ufumbuzi tatu, au mizizi ya kazi ya ujazo\(f(x) = x^3 − 3x^2 − 4x: x = 0\),\(x = −1\) na\(x = 4\).

    Kwa kazi zifuatazo, tafuta zero zote za\(f\) kutumia Mfumo wa Quadratic. Eleza jibu la mwisho kama majibu halisi (kwa fomu kali) na pia kama decimals, iliyozunguka mahali pa elfu.

    1. \(f(t) = 9t^3 − 18t^2 + 6t\)
    2. \(f(x) = x^5 − 4x^4 − 32x^3\)
    3. \(f(x) = 18 − 3x − 2x^2\)
    4. \(f(x) = 12x^2 + 11x − 5\)
    5. \(f(x) = 3x^2 − 6x + 2\)