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

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

Whakaoti mō x

$$x=\frac{2}{5}=0.4$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te x.

$$x\left(2-5x\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 2-5x=0.

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

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.

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

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

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

Tuhia te pūtakerua o te 2^{2}.

$$x=\frac{-2±2}{2\left(-5\right)}$$

Whakareatia 2 ki te -5.

$$x=\frac{-2±2}{-10}$$

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

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

Whakawehe 0 ki te -10.

$$x=0$$

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

$$x=-\frac{4}{-10}$$

Whakahekea te hautanga \frac{-4}{-10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

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.

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

Whakawehea ngā taha e rua ki te -5.

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

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

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

Whakawehe 2 ki te -5.

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

Whakawehe 0 ki te -5.

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

Whakawehea te -\frac{2}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{5}. Nā, tāpiria te pūrua o te -\frac{1}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=\left(-\frac{1}{5}\right)^{2}$$

Pūruatia -\frac{1}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{1}{25}$$

Tauwehea x^{2}-\frac{2}{5}x+\frac{1}{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}.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{1}{25}$$

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

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{1}{25}}$$

Whakarūnātia.

$$x-\frac{1}{5}=\frac{1}{5} x-\frac{1}{5}=-\frac{1}{5}$$

Me tāpiri \frac{1}{5} ki ngā taha e rua o te whārite.

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