Whakaoti i te 2/5x-4x^2=0

$$\frac { 2 } { 5 } x - 4 x ^ { 2 } = 0$$

Whakaoti mō x

$$x=\frac{1}{10}=0.1$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te x.

$$x\left(\frac{2}{5}-4x\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te \frac{2}{5}-4x=0.

$$x=0 x=\frac{1}{10}$$

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.

$$-4x^{2}+\frac{2}{5}x=0$$

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

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

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

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

Whakareatia 2 ki te -4.

$$x=\frac{-\frac{2}{5}±\frac{2}{5}}{-8}$$

Nā, me whakaoti te whārite x=\frac{-\frac{2}{5}±\frac{2}{5}}{-8} ina he tāpiri te ±. Tāpiri -\frac{2}{5} ki te \frac{2}{5} 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.

$$x=\frac{0}{-8}$$

Whakawehe 0 ki te -8.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-\frac{2}{5}±\frac{2}{5}}{-8} ina he tango te ±. Tango \frac{2}{5} mai i -\frac{2}{5} mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x=-\frac{\frac{4}{5}}{-8}$$

Whakawehe -\frac{4}{5} ki te -8.

$$x=\frac{1}{10}$$

Kua oti te whārite te whakatau.

$$x=0 x=\frac{1}{10}$$

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.

$$-4x^{2}+\frac{2}{5}x=0$$

Whakawehea ngā taha e rua ki te -4.

$$\frac{-4x^{2}+\frac{2}{5}x}{-4}=\frac{0}{-4}$$

Mā te whakawehe ki te -4 ka wetekia te whakareanga ki te -4.

$$x^{2}+\frac{\frac{2}{5}}{-4}x=\frac{0}{-4}$$

Whakawehe \frac{2}{5} ki te -4.

$$x^{2}-\frac{1}{10}x=\frac{0}{-4}$$

Whakawehe 0 ki te -4.

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

Whakawehea te -\frac{1}{10}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{20}. Nā, tāpiria te pūrua o te -\frac{1}{20} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{10}x+\left(-\frac{1}{20}\right)^{2}=\left(-\frac{1}{20}\right)^{2}$$

Pūruatia -\frac{1}{20} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{1}{400}$$

Tauwehea x^{2}-\frac{1}{10}x+\frac{1}{400}. 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}.

$$\left(x-\frac{1}{20}\right)^{2}=\frac{1}{400}$$

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

$$\sqrt{\left(x-\frac{1}{20}\right)^{2}}=\sqrt{\frac{1}{400}}$$

Whakarūnātia.

$$x-\frac{1}{20}=\frac{1}{20} x-\frac{1}{20}=-\frac{1}{20}$$

Me tāpiri \frac{1}{20} ki ngā taha e rua o te whārite.

$$x=\frac{1}{10} x=0$$