# 6.6: Ulinganifu wa Polynomial

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##### Malengo ya kujifunza

Mwishoni mwa sehemu hii, utaweza:

• Tumia mali ya Bidhaa ya Zero
• Tatua equations quadratic kwa factoring
• Tatua equations na kazi za polynomial
• Kutatua maombi yanayotokana na equations polynomial

Kabla ya kuanza, fanya jaribio hili la utayari.

1. Kutatua:$$5y−3=0$$.
Ikiwa umekosa tatizo hili, tathmini [kiungo].
2. Sababu kabisa:$$n^3−9n^2−22n$$.
Ikiwa umekosa tatizo hili, tathmini [kiungo].
3. Ikiwa$$f(x)=8x−16$$, tafuta$$f(3)$$ na usuluhishe$$f(x)=0$$.
Ikiwa umekosa tatizo hili, tathmini [kiungo].

Tumetumia muda mwingi kujifunza jinsi ya kuzingatia polynomials. Sasa tutaangalia equations polynomial na kutatua yao kwa kutumia factoring, ikiwa inawezekana.

Equation polynomial ni equation ambayo ina kujieleza polynomial. Kiwango cha equation ya polynomial ni kiwango cha polynomial.

##### MDUARA WA POLYNOMIAL

Equation polynomial ni equation ambayo ina kujieleza polynomial.

Kiwango cha equation ya polynomial ni kiwango cha polynomial.

Tayari tumetatua equations ya polynomial ya shahada moja. Ulinganifu wa polynomial wa shahada moja ni equations linear ni ya fomu$$ax+b=c$$.

Sisi sasa ni kwenda kutatua equations polynomial ya shahada mbili. Equation polynomial ya shahada mbili inaitwa equation quadratic. Imeorodheshwa hapa chini ni baadhi ya mifano ya equations quadratic:

$x^2+5x+6=0 \qquad 3y^2+4y=10 \qquad 64u^2−81=0 \qquad n(n+1)=42 \nonumber$

equation mwisho haionekani kuwa na squared variable, lakini wakati sisi kurahisisha kujieleza upande wa kushoto sisi kupata$$n^2+n$$.

Fomu ya jumla ya equation ya quadratic ni$$ax^2+bx+c=0$$, na$$a\neq 0$$. (Kama$$a=0$$, basi$$0·x^2=0$$ na sisi ni wa kushoto na hakuna muda quadratic.)

Equation ya fomu$$ax^2+bx+c=0$$ inaitwa equation quadratic.

$a,b,\text{ and }c\text{ are real numbers and }a\neq 0\nonumber$

Ili kutatua equations quadratic tunahitaji mbinu tofauti na yale tuliyotumia katika kutatua equations linear. Tutaangalia njia moja hapa na kisha wengine kadhaa katika sura ya baadaye.

## Tumia mali ya Bidhaa ya Zero

Sisi kwanza kutatua equations quadratic kwa kutumia Zero Bidhaa Mali. Mali ya Bidhaa ya Zero inasema kwamba ikiwa bidhaa ya kiasi mbili ni sifuri, basi angalau moja ya kiasi ni sifuri. Njia pekee ya kupata bidhaa sawa na sifuri ni kuzidisha kwa sifuri yenyewe.

##### ZERO BIDHAA MALI

Ikiwa$$a·b=0$$, basi ama$$a=0$$$$b=0$$ au wote wawili.

Sasa tutatumia mali ya Bidhaa ya Zero, ili kutatua equation ya quadratic.

##### Mfano$$\PageIndex{1}$$: How to Solve a Quadratic Equation Using the Zero Product Property

Kutatua:$$(5n−2)(6n−1)=0$$.

Jibu

##### Mfano$$\PageIndex{2}$$

Kutatua:$$(3m−2)(2m+1)=0$$.

Jibu

$$m=\frac{2}{3},\space m=−\frac{1}{2}$$

##### Mfano$$\PageIndex{3}$$

Kutatua:$$(4p+3)(4p−3)=0$$.

