Whakaoti i te 4x(x+2)=4x-2

$$4 x ( x + 2 ) = 4 x - 2$$

Whakaoti mō x (complex solution)

$$x=-\frac{1}{2}+\frac{1}{2}i=-0.5+0.5i$$

$$x=-\frac{1}{2}-\frac{1}{2}i=-0.5-0.5i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te x+2.

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

Tangohia te 4x mai i ngā taha e rua.

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

Pahekotia te 8x me -4x, ka 4x.

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

Me tāpiri te 2 ki ngā taha e rua.

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

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

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

Pūrua 4.

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

Whakareatia -4 ki te 4.

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

Whakareatia -16 ki te 2.

$$x=\frac{-4±\sqrt{16-32}}{2\times 4}$$

Tāpiri 16 ki te -32.

$$x=\frac{-4±\sqrt{-16}}{2\times 4}$$

Tuhia te pūtakerua o te -16.

$$x=\frac{-4±4i}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-4±4i}{8}$$

Nā, me whakaoti te whārite x=\frac{-4±4i}{8} ina he tāpiri te ±. Tāpiri -4 ki te 4i.

$$x=\frac{-4+4i}{8}$$

Whakawehe -4+4i ki te 8.

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

Nā, me whakaoti te whārite x=\frac{-4±4i}{8} ina he tango te ±. Tango 4i mai i -4.

$$x=\frac{-4-4i}{8}$$

Whakawehe -4-4i ki te 8.

$$x=-\frac{1}{2}-\frac{1}{2}i$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te x+2.

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

Tangohia te 4x mai i ngā taha e rua.

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

Pahekotia te 8x me -4x, ka 4x.

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

Whakawehea ngā taha e rua ki te 4.

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

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

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

Whakawehe 4 ki te 4.

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

Whakahekea te hautanga \frac{-2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Whakawehea te 1, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{2}. Nā, tāpiria te pūrua o te \frac{1}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri -\frac{1}{2} ki te \frac{1}{4} 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.

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

Tauwehea x^{2}+x+\frac{1}{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}.

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

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

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

Whakarūnātia.

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

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

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