Whakaoti i te (2x+1)^2+(x+2)(x+1)=x+2

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

Whakaoti mō x

$$x=-\frac{1}{5}=-0.2$$

$$x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te x+1 ka whakakotahi i ngā kupu rite.

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

Pahekotia te 4x^{2} me x^{2}, ka 5x^{2}.

$$5x^{2}+4x+1+3x+2=x+2$$

Pahekotia te 4x me 3x, ka 7x.

$$5x^{2}+7x+1+2=x+2$$

Tāpirihia te 1 ki te 2, ka 3.

$$5x^{2}+7x+3=x+2$$

Tangohia te x mai i ngā taha e rua.

$$5x^{2}+7x+3-x=2$$

Pahekotia te 7x me -x, ka 6x.

$$5x^{2}+6x+3=2$$

Tangohia te 2 mai i ngā taha e rua.

$$5x^{2}+6x+3-2=0$$

Tangohia te 2 i te 3, ka 1.

$$5x^{2}+6x+1=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 5x^{2}+ax+bx+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=6 ab=5\times 1=5$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=1 b=5$$

Tuhia anō te 5x^{2}+6x+1 hei \left(5x^{2}+x\right)+\left(5x+1\right).

$$\left(5x^{2}+x\right)+\left(5x+1\right)$$

Whakatauwehea atu x i te 5x^{2}+x.

$$x\left(5x+1\right)+5x+1$$

Whakatauwehea atu te kīanga pātahi 5x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(5x+1\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te 5x+1=0 me te x+1=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te x+1 ka whakakotahi i ngā kupu rite.

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

Pahekotia te 4x^{2} me x^{2}, ka 5x^{2}.

$$5x^{2}+4x+1+3x+2=x+2$$

Pahekotia te 4x me 3x, ka 7x.

$$5x^{2}+7x+1+2=x+2$$

Tāpirihia te 1 ki te 2, ka 3.

$$5x^{2}+7x+3=x+2$$

Tangohia te x mai i ngā taha e rua.

$$5x^{2}+7x+3-x=2$$

Pahekotia te 7x me -x, ka 6x.

$$5x^{2}+6x+3=2$$

Tangohia te 2 mai i ngā taha e rua.

$$5x^{2}+6x+3-2=0$$

Tangohia te 2 i te 3, ka 1.

$$5x^{2}+6x+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, 6 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5}}{2\times 5}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\times 5}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-6±\sqrt{36-20}}{2\times 5}$$

Tāpiri 36 ki te -20.

$$x=\frac{-6±\sqrt{16}}{2\times 5}$$

Tuhia te pūtakerua o te 16.

$$x=\frac{-6±4}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-6±4}{10}$$

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

$$x=-\frac{2}{10}$$

Whakahekea te hautanga \frac{-2}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

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

$$x=-\frac{10}{10}$$

Whakawehe -10 ki te 10.

$$x=-1$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

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

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te x+1 ka whakakotahi i ngā kupu rite.

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

Pahekotia te 4x^{2} me x^{2}, ka 5x^{2}.

$$5x^{2}+4x+1+3x+2=x+2$$

Pahekotia te 4x me 3x, ka 7x.

$$5x^{2}+7x+1+2=x+2$$

Tāpirihia te 1 ki te 2, ka 3.

$$5x^{2}+7x+3=x+2$$

Tangohia te x mai i ngā taha e rua.

$$5x^{2}+7x+3-x=2$$

Pahekotia te 7x me -x, ka 6x.

$$5x^{2}+6x+3=2$$

Tangohia te 3 mai i ngā taha e rua.

$$5x^{2}+6x=2-3$$

Tangohia te 3 i te 2, ka -1.

$$5x^{2}+6x=-1$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}+6x}{5}=-\frac{1}{5}$$

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

$$x^{2}+\frac{6}{5}x=-\frac{1}{5}$$

Whakawehea te \frac{6}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{5}. Nā, tāpiria te pūrua o te \frac{3}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=-\frac{1}{5}+\left(\frac{3}{5}\right)^{2}$$

Pūruatia \frac{3}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=-\frac{1}{5}+\frac{9}{25}$$

Tāpiri -\frac{1}{5} ki te \frac{9}{25} 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.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{4}{25}$$

Tauwehea x^{2}+\frac{6}{5}x+\frac{9}{25}. 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}.

$$\left(x+\frac{3}{5}\right)^{2}=\frac{4}{25}$$

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

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Whakarūnātia.

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

Me tango \frac{3}{5} mai i ngā taha e rua o te whārite.

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