Whakaoti i te 4x^2+12x+19=0

$$4 x ^ { 2 } + 12 x + 19 = 0$$

Whakaoti mō x (complex solution)

$$x=\frac{-3+\sqrt{10}i}{2}\approx -1.5+1.58113883i$$

$$x=\frac{-\sqrt{10}i-3}{2}\approx -1.5-1.58113883i$$

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.

$$4x^{2}+12x+19=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, 12 mō b, me 19 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 4\times 19}}{2\times 4}$$

Pūrua 12.

$$x=\frac{-12±\sqrt{144-4\times 4\times 19}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-12±\sqrt{144-16\times 19}}{2\times 4}$$

Whakareatia -16 ki te 19.

$$x=\frac{-12±\sqrt{144-304}}{2\times 4}$$

Tāpiri 144 ki te -304.

$$x=\frac{-12±\sqrt{-160}}{2\times 4}$$

Tuhia te pūtakerua o te -160.

$$x=\frac{-12±4\sqrt{10}i}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-12±4\sqrt{10}i}{8}$$

Nā, me whakaoti te whārite x=\frac{-12±4\sqrt{10}i}{8} ina he tāpiri te ±. Tāpiri -12 ki te 4i\sqrt{10}.

$$x=\frac{-12+4\sqrt{10}i}{8}$$

Whakawehe -12+4i\sqrt{10} ki te 8.

$$x=\frac{-3+\sqrt{10}i}{2}$$

Nā, me whakaoti te whārite x=\frac{-12±4\sqrt{10}i}{8} ina he tango te ±. Tango 4i\sqrt{10} mai i -12.

$$x=\frac{-4\sqrt{10}i-12}{8}$$

Whakawehe -12-4i\sqrt{10} ki te 8.

$$x=\frac{-\sqrt{10}i-3}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{-3+\sqrt{10}i}{2} x=\frac{-\sqrt{10}i-3}{2}$$

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.

$$4x^{2}+12x+19=0$$

Me tango 19 mai i ngā taha e rua o te whārite.

$$4x^{2}+12x+19-19=-19$$

Mā te tango i te 19 i a ia ake anō ka toe ko te 0.

$$4x^{2}+12x=-19$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}+12x}{4}=-\frac{19}{4}$$

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

$$x^{2}+\frac{12}{4}x=-\frac{19}{4}$$

Whakawehe 12 ki te 4.

$$x^{2}+3x=-\frac{19}{4}$$

Whakawehea te 3, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{2}. Nā, tāpiria te pūrua o te \frac{3}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-\frac{19}{4}+\left(\frac{3}{2}\right)^{2}$$

Pūruatia \frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+3x+\frac{9}{4}=\frac{-19+9}{4}$$

Tāpiri -\frac{19}{4} ki te \frac{9}{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.

$$x^{2}+3x+\frac{9}{4}=-\frac{5}{2}$$

Tauwehea x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{5}{2}$$

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

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{5}{2}}$$

Whakarūnātia.

$$x+\frac{3}{2}=\frac{\sqrt{10}i}{2} x+\frac{3}{2}=-\frac{\sqrt{10}i}{2}$$

Me tango \frac{3}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{-3+\sqrt{10}i}{2} x=\frac{-\sqrt{10}i-3}{2}$$