Whakaoti i te -4B^2+4B-1=0

$$- 4 B ^ { 2 } + 4 B - 1 = 0$$

Whakaoti mō B

$$B=\frac{1}{2}=0.5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

$$a+b=4 ab=-4\left(-1\right)=4$$

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 -4B^{2}+4B-1 hei \left(-4B^{2}+2B\right)+\left(2B-1\right).

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

Whakatauwehea atu -2B i te -4B^{2}+2B.

$$-2B\left(2B-1\right)+2B-1$$

Whakatauwehea atu te kīanga pātahi 2B-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2B-1\right)\left(-2B+1\right)$$

Hei kimi otinga whārite, me whakaoti te 2B-1=0 me te -2B+1=0.

$$B=\frac{1}{2} B=\frac{1}{2}$$

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.

$$-4B^{2}+4B-1=0$$

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}.

$$B=\frac{-4±\sqrt{4^{2}-4\left(-4\right)\left(-1\right)}}{2\left(-4\right)}$$

Pūrua 4.

$$B=\frac{-4±\sqrt{16-4\left(-4\right)\left(-1\right)}}{2\left(-4\right)}$$

Whakareatia -4 ki te -4.

$$B=\frac{-4±\sqrt{16+16\left(-1\right)}}{2\left(-4\right)}$$

Whakareatia 16 ki te -1.

$$B=\frac{-4±\sqrt{16-16}}{2\left(-4\right)}$$

Tāpiri 16 ki te -16.

$$B=\frac{-4±\sqrt{0}}{2\left(-4\right)}$$

Tuhia te pūtakerua o te 0.

$$B=-\frac{4}{2\left(-4\right)}$$

Whakareatia 2 ki te -4.

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

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

$$B=\frac{1}{2}$$

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.

$$-4B^{2}+4B-1=0$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

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

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

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

Tango -1 mai i 0.

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

Whakawehea ngā taha e rua ki te -4.

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

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

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

Whakawehe 4 ki te -4.

$$B^{2}-B=\frac{1}{-4}$$

Whakawehe 1 ki te -4.

$$B^{2}-B=-\frac{1}{4}$$

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.

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

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

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

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.

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

Tauwehea B^{2}-B+\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}.

$$\left(B-\frac{1}{2}\right)^{2}=0$$

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

$$\sqrt{\left(B-\frac{1}{2}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$B-\frac{1}{2}=0 B-\frac{1}{2}=0$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$B=\frac{1}{2} B=\frac{1}{2}$$

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

$$B=\frac{1}{2}$$