Whakaoti i te left(2xright)^2-4x+5=0

$${ \left(2x \right) }^{ 2 } -4x+5 = 0$$

Whakaoti mō x (complex solution)

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakarohaina te \left(2x\right)^{2}.

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

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

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

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

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

Pūrua -4.

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

Whakareatia -4 ki te 4.

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

Whakareatia -16 ki te 5.

$$x=\frac{-\left(-4\right)±\sqrt{16-80}}{2\times 4}$$

Tāpiri 16 ki te -80.

$$x=\frac{-\left(-4\right)±\sqrt{-64}}{2\times 4}$$

Tuhia te pūtakerua o te -64.

$$x=\frac{-\left(-4\right)±8i}{2\times 4}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±8i}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{4±8i}{8}$$

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

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

Whakawehe 4+8i ki te 8.

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

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

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

Whakawehe 4-8i ki te 8.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakarohaina te \left(2x\right)^{2}.

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

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

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

Tangohia te 5 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Whakawehea ngā taha e rua ki te 4.

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

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

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

Whakawehe -4 ki te 4.

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

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.

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

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

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

Tāpiri -\frac{5}{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.

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

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}.

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

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

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

Whakarūnātia.

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

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

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