Whakaoti i te 25x^2=4?

$$25 x ^ { 2 } = 4 ?$$

Whakaoti mō x

$$x=\frac{2}{5}=0.4$$

$$x=-\frac{2}{5}=-0.4$$

Ngā Hipanga Whakamahi Rerekētanga o ngā Pūrua

Whakawehea ngā taha e rua ki te 25.

$$x^{2}=\frac{4}{25}$$

Tangohia te \frac{4}{25} mai i ngā taha e rua.

$$x^{2}-\frac{4}{25}=0$$

Me whakarea ngā taha e rua ki te 25.

$$25x^{2}-4=0$$

Whakaarohia te 25x^{2}-4. Tuhia anō te 25x^{2}-4 hei \left(5x\right)^{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).

$$\left(5x-2\right)\left(5x+2\right)=0$$

Hei kimi otinga whārite, me whakaoti te 5x-2=0 me te 5x+2=0.

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

Ngā Upane mā te Kimi Pūtakerua

Whakawehea ngā taha e rua ki te 25.

$$x^{2}=\frac{4}{25}$$

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakawehea ngā taha e rua ki te 25.

$$x^{2}=\frac{4}{25}$$

Tangohia te \frac{4}{25} mai i ngā taha e rua.

$$x^{2}-\frac{4}{25}=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 0 mō b, me -\frac{4}{25} mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{4}{25}\right)}}{2}$$

Pūrua 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{4}{25}\right)}}{2}$$

Whakareatia -4 ki te -\frac{4}{25}.

$$x=\frac{0±\sqrt{\frac{16}{25}}}{2}$$

Tuhia te pūtakerua o te \frac{16}{25}.

$$x=\frac{0±\frac{4}{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{0±\frac{4}{5}}{2} ina he tāpiri te ±.

$$x=\frac{2}{5}$$

Nā, me whakaoti te whārite x=\frac{0±\frac{4}{5}}{2} ina he tango te ±.

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

Kua oti te whārite te whakatau.

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