Whakaoti i te 4x^2+2x-8=0

$$4 x ^ { 2 } + 2 x - 8 = 0$$

Whakaoti mō x

$$x = \frac{\sqrt{33} - 1}{4} \approx 1.186140662$$

$$x=\frac{-\sqrt{33}-1}{4}\approx -1.686140662$$

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.

$$4x^{2}+2x-8=0$$

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

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

Pūrua 2.

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

Whakareatia -4 ki te 4.

$$x=\frac{-2±\sqrt{4-16\left(-8\right)}}{2\times 4}$$

Whakareatia -16 ki te -8.

$$x=\frac{-2±\sqrt{4+128}}{2\times 4}$$

Tāpiri 4 ki te 128.

$$x=\frac{-2±\sqrt{132}}{2\times 4}$$

Tuhia te pūtakerua o te 132.

$$x=\frac{-2±2\sqrt{33}}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-2±2\sqrt{33}}{8}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{33}}{8} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{33}.

$$x=\frac{2\sqrt{33}-2}{8}$$

Whakawehe -2+2\sqrt{33} ki te 8.

$$x=\frac{\sqrt{33}-1}{4}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{33}}{8} ina he tango te ±. Tango 2\sqrt{33} mai i -2.

$$x=\frac{-2\sqrt{33}-2}{8}$$

Whakawehe -2-2\sqrt{33} ki te 8.

$$x=\frac{-\sqrt{33}-1}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{33}-1}{4} x=\frac{-\sqrt{33}-1}{4}$$

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.

$$4x^{2}+2x-8=0$$

Me tāpiri 8 ki ngā taha e rua o te whārite.

$$4x^{2}+2x-8-\left(-8\right)=-\left(-8\right)$$

Mā te tango i te -8 i a ia ake anō ka toe ko te 0.

$$4x^{2}+2x=-\left(-8\right)$$

Tango -8 mai i 0.

$$4x^{2}+2x=8$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}+2x}{4}=\frac{8}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$x^{2}+\frac{2}{4}x=\frac{8}{4}$$

Whakahekea te hautanga \frac{2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}+\frac{1}{2}x=\frac{8}{4}$$

Whakawehe 8 ki te 4.

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

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.

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

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

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

Tāpiri 2 ki te \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{33}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{33}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Whakarūnātia.

$$x+\frac{1}{4}=\frac{\sqrt{33}}{4} x+\frac{1}{4}=-\frac{\sqrt{33}}{4}$$

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

$$x=\frac{\sqrt{33}-1}{4} x=\frac{-\sqrt{33}-1}{4}$$