Jibu

$$p=−\frac{3}{4},\space p=\frac{3}{4}$$

##### TUMIA MALI YA BIDHAA YA SIFURI.
1. Weka kila sababu sawa na sifuri.
2. Tatua equations linear.
3. Angalia.

## Tatua Ulinganisho wa Quadratic kwa kuzingatia

Zero Bidhaa Mali kazi vizuri sana kutatua equations quadratic. Equation ya quadratic inapaswa kuzingatiwa, na sifuri pekee upande mmoja. Hivyo sisi kuwa na uhakika wa kuanza na equation quadratic katika hali ya kiwango,$$ax^2+bx+c=0$$. Kisha tunaelezea maneno upande wa kushoto.

Kutatua:$$2y^2=13y+45$$.

Jibu

##### Mfano$$\PageIndex{5}$$

Kutatua:$$3c^2=10c−8$$.

Jibu

$$c=2,\space c=\frac{4}{3}$$

##### Mfano$$\PageIndex{6}$$

Kutatua:$$2d^2−5d=3$$.

Jibu

$$d=3,\space d=−12$$

##### KUTATUA EQUATION QUADRATIC KWA FACTORING.
1. Andika equation quadratic katika fomu ya kawaida,$$ax^2+bx+c=0$$.
3. Tumia mali ya Bidhaa ya Zero.
4. Tatua equations linear.
5. Angalia. Badilisha kila suluhisho tofauti katika equation ya awali.

Kabla ya sisi sababu, ni lazima kuhakikisha equation quadratic ni katika hali ya kawaida.

Kutatua equations quadratic kwa factoring itatumia mbinu zote factoring umejifunza katika sura hii! Je, unatambua muundo maalum wa bidhaa katika mfano unaofuata?

##### Mfano$$\PageIndex{7}$$

Kutatua:$$169q^2=49$$.

Jibu

$$\begin{array} {ll} &169x^2=49 \\ \text{Write the quadratic equation in standard form.} &169x^2−49=0 \\ \text{Factor. It is a difference of squares.} &(13x−7)(13x+7)=0 \\ \text{Use the Zero Product Property to set each factor to }0. & \\ \text{Solve each equation.} &\begin{array} {ll} 13x−7=0 &13x+7=0 \\ 13x=7 &13x=−7 \\ x=\frac{7}{13} &x=−\frac{7}{13} \end{array} \end{array}$$

Angalia:

Tunaacha hundi kwako.

##### Mfano$$\PageIndex{8}$$

Kutatua:$$25p^2=49$$.

Jibu

$$p=\frac{7}{5},p=−\frac{7}{5}$$

##### Mfano$$\PageIndex{9}$$

Kutatua:$$36x^2=121$$.

Jibu

$$x=\frac{11}{6},x=−\frac{11}{6}$$

Katika mfano unaofuata, upande wa kushoto wa equation unafanywa, lakini upande wa kulia sio sifuri. Ili kutumia mali ya Bidhaa ya Zero, upande mmoja wa equation lazima iwe sifuri. Tutaweza kuzidisha sababu na kisha kuandika equation katika hali ya kiwango.

##### Mfano$$\PageIndex{10}$$

Kutatua:$$(3x−8)(x−1)=3x$$.

Jibu

$$\begin{array} {ll} &(3x−8)(x−1)=3x \\ \text{Multiply the binomials.} &3x^2−11x+8=3x \\ \text{Write the quadratic equation in standard form.} &3x^2−14x+8=0 \\ \text{Factor the trinomial.} &(3x−2)(x−4)=0 \\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} \end{array} &\begin{array} {ll} 3x−2=0 &x−4=0 \\ 3x=2 &x=4 \\ x=\frac{2}{3} & \end{array} \\ \text{Check your answers.} &\text{The check is left to you.} \end{array}$$

##### Mfano$$\PageIndex{11}$$

Kutatua:$$(2m+1)(m+3)=12m$$.

Jibu

$$m=1,\space m=\frac{3}{2}$$

##### Mfano$$\PageIndex{12}$$

Kutatua:$$(k+1)(k−1)=8$$.

