Whakaoti i te =x^2-2x-8/(x-1)^2=-1

$$= \frac { x ^ { 2 } - 2 x - 8 } { ( x - 1 ) ^ { 2 } } = - 1$$

Whakaoti mō x

$$x=\frac{3\sqrt{2}}{2}+1\approx 3.121320344$$

$$x=-\frac{3\sqrt{2}}{2}+1\approx -1.121320344$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te \left(x-1\right)^{2}.

$$x^{2}-2x-8=-\left(x-1\right)^{2}$$

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

$$x^{2}-2x-8=-\left(x^{2}-2x+1\right)$$

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

$$x^{2}-2x-8=-x^{2}+2x-1$$

Me tāpiri te x^{2} ki ngā taha e rua.

$$x^{2}-2x-8+x^{2}=2x-1$$

Pahekotia te x^{2} me x^{2}, ka 2x^{2}.

$$2x^{2}-2x-8=2x-1$$

Tangohia te 2x mai i ngā taha e rua.

$$2x^{2}-2x-8-2x=-1$$

Pahekotia te -2x me -2x, ka -4x.

$$2x^{2}-4x-8=-1$$

Me tāpiri te 1 ki ngā taha e rua.

$$2x^{2}-4x-8+1=0$$

Tāpirihia te -8 ki te 1, ka -7.

$$2x^{2}-4x-7=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -4 mō b, me -7 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-7\right)}}{2\times 2}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-7\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\left(-7\right)}}{2\times 2}$$

Whakareatia -8 ki te -7.

$$x=\frac{-\left(-4\right)±\sqrt{16+56}}{2\times 2}$$

Tāpiri 16 ki te 56.

$$x=\frac{-\left(-4\right)±\sqrt{72}}{2\times 2}$$

Tuhia te pūtakerua o te 72.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±6\sqrt{2}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{4±6\sqrt{2}}{4}$$

Nā, me whakaoti te whārite x=\frac{4±6\sqrt{2}}{4} ina he tāpiri te ±. Tāpiri 4 ki te 6\sqrt{2}.

$$x=\frac{6\sqrt{2}+4}{4}$$

Whakawehe 4+6\sqrt{2} ki te 4.

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

Nā, me whakaoti te whārite x=\frac{4±6\sqrt{2}}{4} ina he tango te ±. Tango 6\sqrt{2} mai i 4.

$$x=\frac{4-6\sqrt{2}}{4}$$

Whakawehe 4-6\sqrt{2} ki te 4.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki 1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te \left(x-1\right)^{2}.

$$x^{2}-2x-8=-\left(x-1\right)^{2}$$

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

$$x^{2}-2x-8=-\left(x^{2}-2x+1\right)$$

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

$$x^{2}-2x-8=-x^{2}+2x-1$$

Me tāpiri te x^{2} ki ngā taha e rua.

$$x^{2}-2x-8+x^{2}=2x-1$$

Pahekotia te x^{2} me x^{2}, ka 2x^{2}.

$$2x^{2}-2x-8=2x-1$$

Tangohia te 2x mai i ngā taha e rua.

$$2x^{2}-2x-8-2x=-1$$

Pahekotia te -2x me -2x, ka -4x.

$$2x^{2}-4x-8=-1$$

Me tāpiri te 8 ki ngā taha e rua.

$$2x^{2}-4x=-1+8$$

Tāpirihia te -1 ki te 8, ka 7.

$$2x^{2}-4x=7$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-4x}{2}=\frac{7}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x^{2}+\left(-\frac{4}{2}\right)x=\frac{7}{2}$$

Whakawehe -4 ki te 2.

$$x^{2}-2x=\frac{7}{2}$$

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

$$x^{2}-2x+1=\frac{7}{2}+1$$

Tāpiri \frac{7}{2} ki te 1.

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

Tauwehea x^{2}-2x+1. 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}.

$$\left(x-1\right)^{2}=\frac{9}{2}$$

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

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

Whakarūnātia.

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

Me tāpiri 1 ki ngā taha e rua o te whārite.

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