Whakaoti i te 4x^2+6x+1=0

$$4 x ^ { 2 } + 6 x + 1 = 0$$

Whakaoti mō x

$$x=\frac{\sqrt{5}-3}{4}\approx -0.190983006$$

$$x=\frac{-\sqrt{5}-3}{4}\approx -1.309016994$$

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.

$$4x^{2}+6x+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, 6 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4}}{2\times 4}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\times 4}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-6±\sqrt{36-16}}{2\times 4}$$

Tāpiri 36 ki te -16.

$$x=\frac{-6±\sqrt{20}}{2\times 4}$$

Tuhia te pūtakerua o te 20.

$$x=\frac{-6±2\sqrt{5}}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-6±2\sqrt{5}}{8}$$

Nā, me whakaoti te whārite x=\frac{-6±2\sqrt{5}}{8} ina he tāpiri te ±. Tāpiri -6 ki te 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-6}{8}$$

Whakawehe -6+2\sqrt{5} ki te 8.

$$x=\frac{\sqrt{5}-3}{4}$$

Nā, me whakaoti te whārite x=\frac{-6±2\sqrt{5}}{8} ina he tango te ±. Tango 2\sqrt{5} mai i -6.

$$x=\frac{-2\sqrt{5}-6}{8}$$

Whakawehe -6-2\sqrt{5} ki te 8.

$$x=\frac{-\sqrt{5}-3}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{5}-3}{4} x=\frac{-\sqrt{5}-3}{4}$$

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.

$$4x^{2}+6x+1=0$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$4x^{2}+6x+1-1=-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$4x^{2}+6x=-1$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}+6x}{4}=-\frac{1}{4}$$

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

$$x^{2}+\frac{6}{4}x=-\frac{1}{4}$$

Whakahekea te hautanga \frac{6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}+\frac{3}{2}x=-\frac{1}{4}$$

Whakawehea te \frac{3}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{4}. Nā, tāpiria te pūrua o te \frac{3}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=-\frac{1}{4}+\left(\frac{3}{4}\right)^{2}$$

Pūruatia \frac{3}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{1}{4}+\frac{9}{16}$$

Tāpiri -\frac{1}{4} ki te \frac{9}{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.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{5}{16}$$

Tauwehea te x^{2}+\frac{3}{2}x+\frac{9}{16}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{5}{16}$$

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

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{5}{16}}$$

Whakarūnātia.

$$x+\frac{3}{4}=\frac{\sqrt{5}}{4} x+\frac{3}{4}=-\frac{\sqrt{5}}{4}$$

Me tango \frac{3}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{5}-3}{4} x=\frac{-\sqrt{5}-3}{4}$$