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

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

Whakaoti mō x

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

$$x=1$$

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-3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=2\left(-3\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Tātaihia te tapeke mō ia takirua.

$$-1+6=5 -2+3=1$$

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

$$a=-2 b=3$$

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

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

Tauwehea te 2x i te tuatahi me te 3 i te rōpū tuarua.

$$2x\left(x-1\right)+3\left(x-1\right)$$

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

$$\left(x-1\right)\left(2x+3\right)$$

Hei kimi otinga whārite, me whakaoti te x-1=0 me te 2x+3=0.

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

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 1 mō b, me -3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-3\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-1±\sqrt{1-8\left(-3\right)}}{2\times 2}$$

Whakareatia -8 ki te -3.

$$x=\frac{-1±\sqrt{1+24}}{2\times 2}$$

Tāpiri 1 ki te 24.

$$x=\frac{-1±\sqrt{25}}{2\times 2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-1±5}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-1±5}{4}$$

Nā, me whakaoti te whārite x=\frac{-1±5}{4} ina he tāpiri te ±. Tāpiri -1 ki te 5.

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

Whakawehe 4 ki te 4.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{-1±5}{4} ina he tango te ±. Tango 5 mai i -1.

$$x=-\frac{6}{4}$$

Whakahekea te hautanga \frac{-6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

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.

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

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

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

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

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

Tango -3 mai i 0.

$$2x^{2}+x=3$$

Whakawehea ngā taha e rua ki te 2.

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

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

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

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.

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

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

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

Tāpiri \frac{3}{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.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{25}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{25}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Whakarūnātia.

$$x+\frac{1}{4}=\frac{5}{4} x+\frac{1}{4}=-\frac{5}{4}$$

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

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