Jibu

$$k=3,\space k=−3$$

Katika mfano unaofuata, tunapofanya equation ya quadratic tutapata mambo matatu. Hata hivyo, sababu ya kwanza ni mara kwa mara. Tunajua kwamba sababu haiwezi sawa 0.

##### Mfano$$\PageIndex{13}$$

Kutatua:$$3x^2=12x+63$$.

Jibu

$$\begin{array} {ll} &3x^2=12x+63 \\ \text{Write the quadratic equation in standard form.} &3x^2−12x−63=0 \\ \text{Factor the greatest common factor first.} &3(x^2−4x−21)=0 \\ \text{Factor the trinomial.} &3(x−7)(x+3)=0 \\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} \end{array} &\begin{array} {lll} 3\neq 0 &x−7=0 &x+3=0 \\ 3\neq 0 &x=7 &x=−3 \end{array} \\ \text{Check your answers.} &\text{The check is left to you.} \end{array}$$

##### Mfano$$\PageIndex{14}$$

Kutatua:$$18a^2−30=−33a$$.

Jibu

$$a=−\frac{5}{2},a=\frac{2}{3}$$

##### Mfano$$\PageIndex{15}$$

Kutatua:$$123b=−6−60b^2$$

Jibu

$$b=−2,\space b=−\frac{1}{20}$$

Mali ya Bidhaa ya Zero pia inatumika kwa bidhaa ya mambo matatu au zaidi. Ikiwa bidhaa ni sifuri, angalau moja ya mambo lazima iwe sifuri. Tunaweza kutatua equations baadhi ya shahada kubwa kuliko mbili kwa kutumia Zero Bidhaa Mali, kama sisi kutatuliwa equations quadratic.

##### Mfano$$\PageIndex{16}$$

Kutatua:$$9m^3+100m=60m^2$$

Jibu

$$\begin{array} {ll} & 9m^3+100m=60m^2 \\ \text{Bring all the terms to one side so that the other side is zero.} &9m^3−60m^2+100m=0 \\ \text{Factor the greatest common factor first.} &m(9m^2−60m+100)=0 \\ \text{Factor the trinomial.} &m(3m−10)^2=0 \end{array}\\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} &\begin{array} {lll} m=0 &3m−10=0 &{}\\ m=0 &m=\frac{10}{3} & {} \end{array}\\ \text{Check your answers.} &\text{The check is left to you.} \end{array}$$

##### Mfano$$\PageIndex{17}$$

Kutatua:$$8x^3=24x^2−18x$$.

Jibu

$$x=0,\space x=\frac{3}{2}$$

##### Mfano$$\PageIndex{18}$$

Kutatua:$$16y^2=32y^3+2y$$.

Jibu

$$y=0,\space y=14$$

## Tatua Ulinganisho na Kazi za Polynomial

Kama utafiti wetu wa kazi nyingi unaendelea, mara nyingi itakuwa muhimu kujua wakati kazi itakuwa na thamani fulani au ni pointi gani ziko kwenye grafu ya kazi. Kazi yetu na mali ya Bidhaa ya Zero itatusaidia kupata majibu haya.

##### Mfano$$\PageIndex{19}$$

Kwa kazi$$f(x)=x^2+2x−2$$,

ⓐ kupata$$x$$ wakati$$f(x)=6$$
ⓑ kupata pointi mbili kwamba uongo juu ya grafu ya kazi.

Jibu

$$\begin{array} {ll} &f(x)=x^2+2x−2 \\ \text{Substitute }6\text{ for }f(x). &6=x^2+2x−2 \\ \text{Put the quadratic in standard form.} &x^2+2x−8=0 \\ \text{Factor the trinomial.} &(x+4)(x−2)=0 \\ \begin{array} {l} \text{Use the zero product property.} \\ \text{Solve.} \end{array} &\begin{array} {lll} x+4=0 &\text{or} &x−2=0 \\ x=−4 &\text{or} &x=2 \end{array} \\ \text{Check:} & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ \begin{array} {lll} \quad &\hspace{3mm} f(x)=x^2+2x−2 &f(x)=x^2+2x−2 \\ \quad &f(−4)=(−4)^2+2(−4)−2 &f(2)=2^2+2·2−2 \\ \quad &f(−4)=16−8−2 &f(2)=4+4−2 \\ \quad &f(−4)=6\checkmark &f(2)=6\checkmark \end{array} & \end{array}$$

ⓑ Tangu$$f(−4)=6$$ na$$f(2)=6$$, pointi$$(−4,6)$$ na$$(2,6)$$ uongo juu ya grafu ya kazi.

