Whakaoti i te b^2-4b+13=0

$$b ^ { 2 } - 4 b + 13 = 0$$

Whakaoti mō b

$$b=2+3i$$

$$b=2-3i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$b^{2}-4b+13=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -4 mō b, me 13 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 13}}{2}$$

Pūrua -4.

$$b=\frac{-\left(-4\right)±\sqrt{16-4\times 13}}{2}$$

Whakareatia -4 ki te 13.

$$b=\frac{-\left(-4\right)±\sqrt{16-52}}{2}$$

Tāpiri 16 ki te -52.

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

Tuhia te pūtakerua o te -36.

$$b=\frac{-\left(-4\right)±6i}{2}$$

Ko te tauaro o -4 ko 4.

$$b=\frac{4±6i}{2}$$

Nā, me whakaoti te whārite b=\frac{4±6i}{2} ina he tāpiri te ±. Tāpiri 4 ki te 6i.

$$b=\frac{4+6i}{2}$$

Whakawehe 4+6i ki te 2.

$$b=2+3i$$

Nā, me whakaoti te whārite b=\frac{4±6i}{2} ina he tango te ±. Tango 6i mai i 4.

$$b=\frac{4-6i}{2}$$

Whakawehe 4-6i ki te 2.

$$b=2-3i$$

Kua oti te whārite te whakatau.

$$b=2+3i b=2-3i$$

Ngā Upane mō te Whakaoti i te Pūrua

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$b^{2}-4b+13=0$$

Me tango 13 mai i ngā taha e rua o te whārite.

$$b^{2}-4b+13-13=-13$$

Mā te tango i te 13 i a ia ake anō ka toe ko te 0.

$$b^{2}-4b=-13$$

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.

$$b^{2}-4b+\left(-2\right)^{2}=-13+\left(-2\right)^{2}$$

Pūrua -2.

$$b^{2}-4b+4=-13+4$$

Tāpiri -13 ki te 4.

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

Tauwehea b^{2}-4b+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}.

$$\left(b-2\right)^{2}=-9$$

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

$$\sqrt{\left(b-2\right)^{2}}=\sqrt{-9}$$

Whakarūnātia.

$$b-2=3i b-2=-3i$$

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

$$b=2+3i b=2-3i$$