Whakaoti i te (x-1)^2+(2x+2)^2=16

$$( x - 1 ) ^ { 2 } + ( 2 x + 2 ) ^ { 2 } = 16$$

Whakaoti mō x

$$x = -\frac{11}{5} = -2\frac{1}{5} = -2.2$$

$$x=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

$$x^{2}-2x+1+\left(2x+2\right)^{2}=16$$

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

$$x^{2}-2x+1+4x^{2}+8x+4=16$$

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

$$5x^{2}-2x+1+8x+4=16$$

Pahekotia te -2x me 8x, ka 6x.

$$5x^{2}+6x+1+4=16$$

Tāpirihia te 1 ki te 4, ka 5.

$$5x^{2}+6x+5=16$$

Tangohia te 16 mai i ngā taha e rua.

$$5x^{2}+6x+5-16=0$$

Tangohia te 16 i te 5, ka -11.

$$5x^{2}+6x-11=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 5x^{2}+ax+bx-11. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=6 ab=5\left(-11\right)=-55$$

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 -55.

$$-1,55 -5,11$$

Tātaihia te tapeke mō ia takirua.

$$-1+55=54 -5+11=6$$

Ko te otinga te takirua ka hoatu i te tapeke 6.

$$a=-5 b=11$$

Tuhia anō te 5x^{2}+6x-11 hei \left(5x^{2}-5x\right)+\left(11x-11\right).

$$\left(5x^{2}-5x\right)+\left(11x-11\right)$$

Tauwehea te 5x i te tuatahi me te 11 i te rōpū tuarua.

$$5x\left(x-1\right)+11\left(x-1\right)$$

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

$$\left(x-1\right)\left(5x+11\right)$$

Hei kimi otinga whārite, me whakaoti te x-1=0 me te 5x+11=0.

$$x=1 x=-\frac{11}{5}$$

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(x-1\right)^{2}.

$$x^{2}-2x+1+\left(2x+2\right)^{2}=16$$

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

$$x^{2}-2x+1+4x^{2}+8x+4=16$$

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

$$5x^{2}-2x+1+8x+4=16$$

Pahekotia te -2x me 8x, ka 6x.

$$5x^{2}+6x+1+4=16$$

Tāpirihia te 1 ki te 4, ka 5.

$$5x^{2}+6x+5=16$$

Tangohia te 16 mai i ngā taha e rua.

$$5x^{2}+6x+5-16=0$$

Tangohia te 16 i te 5, ka -11.

$$5x^{2}+6x-11=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, 6 mō b, me -11 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 5\left(-11\right)}}{2\times 5}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\times 5\left(-11\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-6±\sqrt{36-20\left(-11\right)}}{2\times 5}$$

Whakareatia -20 ki te -11.

$$x=\frac{-6±\sqrt{36+220}}{2\times 5}$$

Tāpiri 36 ki te 220.

$$x=\frac{-6±\sqrt{256}}{2\times 5}$$

Tuhia te pūtakerua o te 256.

$$x=\frac{-6±16}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-6±16}{10}$$

Nā, me whakaoti te whārite x=\frac{-6±16}{10} ina he tāpiri te ±. Tāpiri -6 ki te 16.

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

Whakawehe 10 ki te 10.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{-6±16}{10} ina he tango te ±. Tango 16 mai i -6.

$$x=-\frac{22}{10}$$

Whakahekea te hautanga \frac{-22}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{11}{5}$$

Kua oti te whārite te whakatau.

$$x=1 x=-\frac{11}{5}$$

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(x-1\right)^{2}.

$$x^{2}-2x+1+\left(2x+2\right)^{2}=16$$

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

$$x^{2}-2x+1+4x^{2}+8x+4=16$$

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

$$5x^{2}-2x+1+8x+4=16$$

Pahekotia te -2x me 8x, ka 6x.

$$5x^{2}+6x+1+4=16$$

Tāpirihia te 1 ki te 4, ka 5.

$$5x^{2}+6x+5=16$$

Tangohia te 5 mai i ngā taha e rua.

$$5x^{2}+6x=16-5$$

Tangohia te 5 i te 16, ka 11.

$$5x^{2}+6x=11$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}+6x}{5}=\frac{11}{5}$$

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

$$x^{2}+\frac{6}{5}x=\frac{11}{5}$$

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.

$$x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=\frac{11}{5}+\left(\frac{3}{5}\right)^{2}$$

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

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{11}{5}+\frac{9}{25}$$

Tāpiri \frac{11}{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.

$$x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{64}{25}$$

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}.

$$\left(x+\frac{3}{5}\right)^{2}=\frac{64}{25}$$

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

$$\sqrt{\left(x+\frac{3}{5}\right)^{2}}=\sqrt{\frac{64}{25}}$$

Whakarūnātia.

$$x+\frac{3}{5}=\frac{8}{5} x+\frac{3}{5}=-\frac{8}{5}$$

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

$$x=1 x=-\frac{11}{5}$$