##### Mfano$$\PageIndex{20}$$

Kwa kazi$$f(x)=x^2−2x−8$$,

ⓐ kupata$$x$$ wakati$$f(x)=7$$
ⓑ Kupata pointi mbili kwamba uongo juu ya graph ya kazi.

Jibu

$$x=−3$$ au$$x=5$$
$$(−3,7)\space (5,7)$$

##### Mfano$$\PageIndex{21}$$

Kwa kazi$$f(x)=x^2−8x+3$$,

ⓐ kupata$$x$$ wakati$$f(x)=−4$$
ⓑ Kupata pointi mbili kwamba uongo juu ya graph ya kazi.

Jibu

$$x=1$$ au$$x=7$$
$$(1,−4)\space (7,−4)$$

Zero Bidhaa Mali pia hutusaidia kuamua ambapo kazi ni sifuri. Thamani ya$$x$$ ambapo kazi ni$$0$$, inaitwa sifuri ya kazi.

##### ZERO YA KAZI

Kwa kazi yoyote$$f$$, ikiwa$$f(x)=0$$, basi$$x$$ ni sifuri ya kazi.

Wakati$$f(x)=0$$, hatua$$(x,0)$$ ni hatua kwenye grafu. Hatua hii ni$$x$$ - intercept ya grafu. Mara nyingi ni muhimu kujua ambapo grafu ya kazi huvuka axes. Tutaona baadhi ya mifano baadaye.

##### Mfano$$\PageIndex{22}$$

Kwa kazi$$f(x)=3x^2+10x−8$$, tafuta

ⓐ zero za kazi,
ⓑ yoyote$$x$$ -intercepts ya grafu ya kazi
ⓒ yoyote$$y$$ -intercepts ya grafu ya kazi

Jibu

ⓐ Ili kupata zero za kazi, tunahitaji kupata wakati thamani ya kazi ni 0.
$$\begin{array} {ll} &f(x)=3x^2+10x−8 \\ \text{Substitute }0\text{ for}f(x). &0=3x^2+10x−8 \\ \text{Factor the trinomial.} &(x+4)(3x−2)=0 \\ \begin{array} {l} \text{Use the zero product property.} \\ \text{Solve.} \end{array} &\begin{array} {lll} x+4=0 &\text{or} &3x−2=0 \\ x=−4 &\text{or} &x=\frac{2}{3} \end{array} \end{array}$$

$$x$$ -intercept hutokea wakati$$y=0$$. Tangu$$f(−4)=0$$ na$$f(\frac{2}{3})=0$$, pointi$$(−4,0)$$ na$$(\frac{2}{3},0)$$ uongo kwenye grafu. Hizi pointi ni$$x$$ -intercepts ya kazi.

ⓒ A$$y$$ -intercept hutokea wakati$$x=0$$. Ili kupata$$y$$ -intercepts tunahitaji kupata$$f(0)$$.
$$\begin{array} {ll} &f(x)=3x^2+10x−8 \\ \text{Find }f(0)\text{ by substituting }0\text{ for }x. &f(0)=3·0^2+10·0−8 \\ \text{Simplify.} &f(0)=−8 \end{array}$$
Tangu$$f(0)=−8$$, hatua hiyo$$(0,−8)$$ iko kwenye grafu. Hatua hii ni$$y$$ -intercept ya kazi.

##### Mfano$$\PageIndex{23}$$

Kwa kazi$$f(x)=2x^2−7x+5$$, tafuta

ⓐ zero za kazi
ⓑ yoyote$$x$$ -intercepts ya grafu ya kazi
ⓒ yoyote$$y$$ -intercepts ya grafu ya kazi.

