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Ufumbuzi - Kutatua equations za quadratic kwa kukamilisha mraba

p_{1} = 0 + \frac{\sqrt{6}}{3} \cdot i

p_1=0+\frac{\sqrt{6}}{3}\cdoti

p_{2} = 0 - \frac{\sqrt{6}}{3} \cdot i

p_2=0-\frac{\sqrt{6}}{3}\cdoti

Tazama hatua

Maelezo kwa hatua

1. Hamisha masharti yote upande wa kushoto wa equation

$9 p^{2} + 7 = 1$

Punguza -1 kwa pande zote mbili:

$9 p^{2} + 7 - 1 = 1 - 1$

Rahisisha usemi

$9 p^{2} + 6 = 0$

2. Tambua coefficients

Tumia fomu standard ya equation ya quadratic, $a x^{2} + b x + c = 0$ , ili kupata coefficients:

$9 p^{2} + 6 = 0$

$a = 9$
$b = 0$
$c = 6$

3. Fanya coefficient a iwe sawa na 1

Kwa sababu $a = 9$ , gawa coefficients zote na constants kwa pande zote za equation na $9$ :

$9 p^{2} + 0 p + 6 = 0$

$\frac{9}{9} p^{2} + 0 \frac{p}{9} + \frac{6}{9} = \frac{0}{9}$

Rahisisha usemi

$p^{2} + 0 p + \frac{2}{3} = 0$

Vigawanyaji ni:
$a = 1$
$b = 0$
$c = \frac{2}{3}$

4. Hamisha constant upande wa kulia wa equation na unganisha

Ongeza $\frac{2}{3}$ kwa pande zote za equation:

$p^{2} + 0 p + \frac{2}{3} = 0$

$p^{2} + 0 p + \frac{2}{3} - \frac{2}{3} = 0 - \frac{2}{3}$

$p^{2} + 0 p = - \frac{2}{3}$

5. Kamilisha square

Ili kufanya upande wa kushoto wa equation kuwa perfect square trinomial, ongeza constant mpya sawa na ${(\frac{b}{2})}^{2}$ kwa equation:

$b = 0$

${(\frac{b}{2})}^{2} = {(\frac{0}{2})}^{2}$

Tumia kanuni ya kipeo cha sehemu ${(\frac{x}{y})}^{2} = \frac{x^{2}}{y^{2}}$

${(\frac{0}{2})}^{2} = \frac{0^{2}}{2^{2}}$

$\frac{0^{2}}{2^{2}} = \frac{0}{4}$

$\frac{0}{4} = 0$

Ongeza $0$ kwa pande zote za equation:

$p^{2} + 0 p = - \frac{2}{3}$

$p^{2} + 0 p + 0 = - \frac{2}{3} + 0$

Ondoa kuongeza sifuri:

$p^{2} + 0 p + 0 = \frac{- 2}{3}$

Sasa tuna perfect square trinomial, tunaweza kuandika kama fomu ya perfect square kwa kuongezea nusu ya coefficient ya $b$ , $\frac{b}{2}$ :
$b = 0$

$\frac{b}{2} = \frac{0}{2}$

Punguza kigezo kisicho na maana:

$\frac{b}{2} = 0$

$p^{2} + 0 p + 0 = \frac{- 2}{3}$

${(p + 0)}^{2} = - \frac{2}{3}$

6. Suluhisha kwa $x$

Chukua square root ya pande zote za equation: IMPORTANT: Wakati wa kutafuta square root ya constant, tunapata solutions mbili: positive na negative

${(p + 0)}^{2} = - \frac{2}{3}$

$\sqrt{{(p + 0)}^{2}} = \sqrt{- \frac{2}{3}}$

Futa mraba na mzizi mraba kwa upande wa kushoto wa equation:

$p + 0 = \pm \sqrt{- \frac{2}{3}}$

Toa kutoka pande zote mbili

$p + 0 + 0 = \pm \sqrt{- \frac{2}{3}}$

Rahisisha upande wa kushoto:

$p = \pm \sqrt{- \frac{2}{3}}$

Mzizi wa namba hasi haupo katika seti ya Nambari Halisi. Tunaanzisha nambari fikiria "i", ambayo ni mzizi wa moja hasi. $\sqrt{(- 1)} = i$

$p = 0 \pm \sqrt{\frac{2}{3}} \cdot \sqrt{- 1}$

$p = 0 \pm \sqrt{\frac{2}{3}} \cdot i$

$p = 0 \pm \frac{\sqrt{2}}{\sqrt{3}} \cdot i$

$p = 0 \pm \frac{\sqrt{2} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} \cdot i$

$p = 0 \pm \frac{\sqrt{6}}{3} \cdot i$

$p_{1} = 0 + \frac{\sqrt{6}}{3} \cdot i$
$p_{2} = 0 - \frac{\sqrt{6}}{3} \cdot i$

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Kwa nini kujifunza hii

Katika kazi yao msingi zaidi, equation mraba zinafafanua umbo kama duru, elipisi na parabolas. Hizi maumbo zinaweza kutumika kutabiri njia ya kitu katika harakati, kama mpira uliyopigwa na mchezaji wa mpira wa miguu au uliyotupwa kutoka kwenye kanyonya.
Linapokuja suala la harakati ya kitu kwenye anga, mahali pazuri pa kuanza ni anga yenyewe, na mapinduzi ya sayari kuzunguka jua katika mfumo wetu wa jua. Equation mraba ilitumika kuanzisha kuwa mizunguko ya sayari ni elliptical, si ya mduara. Kutambua njia na speed kitu safiri kwenye anga inawezekana hata baada ya kitu kuja kuacha: the equation mraba inaweza kuhesabu speed gari ilikuwa inaendesha wakati ilipata ajali. Kwa informesheni kama hii, sekta ya magari inaweza kubuni breki kuzuia ajali kwa siku za usoni. Viwanda vingi vinatumia equation mraba kutabiri na hivyo kuboresha muda wa maisha ya bidhaa zao na usalama.

Vigezo na mada

Kutatua milinganyo ya quadratiki kwa kukamilisha mraba