Whakaoti i te (3x+1)^2=9

$$( 3 x + 1 ) ^ { 2 } = 9$$

Whakaoti mō x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

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}.

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

Tangohia te 9 mai i ngā taha e rua.

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

Tangohia te 9 i te 1, ka -8.

$$9x^{2}+6x-8=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 9x^{2}+ax+bx-8. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=6 ab=9\left(-8\right)=-72$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Tātaihia te tapeke mō ia takirua.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

Ko te otinga te takirua ka hoatu i te tapeke 6.

$$a=-6 b=12$$

Tuhia anō te 9x^{2}+6x-8 hei \left(9x^{2}-6x\right)+\left(12x-8\right).

$$\left(9x^{2}-6x\right)+\left(12x-8\right)$$

Tauwehea te 3x i te tuatahi me te 4 i te rōpū tuarua.

$$3x\left(3x-2\right)+4\left(3x-2\right)$$

Whakatauwehea atu te kīanga pātahi 3x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-2\right)\left(3x+4\right)$$

Hei kimi otinga whārite, me whakaoti te 3x-2=0 me te 3x+4=0.

$$x=\frac{2}{3} x=-\frac{4}{3}$$

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}.

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

Tangohia te 9 mai i ngā taha e rua.

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

Tangohia te 9 i te 1, ka -8.

$$9x^{2}+6x-8=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, 6 mō b, me -8 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\left(-8\right)}}{2\times 9}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\times 9\left(-8\right)}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$x=\frac{-6±\sqrt{36-36\left(-8\right)}}{2\times 9}$$

Whakareatia -36 ki te -8.

$$x=\frac{-6±\sqrt{36+288}}{2\times 9}$$

Tāpiri 36 ki te 288.

$$x=\frac{-6±\sqrt{324}}{2\times 9}$$

Tuhia te pūtakerua o te 324.

$$x=\frac{-6±18}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{-6±18}{18}$$

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

$$x=\frac{12}{18}$$

Whakahekea te hautanga \frac{12}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$x=\frac{2}{3}$$

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

$$x=-\frac{24}{18}$$

Whakahekea te hautanga \frac{-24}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$x=-\frac{4}{3}$$

Kua oti te whārite te whakatau.

$$x=\frac{2}{3} x=-\frac{4}{3}$$

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}.

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

Tangohia te 1 mai i ngā taha e rua.

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

Tangohia te 1 i te 9, ka 8.

$$9x^{2}+6x=8$$

Whakawehea ngā taha e rua ki te 9.

$$\frac{9x^{2}+6x}{9}=\frac{8}{9}$$

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

$$x^{2}+\frac{6}{9}x=\frac{8}{9}$$

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

$$x^{2}+\frac{2}{3}x=\frac{8}{9}$$

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.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{8}{9}+\left(\frac{1}{3}\right)^{2}$$

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

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{8+1}{9}$$

Tāpiri \frac{8}{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.

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

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}.

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

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

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

Whakarūnātia.

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

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

$$x=\frac{2}{3} x=-\frac{4}{3}$$