Whakaoti i te 16+x^2+left(4-xright)^2+16={left(4sqrt{5}right)}^2

$$16+ { x }^{ 2 } + { \left(4-x \right) }^{ 2 } +16= { \left(4 \sqrt{ 5 } \right) }^{ 2 }$$

Whakaoti mō x

$$x=2\sqrt{5}+2\approx 6.472135955$$

$$x=2-2\sqrt{5}\approx -2.472135955$$

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(4-x\right)^{2}.

$$16+x^{2}+16-8x+x^{2}+16=\left(4\sqrt{5}\right)^{2}$$

Tāpirihia te 16 ki te 16, ka 32.

$$32+x^{2}-8x+x^{2}+16=\left(4\sqrt{5}\right)^{2}$$

Pahekotia te x^{2} me x^{2}, ka 2x^{2}.

$$32+2x^{2}-8x+16=\left(4\sqrt{5}\right)^{2}$$

Tāpirihia te 32 ki te 16, ka 48.

$$48+2x^{2}-8x=\left(4\sqrt{5}\right)^{2}$$

Whakarohaina te \left(4\sqrt{5}\right)^{2}.

$$48+2x^{2}-8x=4^{2}\left(\sqrt{5}\right)^{2}$$

Tātaihia te 4 mā te pū o 2, kia riro ko 16.

$$48+2x^{2}-8x=16\left(\sqrt{5}\right)^{2}$$

Ko te pūrua o \sqrt{5} ko 5.

$$48+2x^{2}-8x=16\times 5$$

Whakareatia te 16 ki te 5, ka 80.

$$48+2x^{2}-8x=80$$

Tangohia te 80 mai i ngā taha e rua.

$$48+2x^{2}-8x-80=0$$

Tangohia te 80 i te 48, ka -32.

$$-32+2x^{2}-8x=0$$

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.

$$2x^{2}-8x-32=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -8 mō b, me -32 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-32\right)}}{2\times 2}$$

Pūrua -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-32\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\left(-32\right)}}{2\times 2}$$

Whakareatia -8 ki te -32.

$$x=\frac{-\left(-8\right)±\sqrt{64+256}}{2\times 2}$$

Tāpiri 64 ki te 256.

$$x=\frac{-\left(-8\right)±\sqrt{320}}{2\times 2}$$

Tuhia te pūtakerua o te 320.

$$x=\frac{-\left(-8\right)±8\sqrt{5}}{2\times 2}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±8\sqrt{5}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{8±8\sqrt{5}}{4}$$

Nā, me whakaoti te whārite x=\frac{8±8\sqrt{5}}{4} ina he tāpiri te ±. Tāpiri 8 ki te 8\sqrt{5}.

$$x=\frac{8\sqrt{5}+8}{4}$$

Whakawehe 8+8\sqrt{5} ki te 4.

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

Nā, me whakaoti te whārite x=\frac{8±8\sqrt{5}}{4} ina he tango te ±. Tango 8\sqrt{5} mai i 8.

$$x=\frac{8-8\sqrt{5}}{4}$$

Whakawehe 8-8\sqrt{5} ki te 4.

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

Kua oti te whārite te whakatau.

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

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(4-x\right)^{2}.

$$16+x^{2}+16-8x+x^{2}+16=\left(4\sqrt{5}\right)^{2}$$

Tāpirihia te 16 ki te 16, ka 32.

$$32+x^{2}-8x+x^{2}+16=\left(4\sqrt{5}\right)^{2}$$

Pahekotia te x^{2} me x^{2}, ka 2x^{2}.

$$32+2x^{2}-8x+16=\left(4\sqrt{5}\right)^{2}$$

Tāpirihia te 32 ki te 16, ka 48.

$$48+2x^{2}-8x=\left(4\sqrt{5}\right)^{2}$$

Whakarohaina te \left(4\sqrt{5}\right)^{2}.

$$48+2x^{2}-8x=4^{2}\left(\sqrt{5}\right)^{2}$$

Tātaihia te 4 mā te pū o 2, kia riro ko 16.

$$48+2x^{2}-8x=16\left(\sqrt{5}\right)^{2}$$

Ko te pūrua o \sqrt{5} ko 5.

$$48+2x^{2}-8x=16\times 5$$

Whakareatia te 16 ki te 5, ka 80.

$$48+2x^{2}-8x=80$$

Tangohia te 48 mai i ngā taha e rua.

$$2x^{2}-8x=80-48$$

Tangohia te 48 i te 80, ka 32.

$$2x^{2}-8x=32$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-8x}{2}=\frac{32}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=\frac{32}{2}$$

Whakawehe -8 ki te 2.

$$x^{2}-4x=\frac{32}{2}$$

Whakawehe 32 ki te 2.

$$x^{2}-4x=16$$

Whakawehea te -4, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -2. Nā, tāpiria te pūrua o te -2 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-4x+\left(-2\right)^{2}=16+\left(-2\right)^{2}$$

Pūrua -2.

$$x^{2}-4x+4=16+4$$

Tāpiri 16 ki te 4.

$$x^{2}-4x+4=20$$

Tauwehea x^{2}-4x+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}.

$$\left(x-2\right)^{2}=20$$

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

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{20}$$

Whakarūnātia.

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

Me tāpiri 2 ki ngā taha e rua o te whārite.

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