Whakaoti i te n^2+4=13

$$n ^ { 2 } + 4 = 13$$

Whakaoti mō n

$$n=3$$

$$n=-3$$

Ngā Hipanga Whakamahi Rerekētanga o ngā Pūrua

Tangohia te 13 mai i ngā taha e rua.

$$n^{2}+4-13=0$$

Tangohia te 13 i te 4, ka -9.

$$n^{2}-9=0$$

Whakaarohia te n^{2}-9. Tuhia anō te n^{2}-9 hei n^{2}-3^{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).

$$\left(n-3\right)\left(n+3\right)=0$$

Hei kimi otinga whārite, me whakaoti te n-3=0 me te n+3=0.

$$n=3 n=-3$$

Ngā Upane mā te Kimi Pūtakerua

Tangohia te 4 mai i ngā taha e rua.

$$n^{2}=13-4$$

Tangohia te 4 i te 13, ka 9.

$$n^{2}=9$$

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

$$n=3 n=-3$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 13 mai i ngā taha e rua.

$$n^{2}+4-13=0$$

Tangohia te 13 i te 4, ka -9.

$$n^{2}-9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 0 mō b, me -9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{0±\sqrt{0^{2}-4\left(-9\right)}}{2}$$

Pūrua 0.

$$n=\frac{0±\sqrt{-4\left(-9\right)}}{2}$$

Whakareatia -4 ki te -9.

$$n=\frac{0±\sqrt{36}}{2}$$

Tuhia te pūtakerua o te 36.

$$n=\frac{0±6}{2}$$

Nā, me whakaoti te whārite n=\frac{0±6}{2} ina he tāpiri te ±. Whakawehe 6 ki te 2.

$$n=3$$

Nā, me whakaoti te whārite n=\frac{0±6}{2} ina he tango te ±. Whakawehe -6 ki te 2.

$$n=-3$$

Kua oti te whārite te whakatau.

$$n=3 n=-3$$