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

$${ x }^{ 2 } - \frac{ x }{ 2 } - \frac{ 3 }{ 2 } = 0$$

Whakaoti mō x

$$x=-1$$

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

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakareatia ngā taha e rua o te whārite ki te 2.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 2x^{2}+ax+bx-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 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 -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.

$$a=-3 b=2$$

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

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

Whakatauwehea atu x i te 2x^{2}-3x.

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

Whakatauwehea atu te kīanga pātahi 2x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia ngā taha e rua o te whārite ki te 2.

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

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}.

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

Whakareatia -4 ki te 2.

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

Whakareatia -8 ki te -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\times 2}$$

Tāpiri 1 ki te 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\times 2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-\left(-1\right)±5}{2\times 2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±5}{2\times 2}$$

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.

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

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

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

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

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

Whakawehe -4 ki te 4.

$$x=-1$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia ngā taha e rua o te whārite ki te 2.

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

Me tāpiri te 3 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

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

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 te x^{2}-\frac{1}{2}x+\frac{1}{16}. 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}.

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

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

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