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

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

Whakaoti mō x (complex solution)

$$x=\frac{-1+\sqrt{11}i}{6}\approx -0.166666667+0.552770798i$$

$$x=\frac{-\sqrt{11}i-1}{6}\approx -0.166666667-0.552770798i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te x^{2} mai i ngā taha e rua.

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

Pahekotia te -5x^{2} me -x^{2}, ka -6x^{2}.

$$-6x^{2}-2=2x$$

Tangohia te 2x mai i ngā taha e rua.

$$-6x^{2}-2-2x=0$$

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.

$$-6x^{2}-2x-2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -6 mō a, -2 mō b, me -2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-6\right)\left(-2\right)}}{2\left(-6\right)}$$

Pūrua -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-6\right)\left(-2\right)}}{2\left(-6\right)}$$

Whakareatia -4 ki te -6.

$$x=\frac{-\left(-2\right)±\sqrt{4+24\left(-2\right)}}{2\left(-6\right)}$$

Whakareatia 24 ki te -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-48}}{2\left(-6\right)}$$

Tāpiri 4 ki te -48.

$$x=\frac{-\left(-2\right)±\sqrt{-44}}{2\left(-6\right)}$$

Tuhia te pūtakerua o te -44.

$$x=\frac{-\left(-2\right)±2\sqrt{11}i}{2\left(-6\right)}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{11}i}{2\left(-6\right)}$$

Whakareatia 2 ki te -6.

$$x=\frac{2±2\sqrt{11}i}{-12}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{11}i}{-12} ina he tāpiri te ±. Tāpiri 2 ki te 2i\sqrt{11}.

$$x=\frac{2+2\sqrt{11}i}{-12}$$

Whakawehe 2+2i\sqrt{11} ki te -12.

$$x=\frac{-\sqrt{11}i-1}{6}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{11}i}{-12} ina he tango te ±. Tango 2i\sqrt{11} mai i 2.

$$x=\frac{-2\sqrt{11}i+2}{-12}$$

Whakawehe 2-2i\sqrt{11} ki te -12.

$$x=\frac{-1+\sqrt{11}i}{6}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{11}i-1}{6} x=\frac{-1+\sqrt{11}i}{6}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te x^{2} mai i ngā taha e rua.

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

Pahekotia te -5x^{2} me -x^{2}, ka -6x^{2}.

$$-6x^{2}-2=2x$$

Tangohia te 2x mai i ngā taha e rua.

$$-6x^{2}-2-2x=0$$

Me tāpiri te 2 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$-6x^{2}-2x=2$$

Whakawehea ngā taha e rua ki te -6.

$$\frac{-6x^{2}-2x}{-6}=\frac{2}{-6}$$

Mā te whakawehe ki te -6 ka wetekia te whakareanga ki te -6.

$$x^{2}+\left(-\frac{2}{-6}\right)x=\frac{2}{-6}$$

Whakahekea te hautanga \frac{-2}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}+\frac{1}{3}x=\frac{2}{-6}$$

Whakahekea te hautanga \frac{2}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Whakawehea te \frac{1}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{6}. Nā, tāpiria te pūrua o te \frac{1}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=-\frac{1}{3}+\left(\frac{1}{6}\right)^{2}$$

Pūruatia \frac{1}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=-\frac{1}{3}+\frac{1}{36}$$

Tāpiri -\frac{1}{3} ki te \frac{1}{36} 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.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=-\frac{11}{36}$$

Tauwehea x^{2}+\frac{1}{3}x+\frac{1}{36}. 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}.

$$\left(x+\frac{1}{6}\right)^{2}=-\frac{11}{36}$$

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

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{-\frac{11}{36}}$$

Whakarūnātia.

$$x+\frac{1}{6}=\frac{\sqrt{11}i}{6} x+\frac{1}{6}=-\frac{\sqrt{11}i}{6}$$

Me tango \frac{1}{6} mai i ngā taha e rua o te whārite.

$$x=\frac{-1+\sqrt{11}i}{6} x=\frac{-\sqrt{11}i-1}{6}$$