Whakaoti i te sqrt{x-1}=x

$$\sqrt { x - 1 } = x$$

Whakaoti mō x (complex solution)

$$x=\frac{1+\sqrt{3}i}{2}\approx 0.5+0.866025404i$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Tātaihia te \sqrt{x-1} mā te pū o 2, kia riro ko x-1.

$$x-1=x^{2}$$

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

$$x-1-x^{2}=0$$

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.

$$-x^{2}+x-1=0$$

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

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

Pūrua 1.

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

Whakareatia -4 ki te -1.

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

Whakareatia 4 ki te -1.

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

Tāpiri 1 ki te -4.

$$x=\frac{-1±\sqrt{-3}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te -3.

$$x=\frac{-1±\sqrt{3}i}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-1±\sqrt{3}i}{-2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{3}i}{-2} ina he tāpiri te ±. Tāpiri -1 ki te i\sqrt{3}.

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

Whakawehe -1+i\sqrt{3} ki te -2.

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

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{3}i}{-2} ina he tango te ±. Tango i\sqrt{3} mai i -1.

$$x=\frac{-\sqrt{3}i-1}{-2}$$

Whakawehe -1-i\sqrt{3} ki te -2.

$$x=\frac{1+\sqrt{3}i}{2}$$

Kua oti te whārite te whakatau.

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

Whakakapia te \frac{-\sqrt{3}i+1}{2} mō te x i te whārite \sqrt{x-1}=x.

$$\sqrt{\frac{-\sqrt{3}i+1}{2}-1}=\frac{-\sqrt{3}i+1}{2}$$

Whakarūnātia. Ko te uara x=\frac{-\sqrt{3}i+1}{2} kāore e ngata ana ki te whārite.

$$-\left(\frac{1}{2}-\frac{1}{2}i\times 3^{\frac{1}{2}}\right)=-\frac{1}{2}i\times 3^{\frac{1}{2}}+\frac{1}{2}$$

Whakakapia te \frac{1+\sqrt{3}i}{2} mō te x i te whārite \sqrt{x-1}=x.

$$\sqrt{\frac{1+\sqrt{3}i}{2}-1}=\frac{1+\sqrt{3}i}{2}$$

Whakarūnātia. Ko te uara x=\frac{1+\sqrt{3}i}{2} kua ngata te whārite.

$$\frac{1}{2}+\frac{1}{2}i\times 3^{\frac{1}{2}}=\frac{1}{2}+\frac{1}{2}i\times 3^{\frac{1}{2}}$$

Ko te whārite \sqrt{x-1}=x he rongoā ahurei.

$$x=\frac{1+\sqrt{3}i}{2}$$