Jibu

$$x=1$$ au$$x=\frac{5}{2}$$
$$(1,0),\space (\frac{5}{2},0)$$$$(0,5)$$

##### Mfano$$\PageIndex{24}$$

Kwa kazi$$f(x)=6x^2+13x−15$$, tafuta

ⓐ zero za kazi
ⓑ yoyote$$x$$ -intercepts ya grafu ya kazi
ⓒ yoyote$$y$$ -intercepts ya grafu ya kazi.

Jibu

$$x=−3$$ au$$x=\frac{5}{6}$$
$$(−3,0),\space (\frac{5}{6},0)$$$$(0,−15)$$

## Kutatua Maombi yanayotokana na Ulinganifu wa Polynomial

Mkakati wa kutatua matatizo tuliyotumia mapema kwa maombi ambayo hutafsiri kwa equations linear itafanya kazi vizuri kwa maombi ambayo hutafsiri kwa milinganyo ya polynomial. Tutaiga mkakati wa kutatua matatizo hapa ili tuweze kuitumia kwa kumbukumbu.

##### KUTUMIA TATIZO KUTATUA MKAKATI WA KUTATUA MATATIZO NENO.
1. Soma tatizo. Hakikisha maneno yote na mawazo yanaeleweka.
2. Tambua kile tunachotafuta.
3. Jina kile tunachotafuta. Chagua variable kuwakilisha kiasi hicho.
4. Tafsiri katika equation. Inaweza kuwa na manufaa kurejesha tatizo katika sentensi moja na taarifa zote muhimu. Kisha, tafsiri sentensi ya Kiingereza kwenye equation ya algebraic.
5. Kutatua equation kutumia mbinu sahihi algebra.
6. Angalia jibu katika tatizo na uhakikishe kuwa ni busara.
7. Jibu swali kwa sentensi kamili.

Tutaanza na tatizo namba kupata mazoezi kutafsiri maneno katika equation polynomial.

##### Mfano$$\PageIndex{25}$$

Bidhaa ya integers mbili isiyo ya kawaida ya mfululizo ni 323. Pata integers.

Jibu

$$\begin{array} {ll} \textbf{Step 1. Read }\text{the problem.} & \\ \textbf{Step 2. Identify }\text{what we are looking for.} &\text{We are looking for two consecutive integers.} \\ \textbf{Step 3. Name}\text{ what we are looking for.} &\text{Let } n=\text{ the first integer.} \\ &n+2= \text{ next consecutive odd integer} \\ \begin{array} {l} \textbf{Step 4. Translate }\text{into an equation. Restate the}\hspace{20mm} \\ \text{problem in a sentence.} \end{array} &\begin{array} {l} \text{The product of the two consecutive odd} \\ \text{integers is }323. \end{array} \\ &\quad n(n+2)=323 \\ \textbf{Step 5. Solve }\text{the equation.} n^2+2n=323 \\ \text{Bring all the terms to one side.} &n^2+2n−323=0 \\ \text{Factor the trinomial.} &(n−17)(n+19)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve the equations.} \end{array} &\begin{array} {ll} n−17=0 \hspace{10mm}&n+19=0 \\ n=17 &n=−19 \end{array} \end{array}$$
Kuna maadili mawili kwa$$n$$ kuwa ni ufumbuzi wa tatizo hili. Hivyo kuna seti mbili ya integers mfululizo isiyo ya kawaida ambayo kazi.

$$\begin{array} {ll} \text{If the first integer is } n=17 \hspace{60mm} &\text{If the first integer is } n=-19 \\ \text{then the next odd integer is} &\text{then the next odd integer is} \\ \hspace{53mm} n+2 &\hspace{53mm} n+2 \\ \hspace{51mm} 17+2 &\hspace{51mm} -19+2 \\ \hspace{55mm} 19 &\hspace{55mm} -17 \\ \hspace{51mm} 17,19 &\hspace{51mm} -17,-19 \\ \textbf{Step 6. Check }\text{the answer.} & \\ \text{The results are consecutive odd integers} & \\ \begin{array} {ll} 17,\space 19\text{ and }−19,\space −17. & \\ 17·19=323\checkmark &−19(−17)=323\checkmark \end{array} & \\ \text{Both pairs of consecutive integers are solutions.} & \\ \textbf{Step 7. Answer }\text{the question} &\text{The consecutive integers are }17, 19\text{ and }−19,−17. \end{array}$$

##### Mfano$$\PageIndex{26}$$

Bidhaa ya integers mbili isiyo ya kawaida ya mfululizo ni 255. Pata integers.

