Whakaoti i te 2x^2-5x+17=0

$$2 { x }^{ 2 } -5x+17=0$$

Whakaoti mō x (complex solution)

$$x=\frac{5+\sqrt{111}i}{4}\approx 1.25+2.633913438i$$

$$x=\frac{-\sqrt{111}i+5}{4}\approx 1.25-2.633913438i$$

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.

$$2x^{2}-5x+17=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -5 mō b, me 17 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\times 17}}{2\times 2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\times 17}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\times 17}}{2\times 2}$$

Whakareatia -8 ki te 17.

$$x=\frac{-\left(-5\right)±\sqrt{25-136}}{2\times 2}$$

Tāpiri 25 ki te -136.

$$x=\frac{-\left(-5\right)±\sqrt{-111}}{2\times 2}$$

Tuhia te pūtakerua o te -111.

$$x=\frac{-\left(-5\right)±\sqrt{111}i}{2\times 2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{111}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{5±\sqrt{111}i}{4}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{111}i}{4} ina he tāpiri te ±. Tāpiri 5 ki te i\sqrt{111}.

$$x=\frac{5+\sqrt{111}i}{4}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{111}i}{4} ina he tango te ±. Tango i\sqrt{111} mai i 5.

$$x=\frac{-\sqrt{111}i+5}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{5+\sqrt{111}i}{4} x=\frac{-\sqrt{111}i+5}{4}$$

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.

$$2x^{2}-5x+17=0$$

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

$$2x^{2}-5x+17-17=-17$$

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

$$2x^{2}-5x=-17$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-5x}{2}=-\frac{17}{2}$$

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

$$x^{2}-\frac{5}{2}x=-\frac{17}{2}$$

Whakawehea te -\frac{5}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{4}. Nā, tāpiria te pūrua o te -\frac{5}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-\frac{17}{2}+\left(-\frac{5}{4}\right)^{2}$$

Pūruatia -\frac{5}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{17}{2}+\frac{25}{16}$$

Tāpiri -\frac{17}{2} ki te \frac{25}{16} 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.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{111}{16}$$

Tauwehea x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=-\frac{111}{16}$$

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

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{-\frac{111}{16}}$$

Whakarūnātia.

$$x-\frac{5}{4}=\frac{\sqrt{111}i}{4} x-\frac{5}{4}=-\frac{\sqrt{111}i}{4}$$

Me tāpiri \frac{5}{4} ki ngā taha e rua o te whārite.

$$x=\frac{5+\sqrt{111}i}{4} x=\frac{-\sqrt{111}i+5}{4}$$