Whakaoti i te 25y^2+y+32=0

$$25 y ^ { 2 } + y + 32 = 0$$

Whakaoti mō y

$$y=\frac{-1+\sqrt{3199}i}{50}\approx -0.02+1.131194059i$$

$$y=\frac{-\sqrt{3199}i-1}{50}\approx -0.02-1.131194059i$$

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.

$$25y^{2}+y+32=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 25 mō a, 1 mō b, me 32 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-1±\sqrt{1^{2}-4\times 25\times 32}}{2\times 25}$$

Pūrua 1.

$$y=\frac{-1±\sqrt{1-4\times 25\times 32}}{2\times 25}$$

Whakareatia -4 ki te 25.

$$y=\frac{-1±\sqrt{1-100\times 32}}{2\times 25}$$

Whakareatia -100 ki te 32.

$$y=\frac{-1±\sqrt{1-3200}}{2\times 25}$$

Tāpiri 1 ki te -3200.

$$y=\frac{-1±\sqrt{-3199}}{2\times 25}$$

Tuhia te pūtakerua o te -3199.

$$y=\frac{-1±\sqrt{3199}i}{2\times 25}$$

Whakareatia 2 ki te 25.

$$y=\frac{-1±\sqrt{3199}i}{50}$$

Nā, me whakaoti te whārite y=\frac{-1±\sqrt{3199}i}{50} ina he tāpiri te ±. Tāpiri -1 ki te i\sqrt{3199}.

$$y=\frac{-1+\sqrt{3199}i}{50}$$

Nā, me whakaoti te whārite y=\frac{-1±\sqrt{3199}i}{50} ina he tango te ±. Tango i\sqrt{3199} mai i -1.

$$y=\frac{-\sqrt{3199}i-1}{50}$$

Kua oti te whārite te whakatau.

$$y=\frac{-1+\sqrt{3199}i}{50} y=\frac{-\sqrt{3199}i-1}{50}$$

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.

$$25y^{2}+y+32=0$$

Me tango 32 mai i ngā taha e rua o te whārite.

$$25y^{2}+y+32-32=-32$$

Mā te tango i te 32 i a ia ake anō ka toe ko te 0.

$$25y^{2}+y=-32$$

Whakawehea ngā taha e rua ki te 25.

$$\frac{25y^{2}+y}{25}=-\frac{32}{25}$$

Mā te whakawehe ki te 25 ka wetekia te whakareanga ki te 25.

$$y^{2}+\frac{1}{25}y=-\frac{32}{25}$$

Whakawehea te \frac{1}{25}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{50}. Nā, tāpiria te pūrua o te \frac{1}{50} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$y^{2}+\frac{1}{25}y+\left(\frac{1}{50}\right)^{2}=-\frac{32}{25}+\left(\frac{1}{50}\right)^{2}$$

Pūruatia \frac{1}{50} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$y^{2}+\frac{1}{25}y+\frac{1}{2500}=-\frac{32}{25}+\frac{1}{2500}$$

Tāpiri -\frac{32}{25} ki te \frac{1}{2500} 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.

$$y^{2}+\frac{1}{25}y+\frac{1}{2500}=-\frac{3199}{2500}$$

Tauwehea y^{2}+\frac{1}{25}y+\frac{1}{2500}. 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}.

$$\left(y+\frac{1}{50}\right)^{2}=-\frac{3199}{2500}$$

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

$$\sqrt{\left(y+\frac{1}{50}\right)^{2}}=\sqrt{-\frac{3199}{2500}}$$

Whakarūnātia.

$$y+\frac{1}{50}=\frac{\sqrt{3199}i}{50} y+\frac{1}{50}=-\frac{\sqrt{3199}i}{50}$$

Me tango \frac{1}{50} mai i ngā taha e rua o te whārite.

$$y=\frac{-1+\sqrt{3199}i}{50} y=\frac{-\sqrt{3199}i-1}{50}$$