Whakaoti i te t^2+3t=3t+4

$$t ^ { 2 } + 3 t = 3 t + 4$$

Whakaoti mō t

$$t=-2$$

$$t=2$$

Ngā Hipanga Whakamahi Rerekētanga o ngā Pūrua

Tangohia te 3t mai i ngā taha e rua.

$$t^{2}+3t-3t=4$$

Pahekotia te 3t me -3t, ka 0.

$$t^{2}=4$$

Tangohia te 4 mai i ngā taha e rua.

$$t^{2}-4=0$$

Whakaarohia te t^{2}-4. Tuhia anō te t^{2}-4 hei t^{2}-2^{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).

$$\left(t-2\right)\left(t+2\right)=0$$

Hei kimi otinga whārite, me whakaoti te t-2=0 me te t+2=0.

$$t=2 t=-2$$

Ngā Upane mā te Kimi Pūtakerua

Tangohia te 3t mai i ngā taha e rua.

$$t^{2}+3t-3t=4$$

Pahekotia te 3t me -3t, ka 0.

$$t^{2}=4$$

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

$$t=2 t=-2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 3t mai i ngā taha e rua.

$$t^{2}+3t-3t=4$$

Pahekotia te 3t me -3t, ka 0.

$$t^{2}=4$$

Tangohia te 4 mai i ngā taha e rua.

$$t^{2}-4=0$$

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

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

Pūrua 0.

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

Whakareatia -4 ki te -4.

$$t=\frac{0±\sqrt{16}}{2}$$

Tuhia te pūtakerua o te 16.

$$t=\frac{0±4}{2}$$

Nā, me whakaoti te whārite t=\frac{0±4}{2} ina he tāpiri te ±. Whakawehe 4 ki te 2.

$$t=2$$

Nā, me whakaoti te whārite t=\frac{0±4}{2} ina he tango te ±. Whakawehe -4 ki te 2.

$$t=-2$$

Kua oti te whārite te whakatau.

$$t=2 t=-2$$