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

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

Whakaoti mō x

$$x = \frac{\sqrt{209} - 3}{10} \approx 1.145683229$$

$$x=\frac{-\sqrt{209}-3}{10}\approx -1.745683229$$

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.

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

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

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

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\times 5\left(-10\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-3±\sqrt{9-20\left(-10\right)}}{2\times 5}$$

Whakareatia -20 ki te -10.

$$x=\frac{-3±\sqrt{9+200}}{2\times 5}$$

Tāpiri 9 ki te 200.

$$x=\frac{-3±\sqrt{209}}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-3±\sqrt{209}}{10}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{209}}{10} ina he tāpiri te ±. Tāpiri -3 ki te \sqrt{209}.

$$x=\frac{\sqrt{209}-3}{10}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{209}}{10} ina he tango te ±. Tango \sqrt{209} mai i -3.

$$x=\frac{-\sqrt{209}-3}{10}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{209}-3}{10} x=\frac{-\sqrt{209}-3}{10}$$

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.

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

Me tāpiri 10 ki ngā taha e rua o te whārite.

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

Mā te tango i te -10 i a ia ake anō ka toe ko te 0.

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

Tango -10 mai i 0.

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

Whakawehea ngā taha e rua ki te 5.

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

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

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

Whakawehe 10 ki te 5.

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

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

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

Pūruatia \frac{3}{10} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=2+\frac{9}{100}$$

Tāpiri 2 ki te \frac{9}{100}.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{209}{100}$$

Tauwehea te x^{2}+\frac{3}{5}x+\frac{9}{100}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{10}\right)^{2}=\frac{209}{100}$$

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

$$\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{\frac{209}{100}}$$

Whakarūnātia.

$$x+\frac{3}{10}=\frac{\sqrt{209}}{10} x+\frac{3}{10}=-\frac{\sqrt{209}}{10}$$

Me tango \frac{3}{10} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{209}-3}{10} x=\frac{-\sqrt{209}-3}{10}$$