Whakaoti i te 5x^2+4x=0

$$5 x ^ { 2 } + 4 x = 0$$

Whakaoti mō x

$$x=-\frac{4}{5}=-0.8$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te x.

$$x\left(5x+4\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 5x+4=0.

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

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.

$$5x^{2}+4x=0$$

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

$$x=\frac{-4±\sqrt{4^{2}}}{2\times 5}$$

Tuhia te pūtakerua o te 4^{2}.

$$x=\frac{-4±4}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-4±4}{10}$$

Nā, me whakaoti te whārite x=\frac{-4±4}{10} ina he tāpiri te ±. Tāpiri -4 ki te 4.

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

Whakawehe 0 ki te 10.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-4±4}{10} ina he tango te ±. Tango 4 mai i -4.

$$x=-\frac{8}{10}$$

Whakahekea te hautanga \frac{-8}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

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.

$$5x^{2}+4x=0$$

Whakawehea ngā taha e rua ki te 5.

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

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

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

Whakawehe 0 ki te 5.

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

Whakawehea te \frac{4}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{2}{5}. Nā, tāpiria te pūrua o te \frac{2}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

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

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

Tauwehea x^{2}+\frac{4}{5}x+\frac{4}{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}.

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

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

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

Whakarūnātia.

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

Me tango \frac{2}{5} mai i ngā taha e rua o te whārite.

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