Whakaoti i te 5x^2-25x=0

$$5 x ^ { 2 } - 25 x = 0$$

Whakaoti mō x

$$x=5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te x.

$$x\left(5x-25\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 5x-25=0.

$$x=0 x=5$$

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.

$$5x^{2}-25x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -25 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}}}{2\times 5}$$

Tuhia te pūtakerua o te \left(-25\right)^{2}.

$$x=\frac{-\left(-25\right)±25}{2\times 5}$$

Ko te tauaro o -25 ko 25.

$$x=\frac{25±25}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{25±25}{10}$$

Nā, me whakaoti te whārite x=\frac{25±25}{10} ina he tāpiri te ±. Tāpiri 25 ki te 25.

$$x=\frac{50}{10}$$

Whakawehe 50 ki te 10.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{25±25}{10} ina he tango te ±. Tango 25 mai i 25.

$$x=\frac{0}{10}$$

Whakawehe 0 ki te 10.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=5 x=0$$

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.

$$5x^{2}-25x=0$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}-25x}{5}=\frac{0}{5}$$

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

$$x^{2}+\left(-\frac{25}{5}\right)x=\frac{0}{5}$$

Whakawehe -25 ki te 5.

$$x^{2}-5x=\frac{0}{5}$$

Whakawehe 0 ki te 5.

$$x^{2}-5x=0$$

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.

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

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

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

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{5}{2} x-\frac{5}{2}=-\frac{5}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=5 x=0$$