Whakaoti i te sqrt{2-x}=x-1

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

Whakaoti mō x

$$x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

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

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

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

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

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

Me tāpiri te 2x ki ngā taha e rua.

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

Pahekotia te -x me 2x, ka x.

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

Tangohia te 1 mai i ngā taha e rua.

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

Tangohia te 1 i te 2, ka 1.

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

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}.

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

Pūrua 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.

$$x=\frac{-1±\sqrt{5}}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-1±\sqrt{5}}{-2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{5}}{-2} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{5}.

$$x=\frac{\sqrt{5}-1}{-2}$$

Whakawehe -1+\sqrt{5} ki te -2.

$$x=\frac{1-\sqrt{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{5}}{-2} ina he tango te ±. Tango \sqrt{5} mai i -1.

$$x=\frac{-\sqrt{5}-1}{-2}$$

Whakawehe -1-\sqrt{5} ki te -2.

$$x=\frac{\sqrt{5}+1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{1-\sqrt{5}}{2} x=\frac{\sqrt{5}+1}{2}$$

Whakakapia te \frac{1-\sqrt{5}}{2} mō te x i te whārite \sqrt{2-x}=x-1.

$$\sqrt{2-\frac{1-\sqrt{5}}{2}}=\frac{1-\sqrt{5}}{2}-1$$

Whakarūnātia. Ko te uara x=\frac{1-\sqrt{5}}{2} kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$\frac{1}{2}+\frac{1}{2}\times 5^{\frac{1}{2}}=-\frac{1}{2}-\frac{1}{2}\times 5^{\frac{1}{2}}$$

Whakakapia te \frac{\sqrt{5}+1}{2} mō te x i te whārite \sqrt{2-x}=x-1.

$$\sqrt{2-\frac{\sqrt{5}+1}{2}}=\frac{\sqrt{5}+1}{2}-1$$

Whakarūnātia. Ko te uara x=\frac{\sqrt{5}+1}{2} kua ngata te whārite.

$$-\left(\frac{1}{2}-\frac{1}{2}\times 5^{\frac{1}{2}}\right)=\frac{1}{2}\times 5^{\frac{1}{2}}-\frac{1}{2}$$

Ko te whārite \sqrt{2-x}=x-1 he rongoā ahurei.

$$x=\frac{\sqrt{5}+1}{2}$$