Whakaoti i te 4x^2+4x+9=0

$$4 x ^ { 2 } + 4 x + 9 = 0$$

Whakaoti mō x (complex solution)

$$x=-\frac{1}{2}+\sqrt{2}i\approx -0.5+1.414213562i$$

$$x=-\sqrt{2}i-\frac{1}{2}\approx -0.5-1.414213562i$$

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.

$$4x^{2}+4x+9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, 4 mō b, me 9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 4\times 9}}{2\times 4}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 4\times 9}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-4±\sqrt{16-16\times 9}}{2\times 4}$$

Whakareatia -16 ki te 9.

$$x=\frac{-4±\sqrt{16-144}}{2\times 4}$$

Tāpiri 16 ki te -144.

$$x=\frac{-4±\sqrt{-128}}{2\times 4}$$

Tuhia te pūtakerua o te -128.

$$x=\frac{-4±8\sqrt{2}i}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-4±8\sqrt{2}i}{8}$$

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

$$x=\frac{-4+2\times 2^{\frac{5}{2}}i}{8}$$

Whakawehe -4+2i\times 2^{\frac{5}{2}} ki te 8.

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

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

$$x=\frac{-2\times 2^{\frac{5}{2}}i-4}{8}$$

Whakawehe -4-2i\times 2^{\frac{5}{2}} ki te 8.

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

Kua oti te whārite te whakatau.

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

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.

$$4x^{2}+4x+9=0$$

Me tango 9 mai i ngā taha e rua o te whārite.

$$4x^{2}+4x+9-9=-9$$

Mā te tango i te 9 i a ia ake anō ka toe ko te 0.

$$4x^{2}+4x=-9$$

Whakawehea ngā taha e rua ki te 4.

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

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

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

Whakawehe 4 ki te 4.

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

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=-\frac{9}{4}+\left(\frac{1}{2}\right)^{2}$$

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri -\frac{9}{4} ki te \frac{1}{4} 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.

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

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

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

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

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

Whakarūnātia.

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

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

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