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

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

Whakaoti mō x

$$x=4$$

$$x=-4$$

Ngā Hipanga Whakamahi Rerekētanga o ngā Pūrua

Whakarohaina te \left(2\sqrt{3}\right)^{2}.

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

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

$$4\left(\sqrt{3}\right)^{2}+2^{2}=x^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$4\times 3+2^{2}=x^{2}$$

Whakareatia te 4 ki te 3, ka 12.

$$12+2^{2}=x^{2}$$

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

$$12+4=x^{2}$$

Tāpirihia te 12 ki te 4, ka 16.

$$16=x^{2}$$

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

$$x^{2}=16$$

Tangohia te 16 mai i ngā taha e rua.

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

Whakaarohia te x^{2}-16. Tuhia anō te x^{2}-16 hei x^{2}-4^{2}. Ka taea te rerekētanga o ngā pūrua te whakatauwehe mā te ture: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-4\right)\left(x+4\right)=0$$

Hei kimi otinga whārite, me whakaoti te x-4=0 me te x+4=0.

$$x=4 x=-4$$

Ngā Upane mā te Kimi Pūtakerua

Whakarohaina te \left(2\sqrt{3}\right)^{2}.

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

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

$$4\left(\sqrt{3}\right)^{2}+2^{2}=x^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$4\times 3+2^{2}=x^{2}$$

Whakareatia te 4 ki te 3, ka 12.

$$12+2^{2}=x^{2}$$

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

$$12+4=x^{2}$$

Tāpirihia te 12 ki te 4, ka 16.

$$16=x^{2}$$

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

$$x^{2}=16$$

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

$$x=4 x=-4$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakarohaina te \left(2\sqrt{3}\right)^{2}.

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

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

$$4\left(\sqrt{3}\right)^{2}+2^{2}=x^{2}$$

Ko te pūrua o \sqrt{3} ko 3.

$$4\times 3+2^{2}=x^{2}$$

Whakareatia te 4 ki te 3, ka 12.

$$12+2^{2}=x^{2}$$

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

$$12+4=x^{2}$$

Tāpirihia te 12 ki te 4, ka 16.

$$16=x^{2}$$

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

$$x^{2}=16$$

Tangohia te 16 mai i ngā taha e rua.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 0 mō b, me -16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-16\right)}}{2}$$

Pūrua 0.

$$x=\frac{0±\sqrt{-4\left(-16\right)}}{2}$$

Whakareatia -4 ki te -16.

$$x=\frac{0±\sqrt{64}}{2}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{0±8}{2}$$

Nā, me whakaoti te whārite x=\frac{0±8}{2} ina he tāpiri te ±. Whakawehe 8 ki te 2.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{0±8}{2} ina he tango te ±. Whakawehe -8 ki te 2.

$$x=-4$$

Kua oti te whārite te whakatau.

$$x=4 x=-4$$