Whakaoti i te 0=x^2-4x+4+5

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

Whakaoti mō x (complex solution)

$$x=2+\sqrt{5}i\approx 2+2.236067977i$$

$$x=-\sqrt{5}i+2\approx 2-2.236067977i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tāpirihia te 4 ki te 5, ka 9.

$$0=x^{2}-4x+9$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{2}-4x+9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -4 mō b, me 9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 9}}{2}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 9}}{2}$$

Whakareatia -4 ki te 9.

$$x=\frac{-\left(-4\right)±\sqrt{16-36}}{2}$$

Tāpiri 16 ki te -36.

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

Tuhia te pūtakerua o te -20.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2\sqrt{5}i}{2}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{5}i}{2} ina he tāpiri te ±. Tāpiri 4 ki te 2i\sqrt{5}.

$$x=\frac{4+2\sqrt{5}i}{2}$$

Whakawehe 4+2i\sqrt{5} ki te 2.

$$x=2+\sqrt{5}i$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{5}i}{2} ina he tango te ±. Tango 2i\sqrt{5} mai i 4.

$$x=\frac{-2\sqrt{5}i+4}{2}$$

Whakawehe 4-2i\sqrt{5} ki te 2.

$$x=-\sqrt{5}i+2$$

Kua oti te whārite te whakatau.

$$x=2+\sqrt{5}i x=-\sqrt{5}i+2$$

Ngā Upane mō te Whakaoti i te Pūrua

Tāpirihia te 4 ki te 5, ka 9.

$$0=x^{2}-4x+9$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{2}-4x+9=0$$

Tangohia te 9 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x^{2}-4x=-9$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=-9+\left(-2\right)^{2}$$

Pūrua -2.

$$x^{2}-4x+4=-9+4$$

Tāpiri -9 ki te 4.

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

Tauwehea te x^{2}-4x+4. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

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

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

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

Whakarūnātia.

$$x-2=\sqrt{5}i x-2=-\sqrt{5}i$$

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

$$x=2+\sqrt{5}i x=-\sqrt{5}i+2$$