Whakaoti i te (3x+1)^2=4

$$( 3 x + 1 ) ^ { 2 } = 4$$

Whakaoti mō x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

$$9x^{2}+6x+1=4$$

Tangohia te 4 mai i ngā taha e rua.

$$9x^{2}+6x+1-4=0$$

Tangohia te 4 i te 1, ka -3.

$$9x^{2}+6x-3=0$$

Whakawehea ngā taha e rua ki te 3.

$$3x^{2}+2x-1=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 3x^{2}+ax+bx-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=2 ab=3\left(-1\right)=-3$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-1 b=3$$

Tuhia anō te 3x^{2}+2x-1 hei \left(3x^{2}-x\right)+\left(3x-1\right).

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

Whakatauwehea atu x i te 3x^{2}-x.

$$x\left(3x-1\right)+3x-1$$

Whakatauwehea atu te kīanga pātahi 3x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-1\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te 3x-1=0 me te x+1=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$9x^{2}+6x+1=4$$

Tangohia te 4 mai i ngā taha e rua.

$$9x^{2}+6x+1-4=0$$

Tangohia te 4 i te 1, ka -3.

$$9x^{2}+6x-3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, 6 mō b, me -3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\left(-3\right)}}{2\times 9}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\times 9\left(-3\right)}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$x=\frac{-6±\sqrt{36-36\left(-3\right)}}{2\times 9}$$

Whakareatia -36 ki te -3.

$$x=\frac{-6±\sqrt{36+108}}{2\times 9}$$

Tāpiri 36 ki te 108.

$$x=\frac{-6±\sqrt{144}}{2\times 9}$$

Tuhia te pūtakerua o te 144.

$$x=\frac{-6±12}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{-6±12}{18}$$

Nā, me whakaoti te whārite x=\frac{-6±12}{18} ina he tāpiri te ±. Tāpiri -6 ki te 12.

$$x=\frac{6}{18}$$

Whakahekea te hautanga \frac{6}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$x=\frac{1}{3}$$

Nā, me whakaoti te whārite x=\frac{-6±12}{18} ina he tango te ±. Tango 12 mai i -6.

$$x=-\frac{18}{18}$$

Whakawehe -18 ki te 18.

$$x=-1$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

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

$$9x^{2}+6x+1=4$$

Tangohia te 1 mai i ngā taha e rua.

$$9x^{2}+6x=4-1$$

Tangohia te 1 i te 4, ka 3.

$$9x^{2}+6x=3$$

Whakawehea ngā taha e rua ki te 9.

$$\frac{9x^{2}+6x}{9}=\frac{3}{9}$$

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

$$x^{2}+\frac{6}{9}x=\frac{3}{9}$$

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

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

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

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

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.

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

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

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

Tāpiri \frac{1}{3} 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.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{4}{9}$$

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

$$\left(x+\frac{1}{3}\right)^{2}=\frac{4}{9}$$

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

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Whakarūnātia.

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

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

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