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

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

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{155}i}{10}-\frac{1}{2}\approx -0.5+1.24498996i$$

$$x=-\frac{\sqrt{155}i}{10}-\frac{1}{2}\approx -0.5-1.24498996i$$

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.

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

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

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

Pūrua 5.

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

Whakareatia -4 ki te 5.

$$x=\frac{-5±\sqrt{25-20\times 9}}{2\times 5}$$

Whakareatia -20 ki te 9.

$$x=\frac{-5±\sqrt{25-180}}{2\times 5}$$

Tāpiri 25 ki te -180.

$$x=\frac{-5±\sqrt{-155}}{2\times 5}$$

Tuhia te pūtakerua o te -155.

$$x=\frac{-5±\sqrt{155}i}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-5±\sqrt{155}i}{10}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{155}i}{10} ina he tāpiri te ±. Tāpiri -5 ki te i\sqrt{155}.

$$x=\frac{-5+\sqrt{155}i}{10}$$

Whakawehe -5+i\sqrt{155} ki te 10.

$$x=\frac{\sqrt{155}i}{10}-\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{155}i}{10} ina he tango te ±. Tango i\sqrt{155} mai i -5.

$$x=\frac{-\sqrt{155}i-5}{10}$$

Whakawehe -5-i\sqrt{155} ki te 10.

$$x=-\frac{\sqrt{155}i}{10}-\frac{1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{155}i}{10}-\frac{1}{2} x=-\frac{\sqrt{155}i}{10}-\frac{1}{2}$$

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.

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

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

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

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

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

Whakawehea ngā taha e rua ki te 5.

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

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

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

Whakawehe 5 ki te 5.

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

Whakawehea te 1, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{2}. Nā, tāpiria te pūrua o te \frac{1}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=-\frac{9}{5}+\left(\frac{1}{2}\right)^{2}$$

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+x+\frac{1}{4}=-\frac{9}{5}+\frac{1}{4}$$

Tāpiri -\frac{9}{5} ki te \frac{1}{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.

$$x^{2}+x+\frac{1}{4}=-\frac{31}{20}$$

Tauwehea x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{31}{20}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{31}{20}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{\sqrt{155}i}{10} x+\frac{1}{2}=-\frac{\sqrt{155}i}{10}$$

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{155}i}{10}-\frac{1}{2} x=-\frac{\sqrt{155}i}{10}-\frac{1}{2}$$