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

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

Whakaoti mō x

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

$$x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

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

Whakaarohia te \left(x-1\right)\left(x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

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

Tangohia te 1 i te 1, ka 0.

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

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

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

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

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

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.

$$a+b=4 ab=3\times 1=3$$

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

$$a=1 b=3$$

Tuhia anō te 3x^{2}+4x+1 hei \left(3x^{2}+x\right)+\left(3x+1\right).

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

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

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

Whakatauwehea atu te kīanga pātahi 3x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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

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

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

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

Whakaarohia te \left(x-1\right)\left(x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

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

Tangohia te 1 i te 1, ka 0.

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

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

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

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

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

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

$$x=\frac{-4±\sqrt{4^{2}-4\times 3}}{2\times 3}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 3}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-4±\sqrt{16-12}}{2\times 3}$$

Tāpiri 16 ki te -12.

$$x=\frac{-4±\sqrt{4}}{2\times 3}$$

Tuhia te pūtakerua o te 4.

$$x=\frac{-4±2}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-4±2}{6}$$

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

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

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{-4±2}{6} ina he tango te ±. Tango 2 mai i -4.

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

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

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

Whakaarohia te \left(x-1\right)\left(x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

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

Tangohia te 1 i te 1, ka 0.

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

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

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

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

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

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

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

Whakawehea ngā taha e rua ki te 3.

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

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

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

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

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

Pūruatia \frac{2}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Tāpiri -\frac{1}{3} ki te \frac{4}{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.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{9}$$

Tauwehea x^{2}+\frac{4}{3}x+\frac{4}{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}.

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

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

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

Whakarūnātia.

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

Me tango \frac{2}{3} mai i ngā taha e rua o te whārite.

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