Whakaoti i te 3^2=(sqrt{3})^2+(3-x)^2

$$3 ^ { 2 } = ( \sqrt { 3 } ) ^ { 2 } + ( 3 - x ) ^ { 2 }$$

Whakaoti mō x

$$x=3-\sqrt{6}\approx 0.550510257$$

$$x=\sqrt{6}+3\approx 5.449489743$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$9=\left(\sqrt{3}\right)^{2}+\left(3-x\right)^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$9=3+\left(3-x\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(3-x\right)^{2}.

$$9=3+9-6x+x^{2}$$

Tāpirihia te 3 ki te 9, ka 12.

$$9=12-6x+x^{2}$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$12-6x+x^{2}=9$$

Tangohia te 9 mai i ngā taha e rua.

$$12-6x+x^{2}-9=0$$

Tangohia te 9 i te 12, ka 3.

$$3-6x+x^{2}=0$$

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.

$$x^{2}-6x+3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -6 mō b, me 3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3}}{2}$$

Pūrua -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 3}}{2}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-6\right)±\sqrt{36-12}}{2}$$

Tāpiri 36 ki te -12.

$$x=\frac{-\left(-6\right)±\sqrt{24}}{2}$$

Tuhia te pūtakerua o te 24.

$$x=\frac{-\left(-6\right)±2\sqrt{6}}{2}$$

Ko te tauaro o -6 ko 6.

$$x=\frac{6±2\sqrt{6}}{2}$$

Nā, me whakaoti te whārite x=\frac{6±2\sqrt{6}}{2} ina he tāpiri te ±. Tāpiri 6 ki te 2\sqrt{6}.

$$x=\frac{2\sqrt{6}+6}{2}$$

Whakawehe 6+2\sqrt{6} ki te 2.

$$x=\sqrt{6}+3$$

Nā, me whakaoti te whārite x=\frac{6±2\sqrt{6}}{2} ina he tango te ±. Tango 2\sqrt{6} mai i 6.

$$x=\frac{6-2\sqrt{6}}{2}$$

Whakawehe 6-2\sqrt{6} ki te 2.

$$x=3-\sqrt{6}$$

Kua oti te whārite te whakatau.

$$x=\sqrt{6}+3 x=3-\sqrt{6}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$9=\left(\sqrt{3}\right)^{2}+\left(3-x\right)^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$9=3+\left(3-x\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(3-x\right)^{2}.

$$9=3+9-6x+x^{2}$$

Tāpirihia te 3 ki te 9, ka 12.

$$9=12-6x+x^{2}$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$12-6x+x^{2}=9$$

Tangohia te 12 mai i ngā taha e rua.

$$-6x+x^{2}=9-12$$

Tangohia te 12 i te 9, ka -3.

$$-6x+x^{2}=-3$$

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.

$$x^{2}-6x=-3$$

Whakawehea te -6, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -3. Nā, tāpiria te pūrua o te -3 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-6x+\left(-3\right)^{2}=-3+\left(-3\right)^{2}$$

Pūrua -3.

$$x^{2}-6x+9=-3+9$$

Tāpiri -3 ki te 9.

$$x^{2}-6x+9=6$$

Tauwehea te x^{2}-6x+9. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=6$$

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

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{6}$$

Whakarūnātia.

$$x-3=\sqrt{6} x-3=-\sqrt{6}$$

Me tāpiri 3 ki ngā taha e rua o te whārite.

$$x=\sqrt{6}+3 x=3-\sqrt{6}$$