Whakaoti i te 2x^2-9x+5=0

$$2 x ^ { 2 } - 9 x + 5 = 0$$

Whakaoti mō x

$$x = \frac{\sqrt{41} + 9}{4} \approx 3.850781059$$

$$x=\frac{9-\sqrt{41}}{4}\approx 0.649218941$$

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.

$$2x^{2}-9x+5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -9 mō b, me 5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\times 5}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\times 5}}{2\times 2}$$

Whakareatia -8 ki te 5.

$$x=\frac{-\left(-9\right)±\sqrt{81-40}}{2\times 2}$$

Tāpiri 81 ki te -40.

$$x=\frac{-\left(-9\right)±\sqrt{41}}{2\times 2}$$

Ko te tauaro o -9 ko 9.

$$x=\frac{9±\sqrt{41}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{9±\sqrt{41}}{4}$$

Nā, me whakaoti te whārite x=\frac{9±\sqrt{41}}{4} ina he tāpiri te ±. Tāpiri 9 ki te \sqrt{41}.

$$x=\frac{\sqrt{41}+9}{4}$$

Nā, me whakaoti te whārite x=\frac{9±\sqrt{41}}{4} ina he tango te ±. Tango \sqrt{41} mai i 9.

$$x=\frac{9-\sqrt{41}}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{41}+9}{4} x=\frac{9-\sqrt{41}}{4}$$

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.

$$2x^{2}-9x+5=0$$

Me tango 5 mai i ngā taha e rua o te whārite.

$$2x^{2}-9x+5-5=-5$$

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

$$2x^{2}-9x=-5$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-9x}{2}=-\frac{5}{2}$$

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

$$x^{2}-\frac{9}{2}x=-\frac{5}{2}$$

Whakawehea te -\frac{9}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{9}{4}. Nā, tāpiria te pūrua o te -\frac{9}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=-\frac{5}{2}+\left(-\frac{9}{4}\right)^{2}$$

Pūruatia -\frac{9}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=-\frac{5}{2}+\frac{81}{16}$$

Tāpiri -\frac{5}{2} ki te \frac{81}{16} 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.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{41}{16}$$

Tauwehea x^{2}-\frac{9}{2}x+\frac{81}{16}. 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}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{41}{16}$$

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

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Whakarūnātia.

$$x-\frac{9}{4}=\frac{\sqrt{41}}{4} x-\frac{9}{4}=-\frac{\sqrt{41}}{4}$$

Me tāpiri \frac{9}{4} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{41}+9}{4} x=\frac{9-\sqrt{41}}{4}$$