Whakaoti i te 0=2(x-1)(x-1)-8

$$0=2(x-1)(x-1)-8$$

Whakaoti mō x

$$x=3$$

$$x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakareatia te x-1 ki te x-1, ka \left(x-1\right)^{2}.

$$0=2\left(x-1\right)^{2}-8$$

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

$$0=2\left(x^{2}-2x+1\right)-8$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x^{2}-2x+1.

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

Tangohia te 8 i te 2, ka -6.

$$0=2x^{2}-4x-6$$

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

$$2x^{2}-4x-6=0$$

Whakawehea ngā taha e rua ki te 2.

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

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

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

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

$$a=-3 b=1$$

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

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

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

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

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

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

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

$$x=3 x=-1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te x-1 ki te x-1, ka \left(x-1\right)^{2}.

$$0=2\left(x-1\right)^{2}-8$$

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

$$0=2\left(x^{2}-2x+1\right)-8$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x^{2}-2x+1.

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

Tangohia te 8 i te 2, ka -6.

$$0=2x^{2}-4x-6$$

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

$$2x^{2}-4x-6=0$$

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

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-6\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\left(-6\right)}}{2\times 2}$$

Whakareatia -8 ki te -6.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 2}$$

Tāpiri 16 ki te 48.

$$x=\frac{-\left(-4\right)±\sqrt{64}}{2\times 2}$$

Tuhia te pūtakerua o te 64.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±8}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{4±8}{4}$$

Nā, me whakaoti te whārite x=\frac{4±8}{4} ina he tāpiri te ±. Tāpiri 4 ki te 8.

$$x=\frac{12}{4}$$

Whakawehe 12 ki te 4.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{4±8}{4} ina he tango te ±. Tango 8 mai i 4.

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

Whakawehe -4 ki te 4.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=3 x=-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te x-1 ki te x-1, ka \left(x-1\right)^{2}.

$$0=2\left(x-1\right)^{2}-8$$

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

$$0=2\left(x^{2}-2x+1\right)-8$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x^{2}-2x+1.

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

Tangohia te 8 i te 2, ka -6.

$$0=2x^{2}-4x-6$$

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

$$2x^{2}-4x-6=0$$

Me tāpiri te 6 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$2x^{2}-4x=6$$

Whakawehea ngā taha e rua ki te 2.

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

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

$$x^{2}+\left(-\frac{4}{2}\right)x=\frac{6}{2}$$

Whakawehe -4 ki te 2.

$$x^{2}-2x=\frac{6}{2}$$

Whakawehe 6 ki te 2.

$$x^{2}-2x=3$$

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.

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

Tāpiri 3 ki te 1.

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

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}.

$$\left(x-1\right)^{2}=4$$

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

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{4}$$

Whakarūnātia.

$$x-1=2 x-1=-2$$

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

$$x=3 x=-1$$