Whakaoti i te x^2+4x=7x-8

$$x ^ { 2 } + 4 x = 7 x - 8$$

Whakaoti mō x (complex solution)

$$x=\frac{3+\sqrt{23}i}{2}\approx 1.5+2.397915762i$$

$$x=\frac{-\sqrt{23}i+3}{2}\approx 1.5-2.397915762i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 7x mai i ngā taha e rua.

$$x^{2}+4x-7x=-8$$

Pahekotia te 4x me -7x, ka -3x.

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

Me tāpiri te 8 ki ngā taha e rua.

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

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

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

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 8}}{2}$$

Whakareatia -4 ki te 8.

$$x=\frac{-\left(-3\right)±\sqrt{9-32}}{2}$$

Tāpiri 9 ki te -32.

$$x=\frac{-\left(-3\right)±\sqrt{-23}}{2}$$

Tuhia te pūtakerua o te -23.

$$x=\frac{-\left(-3\right)±\sqrt{23}i}{2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±\sqrt{23}i}{2}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{23}i}{2} ina he tāpiri te ±. Tāpiri 3 ki te i\sqrt{23}.

$$x=\frac{3+\sqrt{23}i}{2}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{23}i}{2} ina he tango te ±. Tango i\sqrt{23} mai i 3.

$$x=\frac{-\sqrt{23}i+3}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{3+\sqrt{23}i}{2} x=\frac{-\sqrt{23}i+3}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 7x mai i ngā taha e rua.

$$x^{2}+4x-7x=-8$$

Pahekotia te 4x me -7x, ka -3x.

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

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

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

Pūruatia -\frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-3x+\frac{9}{4}=-8+\frac{9}{4}$$

Tāpiri -8 ki te \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=-\frac{23}{4}$$

Tauwehea x^{2}-3x+\frac{9}{4}. 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}.

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{23}{4}$$

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

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{23}{4}}$$

Whakarūnātia.

$$x-\frac{3}{2}=\frac{\sqrt{23}i}{2} x-\frac{3}{2}=-\frac{\sqrt{23}i}{2}$$

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

$$x=\frac{3+\sqrt{23}i}{2} x=\frac{-\sqrt{23}i+3}{2}$$