Whakaoti i te left(2xright)^2-2*(-x)-3=-1

$${ \left(2x \right) }^{ 2 } -2 \times (-x)-3 = -1$$

Whakaoti mō x

$$x=-1$$

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

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakarohaina te \left(2x\right)^{2}.

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

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

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

Me tāpiri te 1 ki ngā taha e rua.

$$4x^{2}-2\left(-x\right)-3+1=0$$

Tāpirihia te -3 ki te 1, ka -2.

$$4x^{2}-2\left(-x\right)-2=0$$

Whakareatia te -1 ki te 2, ka -2.

$$4x^{2}-2\left(-1\right)x-2=0$$

Whakareatia te -2 ki te -1, ka 2.

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

Whakawehea ngā taha e rua ki te 2.

$$2x^{2}+x-1=0$$

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

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

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-1 b=2$$

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

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

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

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

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

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

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakarohaina te \left(2x\right)^{2}.

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

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

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

Me tāpiri te 1 ki ngā taha e rua.

$$4x^{2}-2\left(-x\right)-3+1=0$$

Tāpirihia te -3 ki te 1, ka -2.

$$4x^{2}-2\left(-x\right)-2=0$$

Whakareatia te -1 ki te 2, ka -2.

$$4x^{2}-2\left(-1\right)x-2=0$$

Whakareatia te -2 ki te -1, ka 2.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te 4.

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

Whakareatia -16 ki te -2.

$$x=\frac{-2±\sqrt{4+32}}{2\times 4}$$

Tāpiri 4 ki te 32.

$$x=\frac{-2±\sqrt{36}}{2\times 4}$$

Tuhia te pūtakerua o te 36.

$$x=\frac{-2±6}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-2±6}{8}$$

Nā, me whakaoti te whārite x=\frac{-2±6}{8} ina he tāpiri te ±. Tāpiri -2 ki te 6.

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

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

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

Nā, me whakaoti te whārite x=\frac{-2±6}{8} ina he tango te ±. Tango 6 mai i -2.

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

Whakawehe -8 ki te 8.

$$x=-1$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakarohaina te \left(2x\right)^{2}.

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

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

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

Me tāpiri te 3 ki ngā taha e rua.

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

Tāpirihia te -1 ki te 3, ka 2.

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

Whakareatia te -1 ki te 2, ka -2.

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

Whakareatia te -2 ki te -1, ka 2.

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

Whakawehea ngā taha e rua ki te 4.

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

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

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

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

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

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

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

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.

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

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

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

Tāpiri \frac{1}{2} ki te \frac{1}{16} 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.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Whakarūnātia.

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

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

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