Whakaoti i te 2x^2+5x+29=0

$$2 { x }^{ 2 } +5x+29=0$$

Whakaoti mō x (complex solution)

$$x=\frac{-5+3\sqrt{23}i}{4}\approx -1.25+3.596873642i$$

$$x=\frac{-3\sqrt{23}i-5}{4}\approx -1.25-3.596873642i$$

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.

$$2x^{2}+5x+29=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 5 mō b, me 29 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 29}}{2\times 2}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 2\times 29}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-5±\sqrt{25-8\times 29}}{2\times 2}$$

Whakareatia -8 ki te 29.

$$x=\frac{-5±\sqrt{25-232}}{2\times 2}$$

Tāpiri 25 ki te -232.

$$x=\frac{-5±\sqrt{-207}}{2\times 2}$$

Tuhia te pūtakerua o te -207.

$$x=\frac{-5±3\sqrt{23}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-5±3\sqrt{23}i}{4}$$

Nā, me whakaoti te whārite x=\frac{-5±3\sqrt{23}i}{4} ina he tāpiri te ±. Tāpiri -5 ki te 3i\sqrt{23}.

$$x=\frac{-5+3\sqrt{23}i}{4}$$

Nā, me whakaoti te whārite x=\frac{-5±3\sqrt{23}i}{4} ina he tango te ±. Tango 3i\sqrt{23} mai i -5.

$$x=\frac{-3\sqrt{23}i-5}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{-5+3\sqrt{23}i}{4} x=\frac{-3\sqrt{23}i-5}{4}$$

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.

$$2x^{2}+5x+29=0$$

Me tango 29 mai i ngā taha e rua o te whārite.

$$2x^{2}+5x+29-29=-29$$

Mā te tango i te 29 i a ia ake anō ka toe ko te 0.

$$2x^{2}+5x=-29$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}+5x}{2}=-\frac{29}{2}$$

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

$$x^{2}+\frac{5}{2}x=-\frac{29}{2}$$

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.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=-\frac{29}{2}+\left(\frac{5}{4}\right)^{2}$$

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

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-\frac{29}{2}+\frac{25}{16}$$

Tāpiri -\frac{29}{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.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-\frac{207}{16}$$

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}.

$$\left(x+\frac{5}{4}\right)^{2}=-\frac{207}{16}$$

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

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{-\frac{207}{16}}$$

Whakarūnātia.

$$x+\frac{5}{4}=\frac{3\sqrt{23}i}{4} x+\frac{5}{4}=-\frac{3\sqrt{23}i}{4}$$

Me tango \frac{5}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{-5+3\sqrt{23}i}{4} x=\frac{-3\sqrt{23}i-5}{4}$$