Whakaoti i te 2x^2+10x=0

$$2 { x }^{ 2 } +10x=0$$

Whakaoti mō x

$$x=-5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te x.

$$x\left(2x+10\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 2x+10=0.

$$x=0 x=-5$$

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.

$$2x^{2}+10x=0$$

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

$$x=\frac{-10±\sqrt{10^{2}}}{2\times 2}$$

Tuhia te pūtakerua o te 10^{2}.

$$x=\frac{-10±10}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-10±10}{4}$$

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

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

Whakawehe 0 ki te 4.

$$x=0$$

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

$$x=-\frac{20}{4}$$

Whakawehe -20 ki te 4.

$$x=-5$$

Kua oti te whārite te whakatau.

$$x=0 x=-5$$

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.

$$2x^{2}+10x=0$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}+10x}{2}=\frac{0}{2}$$

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

$$x^{2}+\frac{10}{2}x=\frac{0}{2}$$

Whakawehe 10 ki te 2.

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

Whakawehe 0 ki te 2.

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

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

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

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

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

Tauwehea x^{2}+5x+\frac{25}{4}. 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}.

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

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

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

Whakarūnātia.

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

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

$$x=0 x=-5$$