Whakaoti i te (x-1)^2=4x(x-1)

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

Whakaoti mō x

$$x=1$$

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

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te x-1.

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

Tangohia te 4x^{2} mai i ngā taha e rua.

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

Pahekotia te x^{2} me -4x^{2}, ka -3x^{2}.

$$-3x^{2}-2x+1=-4x$$

Me tāpiri te 4x ki ngā taha e rua.

$$-3x^{2}-2x+1+4x=0$$

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

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

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.

$$a+b=2 ab=-3=-3$$

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.

$$a=3 b=-1$$

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

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

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

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

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

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

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

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

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te x-1.

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

Tangohia te 4x^{2} mai i ngā taha e rua.

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

Pahekotia te x^{2} me -4x^{2}, ka -3x^{2}.

$$-3x^{2}-2x+1=-4x$$

Me tāpiri te 4x ki ngā taha e rua.

$$-3x^{2}-2x+1+4x=0$$

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

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te -3.

$$x=\frac{-2±\sqrt{4+12}}{2\left(-3\right)}$$

Tāpiri 4 ki te 12.

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

Tuhia te pūtakerua o te 16.

$$x=\frac{-2±4}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{-2±4}{-6}$$

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

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

Whakahekea te hautanga \frac{2}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

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

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

Whakawehe -6 ki te -6.

$$x=1$$

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te x-1.

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

Tangohia te 4x^{2} mai i ngā taha e rua.

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

Pahekotia te x^{2} me -4x^{2}, ka -3x^{2}.

$$-3x^{2}-2x+1=-4x$$

Me tāpiri te 4x ki ngā taha e rua.

$$-3x^{2}-2x+1+4x=0$$

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

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

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

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

Whakawehea ngā taha e rua ki te -3.

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

Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.

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

Whakawehe 2 ki te -3.

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

Whakawehe -1 ki 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 tāpiri \frac{1}{3} ki ngā taha e rua o te whārite.

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