Whakaoti i te y^2-6y+25=0

$${ y }^{ 2 } -6y+25=0$$

Whakaoti mō y

$$y=3+4i$$

$$y=3-4i$$

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.

$$y^{2}-6y+25=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -6 mō b, me 25 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 25}}{2}$$

Pūrua -6.

$$y=\frac{-\left(-6\right)±\sqrt{36-4\times 25}}{2}$$

Whakareatia -4 ki te 25.

$$y=\frac{-\left(-6\right)±\sqrt{36-100}}{2}$$

Tāpiri 36 ki te -100.

$$y=\frac{-\left(-6\right)±\sqrt{-64}}{2}$$

Tuhia te pūtakerua o te -64.

$$y=\frac{-\left(-6\right)±8i}{2}$$

Ko te tauaro o -6 ko 6.

$$y=\frac{6±8i}{2}$$

Nā, me whakaoti te whārite y=\frac{6±8i}{2} ina he tāpiri te ±. Tāpiri 6 ki te 8i.

$$y=\frac{6+8i}{2}$$

Whakawehe 6+8i ki te 2.

$$y=3+4i$$

Nā, me whakaoti te whārite y=\frac{6±8i}{2} ina he tango te ±. Tango 8i mai i 6.

$$y=\frac{6-8i}{2}$$

Whakawehe 6-8i ki te 2.

$$y=3-4i$$

Kua oti te whārite te whakatau.

$$y=3+4i y=3-4i$$

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.

$$y^{2}-6y+25=0$$

Me tango 25 mai i ngā taha e rua o te whārite.

$$y^{2}-6y+25-25=-25$$

Mā te tango i te 25 i a ia ake anō ka toe ko te 0.

$$y^{2}-6y=-25$$

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

$$y^{2}-6y+\left(-3\right)^{2}=-25+\left(-3\right)^{2}$$

Pūrua -3.

$$y^{2}-6y+9=-25+9$$

Tāpiri -25 ki te 9.

$$y^{2}-6y+9=-16$$

Tauwehea y^{2}-6y+9. 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}.

$$\left(y-3\right)^{2}=-16$$

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

$$\sqrt{\left(y-3\right)^{2}}=\sqrt{-16}$$

Whakarūnātia.

$$y-3=4i y-3=-4i$$

Me tāpiri 3 ki ngā taha e rua o te whārite.

$$y=3+4i y=3-4i$$