Whakaoti i te left(4xright)^2+4x+4=0

$${ \left(4x \right) }^{ 2 } +4x+4=0$$

Whakaoti mō x (complex solution)

$$x=\frac{-1+\sqrt{15}i}{8}\approx -0.125+0.484122918i$$

$$x=\frac{-\sqrt{15}i-1}{8}\approx -0.125-0.484122918i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakarohaina te \left(4x\right)^{2}.

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

Tātaihia te 4 mā te pū o 2, kia riro ko 16.

$$16x^{2}+4x+4=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 16 mō a, 4 mō b, me 4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua 4.

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

Whakareatia -4 ki te 16.

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

Whakareatia -64 ki te 4.

$$x=\frac{-4±\sqrt{16-256}}{2\times 16}$$

Tāpiri 16 ki te -256.

$$x=\frac{-4±\sqrt{-240}}{2\times 16}$$

Tuhia te pūtakerua o te -240.

$$x=\frac{-4±4\sqrt{15}i}{2\times 16}$$

Whakareatia 2 ki te 16.

$$x=\frac{-4±4\sqrt{15}i}{32}$$

Nā, me whakaoti te whārite x=\frac{-4±4\sqrt{15}i}{32} ina he tāpiri te ±. Tāpiri -4 ki te 4i\sqrt{15}.

$$x=\frac{-4+4\sqrt{15}i}{32}$$

Whakawehe -4+4i\sqrt{15} ki te 32.

$$x=\frac{-1+\sqrt{15}i}{8}$$

Nā, me whakaoti te whārite x=\frac{-4±4\sqrt{15}i}{32} ina he tango te ±. Tango 4i\sqrt{15} mai i -4.

$$x=\frac{-4\sqrt{15}i-4}{32}$$

Whakawehe -4-4i\sqrt{15} ki te 32.

$$x=\frac{-\sqrt{15}i-1}{8}$$

Kua oti te whārite te whakatau.

$$x=\frac{-1+\sqrt{15}i}{8} x=\frac{-\sqrt{15}i-1}{8}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakarohaina te \left(4x\right)^{2}.

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

Tātaihia te 4 mā te pū o 2, kia riro ko 16.

$$16x^{2}+4x+4=0$$

Tangohia te 4 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$16x^{2}+4x=-4$$

Whakawehea ngā taha e rua ki te 16.

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

Mā te whakawehe ki te 16 ka wetekia te whakareanga ki te 16.

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

Whakahekea te hautanga \frac{4}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

Whakahekea te hautanga \frac{-4}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

Whakawehea te \frac{1}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{8}. Nā, tāpiria te pūrua o te \frac{1}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=-\frac{1}{4}+\left(\frac{1}{8}\right)^{2}$$

Pūruatia \frac{1}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{1}{4}+\frac{1}{64}$$

Tāpiri -\frac{1}{4} ki te \frac{1}{64} 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.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{15}{64}$$

Tauwehea x^{2}+\frac{1}{4}x+\frac{1}{64}. 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}.

$$\left(x+\frac{1}{8}\right)^{2}=-\frac{15}{64}$$

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

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{-\frac{15}{64}}$$

Whakarūnātia.

$$x+\frac{1}{8}=\frac{\sqrt{15}i}{8} x+\frac{1}{8}=-\frac{\sqrt{15}i}{8}$$

Me tango \frac{1}{8} mai i ngā taha e rua o te whārite.

$$x=\frac{-1+\sqrt{15}i}{8} x=\frac{-\sqrt{15}i-1}{8}$$