Whakaoti i te x^2+2{left(sqrt{2}div2x-2sqrt{2}right)}^2=8

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

Whakaoti mō x

$$x=2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tuhia te \frac{\sqrt{2}}{2}x hei hautanga kotahi.

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

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

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

Kia whakarewa i te \frac{\sqrt{2}x}{2} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

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

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 4 me te 2.

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

Ko te pūrua o \sqrt{2} ko 2.

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

Whakareatia te 4 ki te 2, ka 8.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te \frac{\left(\sqrt{2}x\right)^{2}}{2^{2}}-2\sqrt{2}x\sqrt{2}+8.

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

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

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

Ko te pūrua o \sqrt{2} ko 2.

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

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

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

Whakawehea te 2x^{2} ki te 4, kia riro ko \frac{1}{2}x^{2}.

$$x^{2}+2\times \frac{1}{2}x^{2}-4x\left(\sqrt{2}\right)^{2}+16=8$$

Whakareatia te 2 ki te \frac{1}{2}, ka 1.

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

Ko te pūrua o \sqrt{2} ko 2.

$$x^{2}+x^{2}-4x\times 2+16=8$$

Whakareatia te -4 ki te 2, ka -8.

$$x^{2}+x^{2}-8x+16=8$$

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

$$2x^{2}-8x+16=8$$

Tangohia te 8 mai i ngā taha e rua.

$$2x^{2}-8x+16-8=0$$

Tangohia te 8 i te 16, ka 8.

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

Whakawehea ngā taha e rua ki te 2.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=1\times 4=4$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$-1,-4 -2,-2$$

Tātaihia te tapeke mō ia takirua.

$$-1-4=-5 -2-2=-4$$

Ko te otinga te takirua ka hoatu i te tapeke -4.

$$a=-2 b=-2$$

Tuhia anō te x^{2}-4x+4 hei \left(x^{2}-2x\right)+\left(-2x+4\right).

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

Tauwehea te x i te tuatahi me te -2 i te rōpū tuarua.

$$x\left(x-2\right)-2\left(x-2\right)$$

Whakatauwehea atu te kīanga pātahi x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-2\right)\left(x-2\right)$$

Tuhia anōtia hei pūrua huarua.

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

Hei kimi i te otinga whārite, whakaotia te x-2=0.

$$x=2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tuhia te \frac{\sqrt{2}}{2}x hei hautanga kotahi.

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

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

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

Kia whakarewa i te \frac{\sqrt{2}x}{2} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

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

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 4 me te 2.

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

Ko te pūrua o \sqrt{2} ko 2.

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

Whakareatia te 4 ki te 2, ka 8.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te \frac{\left(\sqrt{2}x\right)^{2}}{2^{2}}-2\sqrt{2}x\sqrt{2}+8.

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

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

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

Ko te pūrua o \sqrt{2} ko 2.

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

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

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

Whakawehea te 2x^{2} ki te 4, kia riro ko \frac{1}{2}x^{2}.

$$x^{2}+2\times \frac{1}{2}x^{2}-4x\left(\sqrt{2}\right)^{2}+16=8$$

Whakareatia te 2 ki te \frac{1}{2}, ka 1.

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

Ko te pūrua o \sqrt{2} ko 2.

$$x^{2}+x^{2}-4x\times 2+16=8$$

Whakareatia te -4 ki te 2, ka -8.

$$x^{2}+x^{2}-8x+16=8$$

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

$$2x^{2}-8x+16=8$$

Tangohia te 8 mai i ngā taha e rua.

$$2x^{2}-8x+16-8=0$$

Tangohia te 8 i te 16, ka 8.

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

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

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

Pūrua -8.

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

Whakareatia -4 ki te 2.

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

Whakareatia -8 ki te 8.

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

Tāpiri 64 ki te -64.

$$x=\frac{-\left(-8\right)±\sqrt{0}}{2\times 2}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{-8}{2\times 2}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{8}{4}$$

Whakawehe 8 ki te 4.

$$x=2$$

Ngā Upane mō te Whakaoti i te Pūrua

Tuhia te \frac{\sqrt{2}}{2}x hei hautanga kotahi.

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

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

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

Kia whakarewa i te \frac{\sqrt{2}x}{2} ki tētahi taupū, me whakarewa tahi te taurunga me te tauraro ki te taupū kātahi ka whakawehe.

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

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 4 me te 2.

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

Ko te pūrua o \sqrt{2} ko 2.

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

Whakareatia te 4 ki te 2, ka 8.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te \frac{\left(\sqrt{2}x\right)^{2}}{2^{2}}-2\sqrt{2}x\sqrt{2}+8.

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

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

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

Ko te pūrua o \sqrt{2} ko 2.

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

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

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

Whakawehea te 2x^{2} ki te 4, kia riro ko \frac{1}{2}x^{2}.

$$x^{2}+2\times \frac{1}{2}x^{2}-4x\left(\sqrt{2}\right)^{2}+16=8$$

Whakareatia te 2 ki te \frac{1}{2}, ka 1.

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

Ko te pūrua o \sqrt{2} ko 2.

$$x^{2}+x^{2}-4x\times 2+16=8$$

Whakareatia te -4 ki te 2, ka -8.

$$x^{2}+x^{2}-8x+16=8$$

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

$$2x^{2}-8x+16=8$$

Tangohia te 16 mai i ngā taha e rua.

$$2x^{2}-8x=8-16$$

Tangohia te 16 i te 8, ka -8.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe -8 ki te 2.

$$x^{2}-4x=-\frac{8}{2}$$

Whakawehe -8 ki te 2.

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

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.

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

Pūrua -2.

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

Tāpiri -4 ki te 4.

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

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}.

$$\left(x-2\right)^{2}=0$$

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

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x-2=0 x-2=0$$

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

$$x=2 x=2$$

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

$$x=2$$