Whakaoti i te x=x+2-16/x-4

$$x = \frac { x + 2 - 16 } { x - 4 }$$

Whakaoti mō x (complex solution)

$$x=\frac{5+\sqrt{31}i}{2}\approx 2.5+2.783882181i$$

$$x=\frac{-\sqrt{31}i+5}{2}\approx 2.5-2.783882181i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 16 i te 2, ka -14.

$$x=\frac{x-14}{x-4}$$

Tangohia te \frac{x-14}{x-4} mai i ngā taha e rua.

$$x-\frac{x-14}{x-4}=0$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x ki te \frac{x-4}{x-4}.

$$\frac{x\left(x-4\right)}{x-4}-\frac{x-14}{x-4}=0$$

Tā te mea he rite te tauraro o \frac{x\left(x-4\right)}{x-4} me \frac{x-14}{x-4}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{x\left(x-4\right)-\left(x-14\right)}{x-4}=0$$

Mahia ngā whakarea i roto o x\left(x-4\right)-\left(x-14\right).

$$\frac{x^{2}-4x-x+14}{x-4}=0$$

Whakakotahitia ngā kupu rite i x^{2}-4x-x+14.

$$\frac{x^{2}-5x+14}{x-4}=0$$

Tē taea kia ōrite te tāupe x ki 4 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x-4.

$$x^{2}-5x+14=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me 14 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 14}}{2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 14}}{2}$$

Whakareatia -4 ki te 14.

$$x=\frac{-\left(-5\right)±\sqrt{25-56}}{2}$$

Tāpiri 25 ki te -56.

$$x=\frac{-\left(-5\right)±\sqrt{-31}}{2}$$

Tuhia te pūtakerua o te -31.

$$x=\frac{-\left(-5\right)±\sqrt{31}i}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{31}i}{2}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{31}i}{2} ina he tāpiri te ±. Tāpiri 5 ki te i\sqrt{31}.

$$x=\frac{5+\sqrt{31}i}{2}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{31}i}{2} ina he tango te ±. Tango i\sqrt{31} mai i 5.

$$x=\frac{-\sqrt{31}i+5}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{5+\sqrt{31}i}{2} x=\frac{-\sqrt{31}i+5}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 16 i te 2, ka -14.

$$x=\frac{x-14}{x-4}$$

Tangohia te \frac{x-14}{x-4} mai i ngā taha e rua.

$$x-\frac{x-14}{x-4}=0$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x ki te \frac{x-4}{x-4}.

$$\frac{x\left(x-4\right)}{x-4}-\frac{x-14}{x-4}=0$$

Tā te mea he rite te tauraro o \frac{x\left(x-4\right)}{x-4} me \frac{x-14}{x-4}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{x\left(x-4\right)-\left(x-14\right)}{x-4}=0$$

Mahia ngā whakarea i roto o x\left(x-4\right)-\left(x-14\right).

$$\frac{x^{2}-4x-x+14}{x-4}=0$$

Whakakotahitia ngā kupu rite i x^{2}-4x-x+14.

$$\frac{x^{2}-5x+14}{x-4}=0$$

Tē taea kia ōrite te tāupe x ki 4 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x-4.

$$x^{2}-5x+14=0$$

Tangohia te 14 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x^{2}-5x=-14$$

Whakawehea te -5, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{2}. Nā, tāpiria te pūrua o te -\frac{5}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-14+\left(-\frac{5}{2}\right)^{2}$$

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-5x+\frac{25}{4}=-14+\frac{25}{4}$$

Tāpiri -14 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=-\frac{31}{4}$$

Tauwehea x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=-\frac{31}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{-\frac{31}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{\sqrt{31}i}{2} x-\frac{5}{2}=-\frac{\sqrt{31}i}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=\frac{5+\sqrt{31}i}{2} x=\frac{-\sqrt{31}i+5}{2}$$