Whakaoti i te 5x^2+6x+10=0

$$5 x ^ { 2 } + 6 x + 10 = 0$$

Whakaoti mō x (complex solution)

$$x=\frac{-3+\sqrt{41}i}{5}\approx -0.6+1.280624847i$$

$$x=\frac{-\sqrt{41}i-3}{5}\approx -0.6-1.280624847i$$

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.

$$5x^{2}+6x+10=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, 6 mō b, me 10 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5\times 10}}{2\times 5}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\times 5\times 10}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-6±\sqrt{36-20\times 10}}{2\times 5}$$

Whakareatia -20 ki te 10.

$$x=\frac{-6±\sqrt{36-200}}{2\times 5}$$

Tāpiri 36 ki te -200.

$$x=\frac{-6±\sqrt{-164}}{2\times 5}$$

Tuhia te pūtakerua o te -164.

$$x=\frac{-6±2\sqrt{41}i}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-6±2\sqrt{41}i}{10}$$

Nā, me whakaoti te whārite x=\frac{-6±2\sqrt{41}i}{10} ina he tāpiri te ±. Tāpiri -6 ki te 2i\sqrt{41}.

$$x=\frac{-6+2\sqrt{41}i}{10}$$

Whakawehe -6+2i\sqrt{41} ki te 10.

$$x=\frac{-3+\sqrt{41}i}{5}$$

Nā, me whakaoti te whārite x=\frac{-6±2\sqrt{41}i}{10} ina he tango te ±. Tango 2i\sqrt{41} mai i -6.

$$x=\frac{-2\sqrt{41}i-6}{10}$$

Whakawehe -6-2i\sqrt{41} ki te 10.

$$x=\frac{-\sqrt{41}i-3}{5}$$

Kua oti te whārite te whakatau.

$$x=\frac{-3+\sqrt{41}i}{5} x=\frac{-\sqrt{41}i-3}{5}$$

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.

$$5x^{2}+6x+10=0$$

Me tango 10 mai i ngā taha e rua o te whārite.

$$5x^{2}+6x+10-10=-10$$

Mā te tango i te 10 i a ia ake anō ka toe ko te 0.

$$5x^{2}+6x=-10$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}+6x}{5}=-\frac{10}{5}$$

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

$$x^{2}+\frac{6}{5}x=-\frac{10}{5}$$

Whakawehe -10 ki te 5.

$$x^{2}+\frac{6}{5}x=-2$$

Whakawehea te \frac{6}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{5}. Nā, tāpiria te pūrua o te \frac{3}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=-2+\left(\frac{3}{5}\right)^{2}$$

Pūruatia \frac{3}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=-2+\frac{9}{25}$$

Tāpiri -2 ki te \frac{9}{25}.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=-\frac{41}{25}$$

Tauwehea x^{2}+\frac{6}{5}x+\frac{9}{25}. 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}.

$$\left(x+\frac{3}{5}\right)^{2}=-\frac{41}{25}$$

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

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{-\frac{41}{25}}$$

Whakarūnātia.

$$x+\frac{3}{5}=\frac{\sqrt{41}i}{5} x+\frac{3}{5}=-\frac{\sqrt{41}i}{5}$$

Me tango \frac{3}{5} mai i ngā taha e rua o te whārite.

$$x=\frac{-3+\sqrt{41}i}{5} x=\frac{-\sqrt{41}i-3}{5}$$