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.

$$a+b=1 ab=-2=-2$$

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.

$$a=2 b=-1$$

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

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

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

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

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

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

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

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

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.

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

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}.

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

Pūrua 1.

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

Whakareatia -4 ki te -2.

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

Tāpiri 1 ki te 8.

$$x=\frac{-1±\sqrt{9}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-1±3}{2\left(-2\right)}$$

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 tango 1 mai i ngā taha e rua o te whārite.

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

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

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

Whakawehe 1 ki te -2.

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

Whakawehe -1 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 tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

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