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

$$(3x-1)(x-2)+(x+1)(x+2)=1$$

Whakaoti mō x (complex solution)

$$x=\frac{1+\sqrt{2}i}{2}\approx 0.5+0.707106781i$$

$$x=\frac{-\sqrt{2}i+1}{2}\approx 0.5-0.707106781i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 3x-1 ki te x-2 ka whakakotahi i ngā kupu rite.

$$3x^{2}-7x+2+\left(x+1\right)\left(x+2\right)=1$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te x+2 ka whakakotahi i ngā kupu rite.

$$3x^{2}-7x+2+x^{2}+3x+2=1$$

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

$$4x^{2}-7x+2+3x+2=1$$

Pahekotia te -7x me 3x, ka -4x.

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

Tāpirihia te 2 ki te 2, ka 4.

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

Tangohia te 1 mai i ngā taha e rua.

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

Tangohia te 1 i te 4, ka 3.

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

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

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\times 3}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16\times 3}}{2\times 4}$$

Whakareatia -16 ki te 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-48}}{2\times 4}$$

Tāpiri 16 ki te -48.

$$x=\frac{-\left(-4\right)±\sqrt{-32}}{2\times 4}$$

Tuhia te pūtakerua o te -32.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±4\sqrt{2}i}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{4±4\sqrt{2}i}{8}$$

Nā, me whakaoti te whārite x=\frac{4±4\sqrt{2}i}{8} ina he tāpiri te ±. Tāpiri 4 ki te 4i\sqrt{2}.

$$x=\frac{4+4\sqrt{2}i}{8}$$

Whakawehe 4+4i\sqrt{2} ki te 8.

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

Nā, me whakaoti te whārite x=\frac{4±4\sqrt{2}i}{8} ina he tango te ±. Tango 4i\sqrt{2} mai i 4.

$$x=\frac{-4\sqrt{2}i+4}{8}$$

Whakawehe 4-4i\sqrt{2} ki te 8.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 3x-1 ki te x-2 ka whakakotahi i ngā kupu rite.

$$3x^{2}-7x+2+\left(x+1\right)\left(x+2\right)=1$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+1 ki te x+2 ka whakakotahi i ngā kupu rite.

$$3x^{2}-7x+2+x^{2}+3x+2=1$$

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

$$4x^{2}-7x+2+3x+2=1$$

Pahekotia te -7x me 3x, ka -4x.

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

Tāpirihia te 2 ki te 2, ka 4.

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

Tangohia te 4 mai i ngā taha e rua.

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

Tangohia te 4 i te 1, ka -3.

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

Whakawehea ngā taha e rua ki te 4.

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

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

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

Whakawehe -4 ki te 4.

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

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.

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

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

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

Tāpiri -\frac{3}{4} 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.

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

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}.

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

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

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

Whakarūnātia.

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

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

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