Whakaoti i te (x-2)^2=1+x

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

Whakaoti mō x

$$x = \frac{\sqrt{13} + 5}{2} \approx 4.302775638$$

$$x=\frac{5-\sqrt{13}}{2}\approx 0.697224362$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$x^{2}-4x+4=1+x$$

Tangohia te 1 mai i ngā taha e rua.

$$x^{2}-4x+4-1=x$$

Tangohia te 1 i te 4, ka 3.

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

Tangohia te x mai i ngā taha e rua.

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

Pahekotia te -4x me -x, ka -5x.

$$x^{2}-5x+3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me 3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3}}{2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 3}}{2}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-5\right)±\sqrt{25-12}}{2}$$

Tāpiri 25 ki te -12.

$$x=\frac{-\left(-5\right)±\sqrt{13}}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{13}}{2}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{13}}{2} ina he tāpiri te ±. Tāpiri 5 ki te \sqrt{13}.

$$x=\frac{\sqrt{13}+5}{2}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{13}}{2} ina he tango te ±. Tango \sqrt{13} mai i 5.

$$x=\frac{5-\sqrt{13}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{13}+5}{2} x=\frac{5-\sqrt{13}}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$x^{2}-4x+4=1+x$$

Tangohia te x mai i ngā taha e rua.

$$x^{2}-4x+4-x=1$$

Pahekotia te -4x me -x, ka -5x.

$$x^{2}-5x+4=1$$

Tangohia te 4 mai i ngā taha e rua.

$$x^{2}-5x=1-4$$

Tangohia te 4 i te 1, ka -3.

$$x^{2}-5x=-3$$

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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-3+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-3+\frac{25}{4}$$

Tāpiri -3 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{13}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{13}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{\sqrt{13}}{2} x-\frac{5}{2}=-\frac{\sqrt{13}}{2}$$

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

$$x=\frac{\sqrt{13}+5}{2} x=\frac{5-\sqrt{13}}{2}$$