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

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

Whakaoti mō x

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

$$x=-3$$

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}.

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(3x+2\right)^{2}.

$$4x^{2}-4x+1=9x^{2}+12x+4$$

Tangohia te 9x^{2} mai i ngā taha e rua.

$$4x^{2}-4x+1-9x^{2}=12x+4$$

Pahekotia te 4x^{2} me -9x^{2}, ka -5x^{2}.

$$-5x^{2}-4x+1=12x+4$$

Tangohia te 12x mai i ngā taha e rua.

$$-5x^{2}-4x+1-12x=4$$

Pahekotia te -4x me -12x, ka -16x.

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

Tangohia te 4 mai i ngā taha e rua.

$$-5x^{2}-16x+1-4=0$$

Tangohia te 4 i te 1, ka -3.

$$-5x^{2}-16x-3=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -5x^{2}+ax+bx-3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-16 ab=-5\left(-3\right)=15$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 15.

$$-1,-15 -3,-5$$

Tātaihia te tapeke mō ia takirua.

$$-1-15=-16 -3-5=-8$$

Ko te otinga te takirua ka hoatu i te tapeke -16.

$$a=-1 b=-15$$

Tuhia anō te -5x^{2}-16x-3 hei \left(-5x^{2}-x\right)+\left(-15x-3\right).

$$\left(-5x^{2}-x\right)+\left(-15x-3\right)$$

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

$$-x\left(5x+1\right)-3\left(5x+1\right)$$

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

$$\left(5x+1\right)\left(-x-3\right)$$

Hei kimi otinga whārite, me whakaoti te 5x+1=0 me te -x-3=0.

$$x=-\frac{1}{5} x=-3$$

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}.

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(3x+2\right)^{2}.

$$4x^{2}-4x+1=9x^{2}+12x+4$$

Tangohia te 9x^{2} mai i ngā taha e rua.

$$4x^{2}-4x+1-9x^{2}=12x+4$$

Pahekotia te 4x^{2} me -9x^{2}, ka -5x^{2}.

$$-5x^{2}-4x+1=12x+4$$

Tangohia te 12x mai i ngā taha e rua.

$$-5x^{2}-4x+1-12x=4$$

Pahekotia te -4x me -12x, ka -16x.

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

Tangohia te 4 mai i ngā taha e rua.

$$-5x^{2}-16x+1-4=0$$

Tangohia te 4 i te 1, ka -3.

$$-5x^{2}-16x-3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -5 mō a, -16 mō b, me -3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Pūrua -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Whakareatia -4 ki te -5.

$$x=\frac{-\left(-16\right)±\sqrt{256+20\left(-3\right)}}{2\left(-5\right)}$$

Whakareatia 20 ki te -3.

$$x=\frac{-\left(-16\right)±\sqrt{256-60}}{2\left(-5\right)}$$

Tāpiri 256 ki te -60.

$$x=\frac{-\left(-16\right)±\sqrt{196}}{2\left(-5\right)}$$

Tuhia te pūtakerua o te 196.

$$x=\frac{-\left(-16\right)±14}{2\left(-5\right)}$$

Ko te tauaro o -16 ko 16.

$$x=\frac{16±14}{2\left(-5\right)}$$

Whakareatia 2 ki te -5.

$$x=\frac{16±14}{-10}$$

Nā, me whakaoti te whārite x=\frac{16±14}{-10} ina he tāpiri te ±. Tāpiri 16 ki te 14.

$$x=\frac{30}{-10}$$

Whakawehe 30 ki te -10.

$$x=-3$$

Nā, me whakaoti te whārite x=\frac{16±14}{-10} ina he tango te ±. Tango 14 mai i 16.

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

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

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

Kua oti te whārite te whakatau.

$$x=-3 x=-\frac{1}{5}$$

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

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(3x+2\right)^{2}.

$$4x^{2}-4x+1=9x^{2}+12x+4$$

Tangohia te 9x^{2} mai i ngā taha e rua.

$$4x^{2}-4x+1-9x^{2}=12x+4$$

Pahekotia te 4x^{2} me -9x^{2}, ka -5x^{2}.

$$-5x^{2}-4x+1=12x+4$$

Tangohia te 12x mai i ngā taha e rua.

$$-5x^{2}-4x+1-12x=4$$

Pahekotia te -4x me -12x, ka -16x.

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

Tangohia te 1 mai i ngā taha e rua.

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

Tangohia te 1 i te 4, ka 3.

$$-5x^{2}-16x=3$$

Whakawehea ngā taha e rua ki te -5.

$$\frac{-5x^{2}-16x}{-5}=\frac{3}{-5}$$

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

$$x^{2}+\left(-\frac{16}{-5}\right)x=\frac{3}{-5}$$

Whakawehe -16 ki te -5.

$$x^{2}+\frac{16}{5}x=\frac{3}{-5}$$

Whakawehe 3 ki te -5.

$$x^{2}+\frac{16}{5}x=-\frac{3}{5}$$

Whakawehea te \frac{16}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{8}{5}. Nā, tāpiria te pūrua o te \frac{8}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{16}{5}x+\left(\frac{8}{5}\right)^{2}=-\frac{3}{5}+\left(\frac{8}{5}\right)^{2}$$

Pūruatia \frac{8}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{16}{5}x+\frac{64}{25}=-\frac{3}{5}+\frac{64}{25}$$

Tāpiri -\frac{3}{5} ki te \frac{64}{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.

$$x^{2}+\frac{16}{5}x+\frac{64}{25}=\frac{49}{25}$$

Tauwehea x^{2}+\frac{16}{5}x+\frac{64}{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}.

$$\left(x+\frac{8}{5}\right)^{2}=\frac{49}{25}$$

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

$$\sqrt{\left(x+\frac{8}{5}\right)^{2}}=\sqrt{\frac{49}{25}}$$

Whakarūnātia.

$$x+\frac{8}{5}=\frac{7}{5} x+\frac{8}{5}=-\frac{7}{5}$$

Me tango \frac{8}{5} mai i ngā taha e rua o te whārite.

$$x=-\frac{1}{5} x=-3$$