Whakaoti i te 4x^2-2x+1/4=0

$$4 x ^ { 2 } - 2 x + \frac { 1 } { 4 } = 0$$

Whakaoti mō x

$$x=\frac{1}{4}=0.25$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$4x^{2}-2x+\frac{1}{4}=0$$

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

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

Pūrua -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 4\times \frac{1}{4}}}{2\times 4}$$

Whakareatia -4 ki te 4.

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

Whakareatia -16 ki te \frac{1}{4}.

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

Tāpiri 4 ki te -4.

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

Tuhia te pūtakerua o te 0.

$$x=-\frac{-2}{2\times 4}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2}{2\times 4}$$

Whakareatia 2 ki te 4.

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

Whakahekea te hautanga \frac{2}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{1}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$4x^{2}-2x+\frac{1}{4}=0$$

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

$$4x^{2}-2x+\frac{1}{4}-\frac{1}{4}=-\frac{1}{4}$$

Mā te tango i te \frac{1}{4} i a ia ake anō ka toe ko te 0.

$$4x^{2}-2x=-\frac{1}{4}$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}-2x}{4}=-\frac{\frac{1}{4}}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

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

Whakahekea te hautanga \frac{-2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}-\frac{1}{2}x=-\frac{\frac{1}{4}}{4}$$

Whakawehe -\frac{1}{4} ki te 4.

$$x^{2}-\frac{1}{2}x=-\frac{1}{16}$$

Whakawehea te -\frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{4}. Nā, tāpiria te pūrua o te -\frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{-1+1}{16}$$

Tāpiri -\frac{1}{16} ki te \frac{1}{16} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=0$$

Tauwehea x^{2}-\frac{1}{2}x+\frac{1}{16}. 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}.

$$\left(x-\frac{1}{4}\right)^{2}=0$$

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

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x-\frac{1}{4}=0 x-\frac{1}{4}=0$$

Me tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

$$x=\frac{1}{4} x=\frac{1}{4}$$

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

$$x=\frac{1}{4}$$