Whakaoti i te 4x^2+8x+2=0

$$4 x ^ { 2 } + 8 x + 2 = 0$$

Whakaoti mō x

$$x=\frac{\sqrt{2}}{2}-1\approx -0.292893219$$

$$x=-\frac{\sqrt{2}}{2}-1\approx -1.707106781$$

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.

$$4x^{2}+8x+2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, 8 mō b, me 2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 4\times 2}}{2\times 4}$$

Pūrua 8.

$$x=\frac{-8±\sqrt{64-4\times 4\times 2}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-8±\sqrt{64-16\times 2}}{2\times 4}$$

Whakareatia -16 ki te 2.

$$x=\frac{-8±\sqrt{64-32}}{2\times 4}$$

Tāpiri 64 ki te -32.

$$x=\frac{-8±\sqrt{32}}{2\times 4}$$

Tuhia te pūtakerua o te 32.

$$x=\frac{-8±4\sqrt{2}}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-8±4\sqrt{2}}{8}$$

Nā, me whakaoti te whārite x=\frac{-8±4\sqrt{2}}{8} ina he tāpiri te ±. Tāpiri -8 ki te 4\sqrt{2}.

$$x=\frac{4\sqrt{2}-8}{8}$$

Whakawehe -8+4\sqrt{2} ki te 8.

$$x=\frac{\sqrt{2}}{2}-1$$

Nā, me whakaoti te whārite x=\frac{-8±4\sqrt{2}}{8} ina he tango te ±. Tango 4\sqrt{2} mai i -8.

$$x=\frac{-4\sqrt{2}-8}{8}$$

Whakawehe -8-4\sqrt{2} ki te 8.

$$x=-\frac{\sqrt{2}}{2}-1$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{2}}{2}-1 x=-\frac{\sqrt{2}}{2}-1$$

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.

$$4x^{2}+8x+2=0$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$4x^{2}+8x+2-2=-2$$

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

$$4x^{2}+8x=-2$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}+8x}{4}=-\frac{2}{4}$$

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

$$x^{2}+\frac{8}{4}x=-\frac{2}{4}$$

Whakawehe 8 ki te 4.

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

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

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

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

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

Pūrua 1.

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

Tāpiri -\frac{1}{2} ki te 1.

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

Tauwehea x^{2}+2x+1. 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}.

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

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

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

Whakarūnātia.

$$x+1=\frac{\sqrt{2}}{2} x+1=-\frac{\sqrt{2}}{2}$$

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

$$x=\frac{\sqrt{2}}{2}-1 x=-\frac{\sqrt{2}}{2}-1$$