Whakaoti i te 4x^2=-4x-1

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

Whakaoti mō x

$$x=-\frac{1}{2}=-0.5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Me tāpiri te 4x ki ngā taha e rua.

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

Me tāpiri te 1 ki ngā taha e rua.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 4x^{2}+ax+bx+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=4 ab=4\times 1=4$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$1,4 2,2$$

Tātaihia te tapeke mō ia takirua.

$$1+4=5 2+2=4$$

Ko te otinga te takirua ka hoatu i te tapeke 4.

$$a=2 b=2$$

Tuhia anō te 4x^{2}+4x+1 hei \left(4x^{2}+2x\right)+\left(2x+1\right).

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

Whakatauwehea atu 2x i te 4x^{2}+2x.

$$2x\left(2x+1\right)+2x+1$$

Whakatauwehea atu te kīanga pātahi 2x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x+1\right)\left(2x+1\right)$$

Tuhia anōtia hei pūrua huarua.

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

Hei kimi i te otinga whārite, whakaotia te 2x+1=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me tāpiri te 4x ki ngā taha e rua.

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

Me tāpiri te 1 ki ngā taha e rua.

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

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

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

Pūrua 4.

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

Whakareatia -4 ki te 4.

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

Tāpiri 16 ki te -16.

$$x=\frac{-4±\sqrt{0}}{2\times 4}$$

Tuhia te pūtakerua o te 0.

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

Whakareatia 2 ki te 4.

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

Whakahekea te hautanga \frac{-4}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

Ngā Upane mō te Whakaoti i te Pūrua

Me tāpiri te 4x ki ngā taha e rua.

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

Whakawehea ngā taha e rua ki te 4.

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

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

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

Whakawehe 4 ki te 4.

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

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.

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

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

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

Tāpiri -\frac{1}{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.

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

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}.

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

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

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

Whakarūnātia.

$$x+\frac{1}{2}=0 x+\frac{1}{2}=0$$

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

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

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

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