Whakaoti i te 5x^2+x-12=0

$$5 x ^ { 2 } + x - 12 = 0$$

Whakaoti mō x

$$x = \frac{\sqrt{241} - 1}{10} \approx 1.45241747$$

$$x=\frac{-\sqrt{241}-1}{10}\approx -1.65241747$$

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.

$$5x^{2}+x-12=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, 1 mō b, me -12 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 5\left(-12\right)}}{2\times 5}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\times 5\left(-12\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-1±\sqrt{1-20\left(-12\right)}}{2\times 5}$$

Whakareatia -20 ki te -12.

$$x=\frac{-1±\sqrt{1+240}}{2\times 5}$$

Tāpiri 1 ki te 240.

$$x=\frac{-1±\sqrt{241}}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-1±\sqrt{241}}{10}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{241}}{10} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{241}.

$$x=\frac{\sqrt{241}-1}{10}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{241}}{10} ina he tango te ±. Tango \sqrt{241} mai i -1.

$$x=\frac{-\sqrt{241}-1}{10}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{241}-1}{10} x=\frac{-\sqrt{241}-1}{10}$$

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.

$$5x^{2}+x-12=0$$

Me tāpiri 12 ki ngā taha e rua o te whārite.

$$5x^{2}+x-12-\left(-12\right)=-\left(-12\right)$$

Mā te tango i te -12 i a ia ake anō ka toe ko te 0.

$$5x^{2}+x=-\left(-12\right)$$

Tango -12 mai i 0.

$$5x^{2}+x=12$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}+x}{5}=\frac{12}{5}$$

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

$$x^{2}+\frac{1}{5}x=\frac{12}{5}$$

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.

$$x^{2}+\frac{1}{5}x+\left(\frac{1}{10}\right)^{2}=\frac{12}{5}+\left(\frac{1}{10}\right)^{2}$$

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

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{12}{5}+\frac{1}{100}$$

Tāpiri \frac{12}{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.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{241}{100}$$

Tauwehea x^{2}+\frac{1}{5}x+\frac{1}{100}. 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}.

$$\left(x+\frac{1}{10}\right)^{2}=\frac{241}{100}$$

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

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{\frac{241}{100}}$$

Whakarūnātia.

$$x+\frac{1}{10}=\frac{\sqrt{241}}{10} x+\frac{1}{10}=-\frac{\sqrt{241}}{10}$$

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

$$x=\frac{\sqrt{241}-1}{10} x=\frac{-\sqrt{241}-1}{10}$$