Whakaoti i te (x+2)(x+3)=(x-2)

$$(x+2)(x+3)=(x-2)$$

Whakaoti mō x (complex solution)

$$x=-2+2i$$

$$x=-2-2i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te x+3 ka whakakotahi i ngā kupu rite.

$$x^{2}+5x+6=x-2$$

Tangohia te x mai i ngā taha e rua.

$$x^{2}+5x+6-x=-2$$

Pahekotia te 5x me -x, ka 4x.

$$x^{2}+4x+6=-2$$

Me tāpiri te 2 ki ngā taha e rua.

$$x^{2}+4x+6+2=0$$

Tāpirihia te 6 ki te 2, ka 8.

$$x^{2}+4x+8=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 4 mō b, me 8 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 8}}{2}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 8}}{2}$$

Whakareatia -4 ki te 8.

$$x=\frac{-4±\sqrt{16-32}}{2}$$

Tāpiri 16 ki te -32.

$$x=\frac{-4±\sqrt{-16}}{2}$$

Tuhia te pūtakerua o te -16.

$$x=\frac{-4±4i}{2}$$

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

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

Whakawehe -4+4i ki te 2.

$$x=-2+2i$$

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

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

Whakawehe -4-4i ki te 2.

$$x=-2-2i$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te x+3 ka whakakotahi i ngā kupu rite.

$$x^{2}+5x+6=x-2$$

Tangohia te x mai i ngā taha e rua.

$$x^{2}+5x+6-x=-2$$

Pahekotia te 5x me -x, ka 4x.

$$x^{2}+4x+6=-2$$

Tangohia te 6 mai i ngā taha e rua.

$$x^{2}+4x=-2-6$$

Tangohia te 6 i te -2, ka -8.

$$x^{2}+4x=-8$$

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.

$$x^{2}+4x+2^{2}=-8+2^{2}$$

Pūrua 2.

$$x^{2}+4x+4=-8+4$$

Tāpiri -8 ki te 4.

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

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}.

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

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

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

Whakarūnātia.

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

Me tango 2 mai i ngā taha e rua o te whārite.

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