Whakaoti i te 2x^2+x=0

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

Whakaoti mō x

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

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te x.

$$x\left(2x+1\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 2x+1=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.

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

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

$$x=\frac{-1±\sqrt{1^{2}}}{2\times 2}$$

Tuhia te pūtakerua o te 1^{2}.

$$x=\frac{-1±1}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-1±1}{4}$$

Nā, me whakaoti te whārite x=\frac{-1±1}{4} ina he tāpiri te ±. Tāpiri -1 ki te 1.

$$x=\frac{0}{4}$$

Whakawehe 0 ki te 4.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-1±1}{4} ina he tango te ±. Tango 1 mai i -1.

$$x=-\frac{2}{4}$$

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

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

Kua oti te whārite te whakatau.

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

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.

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

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

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

Whakawehe 0 ki te 2.

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

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 tango \frac{1}{4} mai i ngā taha e rua o te whārite.

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