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

$$( x - 1 ) ^ { 2 } = x ( 1 - x )$$

Whakaoti mō x

$$x=\frac{1}{2}=0.5$$

$$x=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te 1-x.

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

Tangohia te x mai i ngā taha e rua.

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

Pahekotia te -2x me -x, ka -3x.

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

Me tāpiri te x^{2} ki ngā taha e rua.

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

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

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

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

$$a+b=-3 ab=2\times 1=2$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-2 b=-1$$

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

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

Tauwehea te 2x i te tuatahi me te -1 i te rōpū tuarua.

$$2x\left(x-1\right)-\left(x-1\right)$$

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

$$\left(x-1\right)\left(2x-1\right)$$

Hei kimi otinga whārite, me whakaoti te x-1=0 me te 2x-1=0.

$$x=1 x=\frac{1}{2}$$

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(x-1\right)^{2}.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te 1-x.

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

Tangohia te x mai i ngā taha e rua.

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

Pahekotia te -2x me -x, ka -3x.

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

Me tāpiri te x^{2} ki ngā taha e rua.

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

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

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -3 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -3.

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

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2\times 2}$$

Tāpiri 9 ki te -8.

$$x=\frac{-\left(-3\right)±\sqrt{1}}{2\times 2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-3\right)±1}{2\times 2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±1}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{3±1}{4}$$

Nā, me whakaoti te whārite x=\frac{3±1}{4} ina he tāpiri te ±. Tāpiri 3 ki te 1.

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

Whakawehe 4 ki te 4.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{3±1}{4} ina he tango te ±. Tango 1 mai i 3.

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

Whakahekea te hautanga \frac{2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{1}{2}$$

Kua oti te whārite te whakatau.

$$x=1 x=\frac{1}{2}$$

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(x-1\right)^{2}.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te 1-x.

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

Tangohia te x mai i ngā taha e rua.

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

Pahekotia te -2x me -x, ka -3x.

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

Me tāpiri te x^{2} ki ngā taha e rua.

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

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

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

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehea te -\frac{3}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{3}{4}. Nā, tāpiria te pūrua o te -\frac{3}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{3}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Tāpiri -\frac{1}{2} ki te \frac{9}{16} 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.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{16}$$

Tauwehea x^{2}-\frac{3}{2}x+\frac{9}{16}. 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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{1}{16}$$

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

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Whakarūnātia.

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

Me tāpiri \frac{3}{4} ki ngā taha e rua o te whārite.

$$x=1 x=\frac{1}{2}$$