Whakaoti i te (x+1)2-(4-2x)=x^2+3

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

Whakaoti mō x (complex solution)

$$x=2-i$$

$$x=2+i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 2.

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

Hei kimi i te tauaro o 4-2x, kimihia te tauaro o ia taurangi.

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

Tangohia te 4 i te 2, ka -2.

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

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

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

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

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

Tangohia te 3 mai i ngā taha e rua.

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

Tangohia te 3 i te -2, ka -5.

$$4x-5-x^{2}=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$-x^{2}+4x-5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, 4 mō b, me -5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-4±\sqrt{16+4\left(-5\right)}}{2\left(-1\right)}$$

Whakareatia 4 ki te -5.

$$x=\frac{-4±\sqrt{16-20}}{2\left(-1\right)}$$

Tāpiri 16 ki te -20.

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

Tuhia te pūtakerua o te -4.

$$x=\frac{-4±2i}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-4±2i}{-2}$$

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

$$x=\frac{-4+2i}{-2}$$

Whakawehe -4+2i ki te -2.

$$x=2-i$$

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

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

Whakawehe -4-2i ki te -2.

$$x=2+i$$

Kua oti te whārite te whakatau.

$$x=2-i x=2+i$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 2.

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

Hei kimi i te tauaro o 4-2x, kimihia te tauaro o ia taurangi.

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

Tangohia te 4 i te 2, ka -2.

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

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

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

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

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

Me tāpiri te 2 ki ngā taha e rua.

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

Tāpirihia te 3 ki te 2, ka 5.

$$4x-x^{2}=5$$

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$-x^{2}+4x=5$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x^{2}+4x}{-1}=\frac{5}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$x^{2}+\frac{4}{-1}x=\frac{5}{-1}$$

Whakawehe 4 ki te -1.

$$x^{2}-4x=\frac{5}{-1}$$

Whakawehe 5 ki te -1.

$$x^{2}-4x=-5$$

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

$$x^{2}-4x+\left(-2\right)^{2}=-5+\left(-2\right)^{2}$$

Pūrua -2.

$$x^{2}-4x+4=-5+4$$

Tāpiri -5 ki te 4.

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

Tauwehea x^{2}-4x+4. 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}.

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

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

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

Whakarūnātia.

$$x-2=i x-2=-i$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$x=2+i x=2-i$$