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

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

Whakaoti mō x

$$x = \frac{\sqrt{41} - 1}{4} \approx 1.350781059$$

$$x=\frac{-\sqrt{41}-1}{4}\approx -1.850781059$$

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.

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

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

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

Pūrua 1.

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

Whakareatia -4 ki te 2.

$$x=\frac{-1±\sqrt{1-8\left(-5\right)}}{2\times 2}$$

Whakareatia -8 ki te -5.

$$x=\frac{-1±\sqrt{1+40}}{2\times 2}$$

Tāpiri 1 ki te 40.

$$x=\frac{-1±\sqrt{41}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-1±\sqrt{41}}{4}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{41}}{4} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{41}.

$$x=\frac{\sqrt{41}-1}{4}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{41}}{4} ina he tango te ±. Tango \sqrt{41} mai i -1.

$$x=\frac{-\sqrt{41}-1}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{41}-1}{4} x=\frac{-\sqrt{41}-1}{4}$$

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.

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

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

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

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

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

Tango -5 mai i 0.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehea te \frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{4}. Nā, tāpiria te pūrua o te \frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{5}{2}+\left(\frac{1}{4}\right)^{2}$$

Pūruatia \frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{5}{2}+\frac{1}{16}$$

Tāpiri \frac{5}{2} ki te \frac{1}{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.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{41}{16}$$

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

$$\left(x+\frac{1}{4}\right)^{2}=\frac{41}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Whakarūnātia.

$$x+\frac{1}{4}=\frac{\sqrt{41}}{4} x+\frac{1}{4}=-\frac{\sqrt{41}}{4}$$

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{41}-1}{4} x=\frac{-\sqrt{41}-1}{4}$$