Whakaoti i te -(-36)/left(1+3xright)^2=1/3

$$- \frac{ (-36) }{ { \left(1+3x \right) }^{ 2 } } = \frac{ 1 }{ 3 }$$

Whakaoti mō x

$$x=-2\sqrt{3}-\frac{1}{3}\approx -3.797434948$$

$$x=2\sqrt{3}-\frac{1}{3}\approx 3.130768282$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki -\frac{1}{3} nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 3\left(3x+1\right)^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o \left(1+3x\right)^{2},3.

$$-3\left(-36\right)=\left(3x+1\right)^{2}$$

Whakareatia te -3 ki te -36, ka 108.

$$108=\left(3x+1\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(3x+1\right)^{2}.

$$108=9x^{2}+6x+1$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$9x^{2}+6x+1=108$$

Tangohia te 108 mai i ngā taha e rua.

$$9x^{2}+6x+1-108=0$$

Tangohia te 108 i te 1, ka -107.

$$9x^{2}+6x-107=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, 6 mō b, me -107 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\left(-107\right)}}{2\times 9}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\times 9\left(-107\right)}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$x=\frac{-6±\sqrt{36-36\left(-107\right)}}{2\times 9}$$

Whakareatia -36 ki te -107.

$$x=\frac{-6±\sqrt{36+3852}}{2\times 9}$$

Tāpiri 36 ki te 3852.

$$x=\frac{-6±\sqrt{3888}}{2\times 9}$$

Tuhia te pūtakerua o te 3888.

$$x=\frac{-6±36\sqrt{3}}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{-6±36\sqrt{3}}{18}$$

Nā, me whakaoti te whārite x=\frac{-6±36\sqrt{3}}{18} ina he tāpiri te ±. Tāpiri -6 ki te 36\sqrt{3}.

$$x=\frac{36\sqrt{3}-6}{18}$$

Whakawehe -6+36\sqrt{3} ki te 18.

$$x=2\sqrt{3}-\frac{1}{3}$$

Nā, me whakaoti te whārite x=\frac{-6±36\sqrt{3}}{18} ina he tango te ±. Tango 36\sqrt{3} mai i -6.

$$x=\frac{-36\sqrt{3}-6}{18}$$

Whakawehe -6-36\sqrt{3} ki te 18.

$$x=-2\sqrt{3}-\frac{1}{3}$$

Kua oti te whārite te whakatau.

$$x=2\sqrt{3}-\frac{1}{3} x=-2\sqrt{3}-\frac{1}{3}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki -\frac{1}{3} nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 3\left(3x+1\right)^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o \left(1+3x\right)^{2},3.

$$-3\left(-36\right)=\left(3x+1\right)^{2}$$

Whakareatia te -3 ki te -36, ka 108.

$$108=\left(3x+1\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(3x+1\right)^{2}.

$$108=9x^{2}+6x+1$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$9x^{2}+6x+1=108$$

Tangohia te 1 mai i ngā taha e rua.

$$9x^{2}+6x=108-1$$

Tangohia te 1 i te 108, ka 107.

$$9x^{2}+6x=107$$

Whakawehea ngā taha e rua ki te 9.

$$\frac{9x^{2}+6x}{9}=\frac{107}{9}$$

Mā te whakawehe ki te 9 ka wetekia te whakareanga ki te 9.

$$x^{2}+\frac{6}{9}x=\frac{107}{9}$$

Whakahekea te hautanga \frac{6}{9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$x^{2}+\frac{2}{3}x=\frac{107}{9}$$

Whakawehea te \frac{2}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{3}. Nā, tāpiria te pūrua o te \frac{1}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{107}{9}+\left(\frac{1}{3}\right)^{2}$$

Pūruatia \frac{1}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{107+1}{9}$$

Tāpiri \frac{107}{9} ki te \frac{1}{9} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=12$$

Tauwehea te x^{2}+\frac{2}{3}x+\frac{1}{9}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=12$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{12}$$

Whakarūnātia.

$$x+\frac{1}{3}=2\sqrt{3} x+\frac{1}{3}=-2\sqrt{3}$$

Me tango \frac{1}{3} mai i ngā taha e rua o te whārite.

$$x=2\sqrt{3}-\frac{1}{3} x=-2\sqrt{3}-\frac{1}{3}$$