Whakaoti i te 7x-1-5-2x=4x-3-1+4x^2

$$7 x - 1 - 5 - 2 x = 4 x - 3 - 1 + 4 x ^ { 2 }$$

Whakaoti mō x (complex solution)

$$x=\frac{-\sqrt{31}i+1}{8}\approx 0.125-0.695970545i$$

$$x=\frac{1+\sqrt{31}i}{8}\approx 0.125+0.695970545i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 5 i te -1, ka -6.

$$7x-6-2x=4x-3-1+4x^{2}$$

Pahekotia te 7x me -2x, ka 5x.

$$5x-6=4x-3-1+4x^{2}$$

Tangohia te 1 i te -3, ka -4.

$$5x-6=4x-4+4x^{2}$$

Tangohia te 4x mai i ngā taha e rua.

$$5x-6-4x=-4+4x^{2}$$

Pahekotia te 5x me -4x, ka x.

$$x-6=-4+4x^{2}$$

Tangohia te -4 mai i ngā taha e rua.

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

Ko te tauaro o -4 ko 4.

$$x-6+4=4x^{2}$$

Tangohia te 4x^{2} mai i ngā taha e rua.

$$x-6+4-4x^{2}=0$$

Tāpirihia te -6 ki te 4, ka -2.

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

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.

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

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}.

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

Pūrua 1.

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

Whakareatia -4 ki te -4.

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

Whakareatia 16 ki te -2.

$$x=\frac{-1±\sqrt{1-32}}{2\left(-4\right)}$$

Tāpiri 1 ki te -32.

$$x=\frac{-1±\sqrt{-31}}{2\left(-4\right)}$$

Tuhia te pūtakerua o te -31.

$$x=\frac{-1±\sqrt{31}i}{2\left(-4\right)}$$

Whakareatia 2 ki te -4.

$$x=\frac{-1±\sqrt{31}i}{-8}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{31}i}{-8} ina he tāpiri te ±. Tāpiri -1 ki te i\sqrt{31}.

$$x=\frac{-1+\sqrt{31}i}{-8}$$

Whakawehe -1+i\sqrt{31} ki te -8.

$$x=\frac{-\sqrt{31}i+1}{8}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{31}i}{-8} ina he tango te ±. Tango i\sqrt{31} mai i -1.

$$x=\frac{-\sqrt{31}i-1}{-8}$$

Whakawehe -1-i\sqrt{31} ki te -8.

$$x=\frac{1+\sqrt{31}i}{8}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{31}i+1}{8} x=\frac{1+\sqrt{31}i}{8}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 5 i te -1, ka -6.

$$7x-6-2x=4x-3-1+4x^{2}$$

Pahekotia te 7x me -2x, ka 5x.

$$5x-6=4x-3-1+4x^{2}$$

Tangohia te 1 i te -3, ka -4.

$$5x-6=4x-4+4x^{2}$$

Tangohia te 4x mai i ngā taha e rua.

$$5x-6-4x=-4+4x^{2}$$

Pahekotia te 5x me -4x, ka x.

$$x-6=-4+4x^{2}$$

Tangohia te 4x^{2} mai i ngā taha e rua.

$$x-6-4x^{2}=-4$$

Me tāpiri te 6 ki ngā taha e rua.

$$x-4x^{2}=-4+6$$

Tāpirihia te -4 ki te 6, ka 2.

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

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.

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

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.

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

Whakawehe 1 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.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=-\frac{31}{64}$$

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}.

$$\left(x-\frac{1}{8}\right)^{2}=-\frac{31}{64}$$

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

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{-\frac{31}{64}}$$

Whakarūnātia.

$$x-\frac{1}{8}=\frac{\sqrt{31}i}{8} x-\frac{1}{8}=-\frac{\sqrt{31}i}{8}$$

Me tāpiri \frac{1}{8} ki ngā taha e rua o te whārite.

$$x=\frac{1+\sqrt{31}i}{8} x=\frac{-\sqrt{31}i+1}{8}$$