Jibu

$$−15,−17$$na$$15, 17$$

##### Mfano$$\PageIndex{27}$$

Bidhaa ya integers mbili za mfululizo isiyo ya kawaida ni 483 Pata integers.

Jibu

$$−23,−21$$na$$21, 23$$

Je, ulishangaa na jozi ya integers hasi ambayo ni moja ya ufumbuzi wa mfano uliopita? Bidhaa ya integers mbili nzuri na bidhaa za integers mbili hasi zote hutoa matokeo mazuri.

Katika baadhi ya programu, ufumbuzi hasi utatokana na algebra, lakini haitakuwa kweli kwa hali hiyo.

##### Mfano$$\PageIndex{28}$$

Chumba cha kulala cha mstatili kina eneo la futi za mraba 117. Urefu wa chumba cha kulala ni miguu minne zaidi ya upana. Pata urefu na upana wa chumba cha kulala.

Jibu
 Hatua ya 1. Soma tatizo. Katika matatizo yanayohusisha takwimu za kijiometri, mchoro unaweza kukusaidia kutazama hali hiyo. Hatua ya 2. Tambua unachotafuta. Tunatafuta urefu na upana. Hatua ya 3. Jina unachotafuta. Hebu$$w=\text{ the width of the bedroom}$$. Urefu ni miguu minne zaidi ya upana. $$w+4=\text{ the length of the garden}$$ Hatua ya 4. Tafsiri katika equation. Rejesha habari muhimu katika sentensi. Eneo la chumba cha kulala ni futi za mraba 117. Tumia formula kwa eneo la mstatili. $$A=l·w$$ Mbadala katika vigezo. $$117=(w+4)w$$ Hatua ya 5. Tatua equation Kusambaza kwanza. $$117=w^2+4w$$ Pata sifuri upande mmoja. $$117=w^2+4w$$ Sababu ya trinomial. $$0=w^2+4w−117$$ Tumia mali ya Bidhaa ya Zero. $$0=(w^2+13)(w−9)$$ Kutatua kila equation. $$0=w+13\quad 0=w−9$$ Kwa kuwa$$w$$ ni upana wa chumba cha kulala, haina maana kwa kuwa hasi. Sisi kuondoa kwamba thamani kwa$$w$$. $$\cancel{w=−13}$$$$\quad w=9$$ $$w=9$$Upana ni miguu 9. Pata thamani ya urefu. $$w+4$$ $$9+4$$ 13 Urefu ni futi 13. Hatua ya 6. Angalia jibu. Je! Jibu lina maana? Ndiyo, hii ina maana. Hatua ya 7. Jibu swali. Upana wa chumba cha kulala ni miguu 9 na urefu ni futi 13.
##### Mfano$$\PageIndex{29}$$

Ishara ya mstatili ina eneo la miguu ya mraba 30. Urefu wa ishara ni mguu mmoja zaidi ya upana. Pata urefu na upana wa ishara.

Jibu

Upana ni futi 5 na urefu ni futi 6.

##### Mfano$$\PageIndex{30}$$

Patio ya mstatili ina eneo la miguu ya mraba 180. Upana wa patio ni miguu mitatu chini ya urefu. Pata urefu na upana wa patio.

Jibu

Urefu wa patio ni futi 12 na upana wa miguu 15.

Katika mfano unaofuata, tutatumia Theorem ya Pythagorean$$(a^2+b^2=c^2)$$. Fomu hii inatoa uhusiano kati ya miguu na hypotenuse ya pembetatu sahihi.

Tutatumia formula hii kwa mfano unaofuata.

##### Mfano$$\PageIndex{31}$$

Meli ya mashua iko katika umbo la pembetatu ya kulia kama inavyoonekana. Hypotenuse itakuwa urefu wa miguu 17. Urefu wa upande mmoja utakuwa chini ya miguu 7 kuliko urefu wa upande mwingine. Pata urefu wa pande za meli.

