Whakaoti i te 4x^2-8x+3=0

$$4 x ^ { 2 } - 8 x + 3 = 0$$

Whakaoti mō x

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

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

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 4x^{2}+ax+bx+3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-8 ab=4\times 3=12$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 12.

$$-1,-12 -2,-6 -3,-4$$

Tātaihia te tapeke mō ia takirua.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

Ko te otinga te takirua ka hoatu i te tapeke -8.

$$a=-6 b=-2$$

Tuhia anō te 4x^{2}-8x+3 hei \left(4x^{2}-6x\right)+\left(-2x+3\right).

$$\left(4x^{2}-6x\right)+\left(-2x+3\right)$$

Tauwehea te 2x i te tuatahi me te -1 i te rōpū tuarua.

$$2x\left(2x-3\right)-\left(2x-3\right)$$

Whakatauwehea atu te kīanga pātahi 2x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-3\right)\left(2x-1\right)$$

Hei kimi otinga whārite, me whakaoti te 2x-3=0 me te 2x-1=0.

$$x=\frac{3}{2} x=\frac{1}{2}$$

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.

$$4x^{2}-8x+3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, -8 mō b, me 3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4\times 3}}{2\times 4}$$

Pūrua -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 4\times 3}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-\left(-8\right)±\sqrt{64-16\times 3}}{2\times 4}$$

Whakareatia -16 ki te 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-48}}{2\times 4}$$

Tāpiri 64 ki te -48.

$$x=\frac{-\left(-8\right)±\sqrt{16}}{2\times 4}$$

Tuhia te pūtakerua o te 16.

$$x=\frac{-\left(-8\right)±4}{2\times 4}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±4}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{8±4}{8}$$

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

$$x=\frac{12}{8}$$

Whakahekea te hautanga \frac{12}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=\frac{3}{2}$$

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

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

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

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

Kua oti te whārite te whakatau.

$$x=\frac{3}{2} x=\frac{1}{2}$$

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.

$$4x^{2}-8x+3=0$$

Me tango 3 mai i ngā taha e rua o te whārite.

$$4x^{2}-8x+3-3=-3$$

Mā te tango i te 3 i a ia ake anō ka toe ko te 0.

$$4x^{2}-8x=-3$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}-8x}{4}=-\frac{3}{4}$$

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

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

Whakawehe -8 ki te 4.

$$x^{2}-2x=-\frac{3}{4}$$

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

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

Tāpiri -\frac{3}{4} ki te 1.

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

Tauwehea x^{2}-2x+1. 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}.

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

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

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

Whakarūnātia.

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

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

$$x=\frac{3}{2} x=\frac{1}{2}$$