Whakaoti i te 2x^2+x-1=0

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

Whakaoti mō x

$$x=-1$$

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

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 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

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.

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

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

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

Pūrua 1.

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

Whakareatia -4 ki te 2.

$$x=\frac{-1±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

Whakareatia -8 ki te -1.

$$x=\frac{-1±\sqrt{1+8}}{2\times 2}$$

Tāpiri 1 ki te 8.

$$x=\frac{-1±\sqrt{9}}{2\times 2}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-1±3}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-1±3}{4}$$

Nā, me whakaoti te whārite x=\frac{-1±3}{4} ina he tāpiri te ±. Tāpiri -1 ki te 3.

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

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

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

Nā, me whakaoti te whārite x=\frac{-1±3}{4} ina he tango te ±. Tango 3 mai i -1.

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

Whakawehe -4 ki te 4.

$$x=-1$$

Kua oti te whārite te whakatau.

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

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.

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

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

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

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

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

Tango -1 mai i 0.

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

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki 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.

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