Whakaoti i te 4x^2+2x=12

$$4 x ^ { 2 } + 2 x = 12$$

Whakaoti mō x

$$x=-2$$

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

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 12 mai i ngā taha e rua.

$$4x^{2}+2x-12=0$$

Whakawehea ngā taha e rua ki te 2.

$$2x^{2}+x-6=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 2x^{2}+ax+bx-6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=2\left(-6\right)=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. 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.

$$-1+12=11 -2+6=4 -3+4=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-3 b=4$$

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

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

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

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

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

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

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

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

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.

$$4x^{2}+2x=12$$

Me tango 12 mai i ngā taha e rua o te whārite.

$$4x^{2}+2x-12=12-12$$

Mā te tango i te 12 i a ia ake anō ka toe ko te 0.

$$4x^{2}+2x-12=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, 2 mō b, me -12 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 4\left(-12\right)}}{2\times 4}$$

Pūrua 2.

$$x=\frac{-2±\sqrt{4-4\times 4\left(-12\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-2±\sqrt{4-16\left(-12\right)}}{2\times 4}$$

Whakareatia -16 ki te -12.

$$x=\frac{-2±\sqrt{4+192}}{2\times 4}$$

Tāpiri 4 ki te 192.

$$x=\frac{-2±\sqrt{196}}{2\times 4}$$

Tuhia te pūtakerua o te 196.

$$x=\frac{-2±14}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-2±14}{8}$$

Nā, me whakaoti te whārite x=\frac{-2±14}{8} ina he tāpiri te ±. Tāpiri -2 ki te 14.

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

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{-2±14}{8} ina he tango te ±. Tango 14 mai i -2.

$$x=-\frac{16}{8}$$

Whakawehe -16 ki te 8.

$$x=-2$$

Kua oti te whārite te whakatau.

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

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.

$$4x^{2}+2x=12$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}+2x}{4}=\frac{12}{4}$$

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

$$x^{2}+\frac{2}{4}x=\frac{12}{4}$$

Whakahekea te hautanga \frac{2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}+\frac{1}{2}x=\frac{12}{4}$$

Whakawehe 12 ki te 4.

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

Whakawehea te \frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{4}. Nā, tāpiria te pūrua o te \frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=3+\frac{1}{16}$$

Tāpiri 3 ki te \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{49}{16}$$

Tauwehea x^{2}+\frac{1}{2}x+\frac{1}{16}. 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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{49}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Whakarūnātia.

$$x+\frac{1}{4}=\frac{7}{4} x+\frac{1}{4}=-\frac{7}{4}$$

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

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