Whakaoti i te 5x^2+x-7=0

$$5 x ^ { 2 } + x - 7 = 0$$

Whakaoti mō x

$$x = \frac{\sqrt{141} - 1}{10} \approx 1.087434209$$

$$x=\frac{-\sqrt{141}-1}{10}\approx -1.287434209$$

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.

$$5x^{2}+x-7=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, 1 mō b, me -7 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 5\left(-7\right)}}{2\times 5}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\times 5\left(-7\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-1±\sqrt{1-20\left(-7\right)}}{2\times 5}$$

Whakareatia -20 ki te -7.

$$x=\frac{-1±\sqrt{1+140}}{2\times 5}$$

Tāpiri 1 ki te 140.

$$x=\frac{-1±\sqrt{141}}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-1±\sqrt{141}}{10}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{141}}{10} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{141}.

$$x=\frac{\sqrt{141}-1}{10}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{141}}{10} ina he tango te ±. Tango \sqrt{141} mai i -1.

$$x=\frac{-\sqrt{141}-1}{10}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{141}-1}{10} x=\frac{-\sqrt{141}-1}{10}$$

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.

$$5x^{2}+x-7=0$$

Me tāpiri 7 ki ngā taha e rua o te whārite.

$$5x^{2}+x-7-\left(-7\right)=-\left(-7\right)$$

Mā te tango i te -7 i a ia ake anō ka toe ko te 0.

$$5x^{2}+x=-\left(-7\right)$$

Tango -7 mai i 0.

$$5x^{2}+x=7$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}+x}{5}=\frac{7}{5}$$

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

$$x^{2}+\frac{1}{5}x=\frac{7}{5}$$

Whakawehea te \frac{1}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{10}. Nā, tāpiria te pūrua o te \frac{1}{10} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{5}x+\left(\frac{1}{10}\right)^{2}=\frac{7}{5}+\left(\frac{1}{10}\right)^{2}$$

Pūruatia \frac{1}{10} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{7}{5}+\frac{1}{100}$$

Tāpiri \frac{7}{5} ki te \frac{1}{100} 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.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{141}{100}$$

Tauwehea te x^{2}+\frac{1}{5}x+\frac{1}{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}.

$$\left(x+\frac{1}{10}\right)^{2}=\frac{141}{100}$$

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

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{\frac{141}{100}}$$

Whakarūnātia.

$$x+\frac{1}{10}=\frac{\sqrt{141}}{10} x+\frac{1}{10}=-\frac{\sqrt{141}}{10}$$

Me tango \frac{1}{10} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{141}-1}{10} x=\frac{-\sqrt{141}-1}{10}$$