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

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

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.

$$a+b=-4 ab=4\left(-3\right)=-12$$

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 -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 -4.

$$a=-6 b=2$$

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

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

Whakatauwehea atu 2x i te 4x^{2}-6x.

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

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.

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

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

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

Pūrua -4.

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

Whakareatia -4 ki te 4.

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

Whakareatia -16 ki te -3.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 4}$$

Tāpiri 16 ki te 48.

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

Tuhia te pūtakerua o te 64.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±8}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{4±8}{8}$$

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

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

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{4±8}{8} ina he tango te ±. Tango 8 mai i 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.

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

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.

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

Me tāpiri 3 ki ngā taha e rua o te whārite.

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

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

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

Tango -3 mai i 0.

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

Whakawehea ngā taha e rua ki te 4.

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

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

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

Whakawehe -4 ki te 4.

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

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.

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

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

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

Tāpiri \frac{3}{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.

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

Tauwehea te x^{2}-x+\frac{1}{4}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

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

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

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

Whakarūnātia.

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

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

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