Whakaoti i te left(x+1right)^2=4

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

Whakaoti mō x

$$x=1$$

$$x=-3$$

Ngā Upane Whakamahi Tauwehe

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

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

Tangohia te 4 mai i ngā taha e rua.

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

Tangohia te 4 i te 1, ka -3.

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

Hei whakaoti i te whārite, whakatauwehea te x^{2}+2x-3 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

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

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.

$$a=-1 b=3$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

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

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

$$x=1 x=-3$$

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}.

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

Tangohia te 4 mai i ngā taha e rua.

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

Tangohia te 4 i te 1, ka -3.

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

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 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.

$$a=-1 b=3$$

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

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

Tauwehea te x i te tuatahi me te 3 i te rōpū tuarua.

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

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

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

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

$$x=1 x=-3$$

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}.

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

Tangohia te 4 mai i ngā taha e rua.

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

Tangohia te 4 i te 1, ka -3.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te -3.

$$x=\frac{-2±\sqrt{4+12}}{2}$$

Tāpiri 4 ki te 12.

$$x=\frac{-2±\sqrt{16}}{2}$$

Tuhia te pūtakerua o te 16.

$$x=\frac{-2±4}{2}$$

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

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

Whakawehe 2 ki te 2.

$$x=1$$

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

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

Whakawehe -6 ki te 2.

$$x=-3$$

Kua oti te whārite te whakatau.

$$x=1 x=-3$$

Ngā Upane mō te Whakaoti i te Pūrua

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 tango 1 mai i ngā taha e rua o te whārite.

$$x=1 x=-3$$