Whakaoti i te 8x^2+x-3=0

$$8 x ^ { 2 } + x - 3 = 0$$

Whakaoti mō x

$$x=\frac{\sqrt{97}-1}{16}\approx 0.553053613$$

$$x=\frac{-\sqrt{97}-1}{16}\approx -0.678053613$$

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.

$$8x^{2}+x-3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 8 mō a, 1 mō b, me -3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\times 8\left(-3\right)}}{2\times 8}$$

Whakareatia -4 ki te 8.

$$x=\frac{-1±\sqrt{1-32\left(-3\right)}}{2\times 8}$$

Whakareatia -32 ki te -3.

$$x=\frac{-1±\sqrt{1+96}}{2\times 8}$$

Tāpiri 1 ki te 96.

$$x=\frac{-1±\sqrt{97}}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{-1±\sqrt{97}}{16}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{97}}{16} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{97}.

$$x=\frac{\sqrt{97}-1}{16}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{97}}{16} ina he tango te ±. Tango \sqrt{97} mai i -1.

$$x=\frac{-\sqrt{97}-1}{16}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{97}-1}{16} x=\frac{-\sqrt{97}-1}{16}$$

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.

$$8x^{2}+x-3=0$$

Me tāpiri 3 ki ngā taha e rua o te whārite.

$$8x^{2}+x-3-\left(-3\right)=-\left(-3\right)$$

Mā te tango i te -3 i a ia ake anō ka toe ko te 0.

$$8x^{2}+x=-\left(-3\right)$$

Tango -3 mai i 0.

$$8x^{2}+x=3$$

Whakawehea ngā taha e rua ki te 8.

$$\frac{8x^{2}+x}{8}=\frac{3}{8}$$

Mā te whakawehe ki te 8 ka wetekia te whakareanga ki te 8.

$$x^{2}+\frac{1}{8}x=\frac{3}{8}$$

Whakawehea te \frac{1}{8}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{16}. Nā, tāpiria te pūrua o te \frac{1}{16} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{8}x+\left(\frac{1}{16}\right)^{2}=\frac{3}{8}+\left(\frac{1}{16}\right)^{2}$$

Pūruatia \frac{1}{16} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{3}{8}+\frac{1}{256}$$

Tāpiri \frac{3}{8} ki te \frac{1}{256} 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.

$$x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{97}{256}$$

Tauwehea x^{2}+\frac{1}{8}x+\frac{1}{256}. 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}.

$$\left(x+\frac{1}{16}\right)^{2}=\frac{97}{256}$$

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

$$\sqrt{\left(x+\frac{1}{16}\right)^{2}}=\sqrt{\frac{97}{256}}$$

Whakarūnātia.

$$x+\frac{1}{16}=\frac{\sqrt{97}}{16} x+\frac{1}{16}=-\frac{\sqrt{97}}{16}$$

Me tango \frac{1}{16} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{97}-1}{16} x=\frac{-\sqrt{97}-1}{16}$$