Whakaoti i te 2x^2+3x+17=1

$$2 x ^ { 2 } + 3 x + 17 = 1$$

Whakaoti mō x (complex solution)

$$x=\frac{-3+\sqrt{119}i}{4}\approx -0.75+2.727178029i$$

$$x=\frac{-\sqrt{119}i-3}{4}\approx -0.75-2.727178029i$$

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.

$$2x^{2}+3x+17=1$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$2x^{2}+3x+17-1=1-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$2x^{2}+3x+17-1=0$$

Tango 1 mai i 17.

$$2x^{2}+3x+16=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 3 mō b, me 16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 2\times 16}}{2\times 2}$$

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\times 2\times 16}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-3±\sqrt{9-8\times 16}}{2\times 2}$$

Whakareatia -8 ki te 16.

$$x=\frac{-3±\sqrt{9-128}}{2\times 2}$$

Tāpiri 9 ki te -128.

$$x=\frac{-3±\sqrt{-119}}{2\times 2}$$

Tuhia te pūtakerua o te -119.

$$x=\frac{-3±\sqrt{119}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-3±\sqrt{119}i}{4}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{119}i}{4} ina he tāpiri te ±. Tāpiri -3 ki te i\sqrt{119}.

$$x=\frac{-3+\sqrt{119}i}{4}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{119}i}{4} ina he tango te ±. Tango i\sqrt{119} mai i -3.

$$x=\frac{-\sqrt{119}i-3}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{-3+\sqrt{119}i}{4} x=\frac{-\sqrt{119}i-3}{4}$$

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.

$$2x^{2}+3x+17=1$$

Me tango 17 mai i ngā taha e rua o te whārite.

$$2x^{2}+3x+17-17=1-17$$

Mā te tango i te 17 i a ia ake anō ka toe ko te 0.

$$2x^{2}+3x=1-17$$

Tango 17 mai i 1.

$$2x^{2}+3x=-16$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}+3x}{2}=-\frac{16}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x^{2}+\frac{3}{2}x=-\frac{16}{2}$$

Whakawehe -16 ki te 2.

$$x^{2}+\frac{3}{2}x=-8$$

Whakawehea te \frac{3}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{4}. Nā, tāpiria te pūrua o te \frac{3}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=-8+\left(\frac{3}{4}\right)^{2}$$

Pūruatia \frac{3}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-8+\frac{9}{16}$$

Tāpiri -8 ki te \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{119}{16}$$

Tauwehea x^{2}+\frac{3}{2}x+\frac{9}{16}. 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}.

$$\left(x+\frac{3}{4}\right)^{2}=-\frac{119}{16}$$

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

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{-\frac{119}{16}}$$

Whakarūnātia.

$$x+\frac{3}{4}=\frac{\sqrt{119}i}{4} x+\frac{3}{4}=-\frac{\sqrt{119}i}{4}$$

Me tango \frac{3}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{-3+\sqrt{119}i}{4} x=\frac{-\sqrt{119}i-3}{4}$$