Jibu
 Hatua ya 1. Soma tatizo Hatua ya 2. Tambua unachotafuta. Tunatafuta urefu wa pande za meli. Hatua ya 3. Jina unachotafuta. Upande mmoja ni 7 chini ya nyingine. Hebu$$x=\text{ length of a side of the sail}$$. $$x−7=\text{ length of other side}$$ Hatua ya 4. Tafsiri katika equation. Kwa kuwa hii ni pembetatu sahihi tunaweza kutumia Theorem ya Pythagorean. $$a^2+b^2=c^2$$ Mbadala katika vigezo. $$x^2+(x−7)^2=17^2$$ Hatua ya 5. Kutatua equation Kurahisisha. $$x^2+x^2−14x+49=289$$ $$2x^2−14x+49=289$$ Ni equation quadratic, hivyo kupata sifuri upande mmoja. $$2x^2−14x−240=0$$ Factor sababu kubwa ya kawaida. $$2(x^2−7x−120)=0$$ Sababu ya trinomial. $$2(x−15)(x+8)=0$$ Tumia mali ya Bidhaa ya Zero. $$2\neq 0\quad x−15=0\quad x+8=0$$ Kutatua. $$2\neq 0\quad x=15\quad x=−8$$ Kwa kuwa$$x$$ ni upande wa pembetatu,$$x=−8$$ haina maana. $$2\neq 0\quad x=15\quad \cancel{x=−8}$$ Pata urefu wa upande mwingine. Ikiwa urefu wa upande mmoja ni basi urefu wa upande mwingine ni 8 ni urefu wa upande mwingine. Hatua ya 6. Angalia jibu katika tatizo Je! Nambari hizi zina maana? Hatua ya 7. Jibu swali Pande za meli ni futi 8, 15 na 17.
##### Mfano$$\PageIndex{32}$$

Justine anataka kuweka staha katika kona ya mashamba yake kwa umbo la pembetatu ya kulia. Urefu wa upande mmoja wa staha ni miguu 7 zaidi kuliko upande mwingine. Hypotenuse ni 13. Pata urefu wa pande mbili za staha.

Jibu

Miguu 5 na miguu 12

##### Mfano$$\PageIndex{33}$$

Bustani ya kutafakari iko katika sura ya pembetatu ya kulia, na mguu mmoja wa miguu 7. Urefu wa hypotenuse ni moja zaidi ya urefu wa mguu mwingine. Pata urefu wa hypotenuse na mguu mwingine.

Jibu

Miguu 24 na miguu 25

Mfano unaofuata unatumia kazi inayotoa urefu wa kitu kama kazi ya wakati unapotupwa kutoka futi 80 juu ya ardhi.

##### Mfano$$\PageIndex{34}$$

Dennis ni kwenda kutupa mpira wake mpira bendi ya juu kutoka juu ya jengo chuo. Wakati yeye kumtupia mpira bendi mpira kutoka 80 miguu juu ya ardhi, kazi$$h(t)=−16t^2+64t+80$$ mifano urefu$$h$$,, ya mpira juu ya ardhi kama kazi ya muda,$$t$$. Kupata:

ⓐ zero za kazi hii ambayo inatuambia wakati mpira unapopiga ardhi
ⓑ wakati mpira utakuwa na miguu 80 juu ya ardhi
ⓒ urefu wa mpira kwa$$t=2$$ sekunde.

Jibu

ⓐ Zero za kazi hii zinapatikana kwa kutatua$$h(t)=0$$. Hii itatuambia wakati mpira utapiga chini.
$$\begin{array} {ll} &h(t)=0 \\ \text{Substitute in the polynomial for }h(t). &−16t^2+64t+80=0 \\ \text{Factor the GCF, }−16. &−16(t^2−4t−5)=0 \\ \text{Factor the trinomial.} &−16(t−5)(t+1)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve.} \end{array} &\begin{array} {ll} t−5=0 &t+1=0 \\ t=5 &t=−1 \end{array} \end{array}$$

Matokeo$$t=5$$ inatuambia mpira utapiga chini sekunde 5 baada ya kutupwa. Tangu wakati hauwezi kuwa hasi, matokeo$$t=−1$$ yameondolewa.

