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

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

Whakaoti mō x

$$x=\frac{\sqrt{33}-1}{8}\approx 0.593070331$$

$$x=\frac{-\sqrt{33}-1}{8}\approx -0.843070331$$

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.

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

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

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

Pūrua 1.

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

Whakareatia -4 ki te 4.

$$x=\frac{-1±\sqrt{1-16\left(-2\right)}}{2\times 4}$$

Whakareatia -16 ki te -2.

$$x=\frac{-1±\sqrt{1+32}}{2\times 4}$$

Tāpiri 1 ki te 32.

$$x=\frac{-1±\sqrt{33}}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-1±\sqrt{33}}{8}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{33}}{8} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{33}.

$$x=\frac{\sqrt{33}-1}{8}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{33}}{8} ina he tango te ±. Tango \sqrt{33} mai i -1.

$$x=\frac{-\sqrt{33}-1}{8}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{33}-1}{8} x=\frac{-\sqrt{33}-1}{8}$$

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.

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

Me tāpiri 2 ki ngā taha e rua o te whārite.

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

Mā te tango i te -2 i a ia ake anō ka toe ko te 0.

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

Tango -2 mai i 0.

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

Whakawehea ngā taha e rua ki te 4.

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

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

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

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

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

Whakawehea te \frac{1}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{8}. Nā, tāpiria te pūrua o te \frac{1}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\frac{1}{2}+\left(\frac{1}{8}\right)^{2}$$

Pūruatia \frac{1}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{1}{2}+\frac{1}{64}$$

Tāpiri \frac{1}{2} ki te \frac{1}{64} 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.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{33}{64}$$

Tauwehea x^{2}+\frac{1}{4}x+\frac{1}{64}. 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}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{33}{64}$$

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

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{33}{64}}$$

Whakarūnātia.

$$x+\frac{1}{8}=\frac{\sqrt{33}}{8} x+\frac{1}{8}=-\frac{\sqrt{33}}{8}$$

Me tango \frac{1}{8} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{33}-1}{8} x=\frac{-\sqrt{33}-1}{8}$$