Whakaoti i te x^2+9=x

$$x ^ { 2 } + 9 = x$$

Whakaoti mō x (complex solution)

$$x=\frac{1+\sqrt{35}i}{2}\approx 0.5+2.958039892i$$

$$x=\frac{-\sqrt{35}i+1}{2}\approx 0.5-2.958039892i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te x mai i ngā taha e rua.

$$x^{2}+9-x=0$$

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.

$$x^{2}-x+9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -1 mō b, me 9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 9}}{2}$$

Whakareatia -4 ki te 9.

$$x=\frac{-\left(-1\right)±\sqrt{1-36}}{2}$$

Tāpiri 1 ki te -36.

$$x=\frac{-\left(-1\right)±\sqrt{-35}}{2}$$

Tuhia te pūtakerua o te -35.

$$x=\frac{-\left(-1\right)±\sqrt{35}i}{2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±\sqrt{35}i}{2}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{35}i}{2} ina he tāpiri te ±. Tāpiri 1 ki te i\sqrt{35}.

$$x=\frac{1+\sqrt{35}i}{2}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{35}i}{2} ina he tango te ±. Tango i\sqrt{35} mai i 1.

$$x=\frac{-\sqrt{35}i+1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{1+\sqrt{35}i}{2} x=\frac{-\sqrt{35}i+1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te x mai i ngā taha e rua.

$$x^{2}+9-x=0$$

Tangohia te 9 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x^{2}-x=-9$$

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-9+\left(-\frac{1}{2}\right)^{2}$$

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-x+\frac{1}{4}=-9+\frac{1}{4}$$

Tāpiri -9 ki te \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=-\frac{35}{4}$$

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

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{35}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{35}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{\sqrt{35}i}{2} x-\frac{1}{2}=-\frac{\sqrt{35}i}{2}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=\frac{1+\sqrt{35}i}{2} x=\frac{-\sqrt{35}i+1}{2}$$