Whakaoti i te 3x^2+10x-5=0

$$3 { x }^{ 2 } +10x-5=0$$

Whakaoti mō x

$$x=\frac{2\sqrt{10}-5}{3}\approx 0.44151844$$

$$x=\frac{-2\sqrt{10}-5}{3}\approx -3.774851773$$

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.

$$3x^{2}+10x-5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 10 mō b, me -5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\times 3\left(-5\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-10±\sqrt{100-12\left(-5\right)}}{2\times 3}$$

Whakareatia -12 ki te -5.

$$x=\frac{-10±\sqrt{100+60}}{2\times 3}$$

Tāpiri 100 ki te 60.

$$x=\frac{-10±\sqrt{160}}{2\times 3}$$

Tuhia te pūtakerua o te 160.

$$x=\frac{-10±4\sqrt{10}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-10±4\sqrt{10}}{6}$$

Nā, me whakaoti te whārite x=\frac{-10±4\sqrt{10}}{6} ina he tāpiri te ±. Tāpiri -10 ki te 4\sqrt{10}.

$$x=\frac{4\sqrt{10}-10}{6}$$

Whakawehe -10+4\sqrt{10} ki te 6.

$$x=\frac{2\sqrt{10}-5}{3}$$

Nā, me whakaoti te whārite x=\frac{-10±4\sqrt{10}}{6} ina he tango te ±. Tango 4\sqrt{10} mai i -10.

$$x=\frac{-4\sqrt{10}-10}{6}$$

Whakawehe -10-4\sqrt{10} ki te 6.

$$x=\frac{-2\sqrt{10}-5}{3}$$

Kua oti te whārite te whakatau.

$$x=\frac{2\sqrt{10}-5}{3} x=\frac{-2\sqrt{10}-5}{3}$$

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.

$$3x^{2}+10x-5=0$$

Me tāpiri 5 ki ngā taha e rua o te whārite.

$$3x^{2}+10x-5-\left(-5\right)=-\left(-5\right)$$

Mā te tango i te -5 i a ia ake anō ka toe ko te 0.

$$3x^{2}+10x=-\left(-5\right)$$

Tango -5 mai i 0.

$$3x^{2}+10x=5$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}+10x}{3}=\frac{5}{3}$$

Mā te whakawehe ki te 3 ka wetekia te whakareanga ki te 3.

$$x^{2}+\frac{10}{3}x=\frac{5}{3}$$

Whakawehea te \frac{10}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{5}{3}. Nā, tāpiria te pūrua o te \frac{5}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=\frac{5}{3}+\left(\frac{5}{3}\right)^{2}$$

Pūruatia \frac{5}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{5}{3}+\frac{25}{9}$$

Tāpiri \frac{5}{3} ki te \frac{25}{9} 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.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{40}{9}$$

Tauwehea x^{2}+\frac{10}{3}x+\frac{25}{9}. 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}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{40}{9}$$

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

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{40}{9}}$$

Whakarūnātia.

$$x+\frac{5}{3}=\frac{2\sqrt{10}}{3} x+\frac{5}{3}=-\frac{2\sqrt{10}}{3}$$

Me tango \frac{5}{3} mai i ngā taha e rua o te whārite.

$$x=\frac{2\sqrt{10}-5}{3} x=\frac{-2\sqrt{10}-5}{3}$$