Whakaoti i te 2x+3/x=4

$$2 x + \frac { 3 } { x } = 4$$

Whakaoti mō x (complex solution)

$$x=\frac{\sqrt{2}i}{2}+1\approx 1+0.707106781i$$

$$x=-\frac{\sqrt{2}i}{2}+1\approx 1-0.707106781i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x.

$$2xx+3=4x$$

Whakareatia te x ki te x, ka x^{2}.

$$2x^{2}+3=4x$$

Tangohia te 4x mai i ngā taha e rua.

$$2x^{2}+3-4x=0$$

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.

$$2x^{2}-4x+3=0$$

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

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\times 3}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\times 3}}{2\times 2}$$

Whakareatia -8 ki te 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-24}}{2\times 2}$$

Tāpiri 16 ki te -24.

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

Tuhia te pūtakerua o te -8.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2\sqrt{2}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{4±2\sqrt{2}i}{4}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{2}i}{4} ina he tāpiri te ±. Tāpiri 4 ki te 2i\sqrt{2}.

$$x=\frac{4+2\sqrt{2}i}{4}$$

Whakawehe 4+2i\sqrt{2} ki te 4.

$$x=\frac{\sqrt{2}i}{2}+1$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{2}i}{4} ina he tango te ±. Tango 2i\sqrt{2} mai i 4.

$$x=\frac{-2\sqrt{2}i+4}{4}$$

Whakawehe 4-2i\sqrt{2} ki te 4.

$$x=-\frac{\sqrt{2}i}{2}+1$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{2}i}{2}+1 x=-\frac{\sqrt{2}i}{2}+1$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x.

$$2xx+3=4x$$

Whakareatia te x ki te x, ka x^{2}.

$$2x^{2}+3=4x$$

Tangohia te 4x mai i ngā taha e rua.

$$2x^{2}+3-4x=0$$

Tangohia te 3 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$2x^{2}-4x=-3$$

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

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

Whakawehe -4 ki te 2.

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

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

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

Tāpiri -\frac{3}{2} ki te 1.

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

Tauwehea x^{2}-2x+1. 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}.

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

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

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

Whakarūnātia.

$$x-1=\frac{\sqrt{2}i}{2} x-1=-\frac{\sqrt{2}i}{2}$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{2}i}{2}+1 x=-\frac{\sqrt{2}i}{2}+1$$