Whakaoti i te 10x^2-5x=0

$$10 x ^ { 2 } - 5 x = 0$$

Whakaoti mō x

$$x=\frac{1}{2}=0.5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te x.

$$x\left(10x-5\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 10x-5=0.

$$x=0 x=\frac{1}{2}$$

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.

$$10x^{2}-5x=0$$

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

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2\times 10}$$

Tuhia te pūtakerua o te \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2\times 10}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±5}{2\times 10}$$

Whakareatia 2 ki te 10.

$$x=\frac{5±5}{20}$$

Nā, me whakaoti te whārite x=\frac{5±5}{20} ina he tāpiri te ±. Tāpiri 5 ki te 5.

$$x=\frac{10}{20}$$

Whakahekea te hautanga \frac{10}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$x=\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{5±5}{20} ina he tango te ±. Tango 5 mai i 5.

$$x=\frac{0}{20}$$

Whakawehe 0 ki te 20.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=\frac{1}{2} x=0$$

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.

$$10x^{2}-5x=0$$

Whakawehea ngā taha e rua ki te 10.

$$\frac{10x^{2}-5x}{10}=\frac{0}{10}$$

Mā te whakawehe ki te 10 ka wetekia te whakareanga ki te 10.

$$x^{2}+\left(-\frac{5}{10}\right)x=\frac{0}{10}$$

Whakahekea te hautanga \frac{-5}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 5.

$$x^{2}-\frac{1}{2}x=\frac{0}{10}$$

Whakawehe 0 ki te 10.

$$x^{2}-\frac{1}{2}x=0$$

Whakawehea te -\frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{4}. Nā, tāpiria te pūrua o te -\frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\left(-\frac{1}{4}\right)^{2}$$

Pūruatia -\frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Tauwehea x^{2}-\frac{1}{2}x+\frac{1}{16}. 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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

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

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Whakarūnātia.

$$x-\frac{1}{4}=\frac{1}{4} x-\frac{1}{4}=-\frac{1}{4}$$

Me tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

$$x=\frac{1}{2} x=0$$