Whakaoti i te -4a^2-3a+1=0

$$-4 { a }^{ 2 } -3a+1 = 0$$

Whakaoti mō a

$$a=\frac{1}{4}=0.25$$

$$a=-1$$

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 -4a^{2}+aa+ba+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=-4=-4$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -4.

$$1,-4 2,-2$$

Tātaihia te tapeke mō ia takirua.

$$1-4=-3 2-2=0$$

Ko te otinga te takirua ka hoatu i te tapeke -3.

$$a=1 b=-4$$

Tuhia anō te -4a^{2}-3a+1 hei \left(-4a^{2}+a\right)+\left(-4a+1\right).

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

Tauwehea te -a i te tuatahi me te -1 i te rōpū tuarua.

$$-a\left(4a-1\right)-\left(4a-1\right)$$

Whakatauwehea atu te kīanga pātahi 4a-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(4a-1\right)\left(-a-1\right)$$

Hei kimi otinga whārite, me whakaoti te 4a-1=0 me te -a-1=0.

$$a=\frac{1}{4} a=-1$$

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.

$$-4a^{2}-3a+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -4 mō a, -3 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-4\right)}}{2\left(-4\right)}$$

Pūrua -3.

$$a=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)}}{2\left(-4\right)}$$

Whakareatia -4 ki te -4.

$$a=\frac{-\left(-3\right)±\sqrt{9+16}}{2\left(-4\right)}$$

Tāpiri 9 ki te 16.

$$a=\frac{-\left(-3\right)±\sqrt{25}}{2\left(-4\right)}$$

Tuhia te pūtakerua o te 25.

$$a=\frac{-\left(-3\right)±5}{2\left(-4\right)}$$

Ko te tauaro o -3 ko 3.

$$a=\frac{3±5}{2\left(-4\right)}$$

Whakareatia 2 ki te -4.

$$a=\frac{3±5}{-8}$$

Nā, me whakaoti te whārite a=\frac{3±5}{-8} ina he tāpiri te ±. Tāpiri 3 ki te 5.

$$a=\frac{8}{-8}$$

Whakawehe 8 ki te -8.

$$a=-1$$

Nā, me whakaoti te whārite a=\frac{3±5}{-8} ina he tango te ±. Tango 5 mai i 3.

$$a=-\frac{2}{-8}$$

Whakahekea te hautanga \frac{-2}{-8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$a=\frac{1}{4}$$

Kua oti te whārite te whakatau.

$$a=-1 a=\frac{1}{4}$$

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.

$$-4a^{2}-3a+1=0$$

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

$$-4a^{2}-3a+1-1=-1$$

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

$$-4a^{2}-3a=-1$$

Whakawehea ngā taha e rua ki te -4.

$$\frac{-4a^{2}-3a}{-4}=-\frac{1}{-4}$$

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

$$a^{2}+\left(-\frac{3}{-4}\right)a=-\frac{1}{-4}$$

Whakawehe -3 ki te -4.

$$a^{2}+\frac{3}{4}a=-\frac{1}{-4}$$

Whakawehe -1 ki te -4.

$$a^{2}+\frac{3}{4}a=\frac{1}{4}$$

Whakawehea te \frac{3}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{8}. Nā, tāpiria te pūrua o te \frac{3}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$a^{2}+\frac{3}{4}a+\left(\frac{3}{8}\right)^{2}=\frac{1}{4}+\left(\frac{3}{8}\right)^{2}$$

Pūruatia \frac{3}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$a^{2}+\frac{3}{4}a+\frac{9}{64}=\frac{1}{4}+\frac{9}{64}$$

Tāpiri \frac{1}{4} ki te \frac{9}{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.

$$a^{2}+\frac{3}{4}a+\frac{9}{64}=\frac{25}{64}$$

Tauwehea a^{2}+\frac{3}{4}a+\frac{9}{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}.

$$\left(a+\frac{3}{8}\right)^{2}=\frac{25}{64}$$

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

$$\sqrt{\left(a+\frac{3}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Whakarūnātia.

$$a+\frac{3}{8}=\frac{5}{8} a+\frac{3}{8}=-\frac{5}{8}$$

Me tango \frac{3}{8} mai i ngā taha e rua o te whārite.

$$a=\frac{1}{4} a=-1$$