Whakaoti i te x(2x+1)-(x-1)^2/2=3

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

Whakaoti mō x

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

$$x=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

$$2x\left(2x+1\right)-\left(x-1\right)^{2}=6$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2x ki te 2x+1.

$$4x^{2}+2x-\left(x-1\right)^{2}=6$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

$$4x^{2}+2x-\left(x^{2}-2x+1\right)=6$$

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

$$4x^{2}+2x-x^{2}+2x-1=6$$

Pahekotia te 4x^{2} me -x^{2}, ka 3x^{2}.

$$3x^{2}+2x+2x-1=6$$

Pahekotia te 2x me 2x, ka 4x.

$$3x^{2}+4x-1=6$$

Tangohia te 6 mai i ngā taha e rua.

$$3x^{2}+4x-1-6=0$$

Tangohia te 6 i te -1, ka -7.

$$3x^{2}+4x-7=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 3x^{2}+ax+bx-7. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=4 ab=3\left(-7\right)=-21$$

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 -21.

$$-1,21 -3,7$$

Tātaihia te tapeke mō ia takirua.

$$-1+21=20 -3+7=4$$

Ko te otinga te takirua ka hoatu i te tapeke 4.

$$a=-3 b=7$$

Tuhia anō te 3x^{2}+4x-7 hei \left(3x^{2}-3x\right)+\left(7x-7\right).

$$\left(3x^{2}-3x\right)+\left(7x-7\right)$$

Tauwehea te 3x i te tuatahi me te 7 i te rōpū tuarua.

$$3x\left(x-1\right)+7\left(x-1\right)$$

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

$$\left(x-1\right)\left(3x+7\right)$$

Hei kimi otinga whārite, me whakaoti te x-1=0 me te 3x+7=0.

$$x=1 x=-\frac{7}{3}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$2x\left(2x+1\right)-\left(x-1\right)^{2}=6$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2x ki te 2x+1.

$$4x^{2}+2x-\left(x-1\right)^{2}=6$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

$$4x^{2}+2x-\left(x^{2}-2x+1\right)=6$$

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

$$4x^{2}+2x-x^{2}+2x-1=6$$

Pahekotia te 4x^{2} me -x^{2}, ka 3x^{2}.

$$3x^{2}+2x+2x-1=6$$

Pahekotia te 2x me 2x, ka 4x.

$$3x^{2}+4x-1=6$$

Tangohia te 6 mai i ngā taha e rua.

$$3x^{2}+4x-1-6=0$$

Tangohia te 6 i te -1, ka -7.

$$3x^{2}+4x-7=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 4 mō b, me -7 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3\left(-7\right)}}{2\times 3}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 3\left(-7\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-4±\sqrt{16-12\left(-7\right)}}{2\times 3}$$

Whakareatia -12 ki te -7.

$$x=\frac{-4±\sqrt{16+84}}{2\times 3}$$

Tāpiri 16 ki te 84.

$$x=\frac{-4±\sqrt{100}}{2\times 3}$$

Tuhia te pūtakerua o te 100.

$$x=\frac{-4±10}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-4±10}{6}$$

Nā, me whakaoti te whārite x=\frac{-4±10}{6} ina he tāpiri te ±. Tāpiri -4 ki te 10.

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

Whakawehe 6 ki te 6.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{-4±10}{6} ina he tango te ±. Tango 10 mai i -4.

$$x=-\frac{14}{6}$$

Whakahekea te hautanga \frac{-14}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{7}{3}$$

Kua oti te whārite te whakatau.

$$x=1 x=-\frac{7}{3}$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$2x\left(2x+1\right)-\left(x-1\right)^{2}=6$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2x ki te 2x+1.

$$4x^{2}+2x-\left(x-1\right)^{2}=6$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-1\right)^{2}.

$$4x^{2}+2x-\left(x^{2}-2x+1\right)=6$$

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

$$4x^{2}+2x-x^{2}+2x-1=6$$

Pahekotia te 4x^{2} me -x^{2}, ka 3x^{2}.

$$3x^{2}+2x+2x-1=6$$

Pahekotia te 2x me 2x, ka 4x.

$$3x^{2}+4x-1=6$$

Me tāpiri te 1 ki ngā taha e rua.

$$3x^{2}+4x=6+1$$

Tāpirihia te 6 ki te 1, ka 7.

$$3x^{2}+4x=7$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}+4x}{3}=\frac{7}{3}$$

Mā te whakawehe ki te 3 ka wetekia te whakareanga ki te 3.

$$x^{2}+\frac{4}{3}x=\frac{7}{3}$$

Whakawehea te \frac{4}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{2}{3}. Nā, tāpiria te pūrua o te \frac{2}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{7}{3}+\left(\frac{2}{3}\right)^{2}$$

Pūruatia \frac{2}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{7}{3}+\frac{4}{9}$$

Tāpiri \frac{7}{3} ki te \frac{4}{9} 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.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{25}{9}$$

Tauwehea x^{2}+\frac{4}{3}x+\frac{4}{9}. 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}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{25}{9}$$

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

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Whakarūnātia.

$$x+\frac{2}{3}=\frac{5}{3} x+\frac{2}{3}=-\frac{5}{3}$$

Me tango \frac{2}{3} mai i ngā taha e rua o te whārite.

$$x=1 x=-\frac{7}{3}$$