Whakaoti i te x^2+5x+9=0

$${ x }^{ 2 } +5x+9=0$$

Whakaoti mō x (complex solution)

$$x=\frac{-5+\sqrt{11}i}{2}\approx -2.5+1.658312395i$$

$$x=\frac{-\sqrt{11}i-5}{2}\approx -2.5-1.658312395i$$

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.

$$x^{2}+5x+9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 5 mō b, me 9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 9}}{2}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 9}}{2}$$

Whakareatia -4 ki te 9.

$$x=\frac{-5±\sqrt{25-36}}{2}$$

Tāpiri 25 ki te -36.

$$x=\frac{-5±\sqrt{-11}}{2}$$

Tuhia te pūtakerua o te -11.

$$x=\frac{-5±\sqrt{11}i}{2}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{11}i}{2} ina he tāpiri te ±. Tāpiri -5 ki te i\sqrt{11}.

$$x=\frac{-5+\sqrt{11}i}{2}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{11}i}{2} ina he tango te ±. Tango i\sqrt{11} mai i -5.

$$x=\frac{-\sqrt{11}i-5}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{-5+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i-5}{2}$$

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.

$$x^{2}+5x+9=0$$

Me tango 9 mai i ngā taha e rua o te whārite.

$$x^{2}+5x+9-9=-9$$

Mā te tango i te 9 i a ia ake anō ka toe ko te 0.

$$x^{2}+5x=-9$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-9+\left(\frac{5}{2}\right)^{2}$$

Pūruatia \frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+5x+\frac{25}{4}=-9+\frac{25}{4}$$

Tāpiri -9 ki te \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=-\frac{11}{4}$$

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

$$\left(x+\frac{5}{2}\right)^{2}=-\frac{11}{4}$$

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

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{-\frac{11}{4}}$$

Whakarūnātia.

$$x+\frac{5}{2}=\frac{\sqrt{11}i}{2} x+\frac{5}{2}=-\frac{\sqrt{11}i}{2}$$

Me tango \frac{5}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{-5+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i-5}{2}$$