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

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

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

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

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

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

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

Tangohia te 4 mai i ngā taha e rua.

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

Tangohia te 4 i te 1, ka -3.

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

Me tāpiri te 4x ki ngā taha e rua.

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

Pahekotia te x me 4x, ka 5x.

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

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

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

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.

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

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}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-5±\sqrt{25+4\left(-3\right)}}{2\left(-1\right)}$$

Whakareatia 4 ki te -3.

$$x=\frac{-5±\sqrt{25-12}}{2\left(-1\right)}$$

Tāpiri 25 ki te -12.

$$x=\frac{-5±\sqrt{13}}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{-5±\sqrt{13}}{-2}$$

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

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

Whakawehe -5+\sqrt{13} ki te -2.

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

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

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

Whakawehe -5-\sqrt{13} ki te -2.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

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

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

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

Me tāpiri te 4x ki ngā taha e rua.

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

Pahekotia te x me 4x, ka 5x.

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

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

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

Tangohia te 1 mai i ngā taha e rua.

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

Tangohia te 1 i te 4, ka 3.

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

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

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

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x^{2}+5x}{-1}=\frac{3}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$x^{2}+\frac{5}{-1}x=\frac{3}{-1}$$

Whakawehe 5 ki te -1.

$$x^{2}-5x=\frac{3}{-1}$$

Whakawehe 3 ki te -1.

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

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.

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