Whakaoti i te 9y^2-12y+4=y^2

$$9 y ^ { 2 } - 12 y + 4 = y ^ { 2 }$$

Whakaoti mō y

$$y=\frac{1}{2}=0.5$$

$$y=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

$$9y^{2}-12y+4-y^{2}=0$$

Pahekotia te 9y^{2} me -y^{2}, ka 8y^{2}.

$$8y^{2}-12y+4=0$$

Whakawehea ngā taha e rua ki te 4.

$$2y^{2}-3y+1=0$$

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

$$a+b=-3 ab=2\times 1=2$$

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-2 b=-1$$

Tuhia anō te 2y^{2}-3y+1 hei \left(2y^{2}-2y\right)+\left(-y+1\right).

$$\left(2y^{2}-2y\right)+\left(-y+1\right)$$

Tauwehea te 2y i te tuatahi me te -1 i te rōpū tuarua.

$$2y\left(y-1\right)-\left(y-1\right)$$

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

$$\left(y-1\right)\left(2y-1\right)$$

Hei kimi otinga whārite, me whakaoti te y-1=0 me te 2y-1=0.

$$y=1 y=\frac{1}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$9y^{2}-12y+4-y^{2}=0$$

Pahekotia te 9y^{2} me -y^{2}, ka 8y^{2}.

$$8y^{2}-12y+4=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 8 mō a, -12 mō b, me 4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 8\times 4}}{2\times 8}$$

Pūrua -12.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 8\times 4}}{2\times 8}$$

Whakareatia -4 ki te 8.

$$y=\frac{-\left(-12\right)±\sqrt{144-32\times 4}}{2\times 8}$$

Whakareatia -32 ki te 4.

$$y=\frac{-\left(-12\right)±\sqrt{144-128}}{2\times 8}$$

Tāpiri 144 ki te -128.

$$y=\frac{-\left(-12\right)±\sqrt{16}}{2\times 8}$$

Tuhia te pūtakerua o te 16.

$$y=\frac{-\left(-12\right)±4}{2\times 8}$$

Ko te tauaro o -12 ko 12.

$$y=\frac{12±4}{2\times 8}$$

Whakareatia 2 ki te 8.

$$y=\frac{12±4}{16}$$

Nā, me whakaoti te whārite y=\frac{12±4}{16} ina he tāpiri te ±. Tāpiri 12 ki te 4.

$$y=\frac{16}{16}$$

Whakawehe 16 ki te 16.

$$y=1$$

Nā, me whakaoti te whārite y=\frac{12±4}{16} ina he tango te ±. Tango 4 mai i 12.

$$y=\frac{8}{16}$$

Whakahekea te hautanga \frac{8}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

$$y=\frac{1}{2}$$

Kua oti te whārite te whakatau.

$$y=1 y=\frac{1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$9y^{2}-12y+4-y^{2}=0$$

Pahekotia te 9y^{2} me -y^{2}, ka 8y^{2}.

$$8y^{2}-12y+4=0$$

Tangohia te 4 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$8y^{2}-12y=-4$$

Whakawehea ngā taha e rua ki te 8.

$$\frac{8y^{2}-12y}{8}=-\frac{4}{8}$$

Mā te whakawehe ki te 8 ka wetekia te whakareanga ki te 8.

$$y^{2}+\left(-\frac{12}{8}\right)y=-\frac{4}{8}$$

Whakahekea te hautanga \frac{-12}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$y^{2}-\frac{3}{2}y=-\frac{4}{8}$$

Whakahekea te hautanga \frac{-4}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$y^{2}-\frac{3}{2}y=-\frac{1}{2}$$

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

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

Pūruatia -\frac{3}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$y^{2}-\frac{3}{2}y+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Tāpiri -\frac{1}{2} ki te \frac{9}{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.

$$y^{2}-\frac{3}{2}y+\frac{9}{16}=\frac{1}{16}$$

Tauwehea y^{2}-\frac{3}{2}y+\frac{9}{16}. 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}.

$$\left(y-\frac{3}{4}\right)^{2}=\frac{1}{16}$$

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

$$\sqrt{\left(y-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Whakarūnātia.

$$y-\frac{3}{4}=\frac{1}{4} y-\frac{3}{4}=-\frac{1}{4}$$

Me tāpiri \frac{3}{4} ki ngā taha e rua o te whārite.

$$y=1 y=\frac{1}{2}$$