ⓑ mpira itakuwa 80 miguu juu ya ardhi wakati$$h(t)=80$$.
$$\begin{array} {ll} &h(t)=80 \\ \text{Substitute in the polynomial for }h(t). &−16t^2+64t+80=80 \\ \text{Subtract 80 from both sides.} &−16t^2+64t=0 \\ \text{Factor the GCF, }−16t. &−16t(t−4)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve.}\end{array} &\begin{array} {ll} −16t=0 &t−4=0 \\ t=0 &t=4 \end{array} \\ &\text{The ball will be at 80 feet the moment Dennis} \\ &\text{tosses the ball and then 4 seconds later, when} \\ &\text{the ball is falling.} \end{array}$$

ⓒ Ili kupata mpira wa urefu kwa$$t=2$$ sekunde tunapata$$h(2)$$.
$$\begin{array} {ll} &h(t)=−16t^2+64t+80 \\ \text{To find }h(2)\text{ substitute }2\text{ for }t. &h(2)=−16(2)^2+64·2+80 \\ \text{Simplify.} &h(2)=144 \\ &\text{After 2 seconds, the ball will be at 144 feet.} \end{array}$$

##### Mfano$$\PageIndex{35}$$

Genevieve ni kwenda kutupa mwamba kutoka juu uchaguzi unaoelekea bahari. Wakati yeye throws mwamba zaidi kutoka 160 miguu juu ya bahari, kazi$$h(t)=−16t^2+48t+160$$ mifano urefu$$h$$,, ya mwamba juu ya bahari kama kazi ya muda,$$t$$. Kupata:

ⓐ zeros ya kazi hii ambayo kutuambia wakati mwamba hit bahari
ⓑ wakati mwamba itakuwa 160 miguu juu ya bahari.
ⓒ urefu wa mwamba kwa$$t=1.5$$ sekunde.

Jibu

ⓐ 5 ⓑ 0; 3 ⓒ 196

##### Mfano$$\PageIndex{36}$$

Calib ni kwenda kutupa senti yake bahati kutoka balcony yake juu ya meli cruise. Wakati yeye kumtupia senti zaidi kutoka 128 miguu juu ya ardhi, kazi$$h(t)=−16t^2+32t+128$$ mifano urefu$$h$$,, ya senti juu ya bahari kama kazi ya muda,$$t$$. Kupata:

ⓐ zeros ya kazi hii ambayo ni wakati senti hit bahari
ⓑ wakati senti itakuwa 128 miguu juu ya bahari.
ⓒ urefu senti itakuwa$$t=1$$ sekunde ambayo ni wakati senti itakuwa katika hatua yake ya juu.

Jibu

ⓐ 4 ⓑ 0; 2 ⓒ 144

Fikia rasilimali hii ya mtandaoni kwa maelekezo ya ziada na mazoezi na usawa wa quadratic.

## Dhana muhimu

• Ulinganifu wa Polynomial: equation ya polynomial ni equation ambayo ina usemi wa polynomial. Kiwango cha equation ya polynomial ni kiwango cha polynomial.
• Quadratic Equation: Equation ya fomu$$ax^2+bx+c=0$$ inaitwa equation quadratic.

$a,b,c\text{ are real numbers and } a\neq 0\nonumber$

• Zero Bidhaa Mali: Kama$$a·b=0$$, basi ama$$a=0$$$$b=0$$ au wote wawili.
• Jinsi ya kutumia mali ya Bidhaa Zero
1. Weka kila sababu sawa na sifuri.
2. Tatua equations linear.
3. Angalia.
• Jinsi ya kutatua equation quadratic kwa factoring.
1. Andika equation quadratic katika fomu ya kawaida,$$ax^2+bx+c=0$$.
• Zero ya Kazi: Kwa kazi yoyote$$f$$, kama$$f(x)=0$$, basi$$x$$ ni sifuri ya kazi.