Whakaoti i te 6x-1=9x^2

$$6x-1=9 { x }^{ 2 }$$

Whakaoti mō x

$$x=\frac{1}{3}\approx 0.333333333$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

$$6x-1-9x^{2}=0$$

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-9x^{2}+6x-1=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -9x^{2}+ax+bx-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=6 ab=-9\left(-1\right)=9$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 9.

$$1,9 3,3$$

Tātaihia te tapeke mō ia takirua.

$$1+9=10 3+3=6$$

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

$$a=3 b=3$$

Tuhia anō te -9x^{2}+6x-1 hei \left(-9x^{2}+3x\right)+\left(3x-1\right).

$$\left(-9x^{2}+3x\right)+\left(3x-1\right)$$

Whakatauwehea atu -3x i te -9x^{2}+3x.

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

Whakatauwehea atu te kīanga pātahi 3x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$6x-1-9x^{2}=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$-9x^{2}+6x-1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -9 mō a, 6 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Whakareatia -4 ki te -9.

$$x=\frac{-6±\sqrt{36+36\left(-1\right)}}{2\left(-9\right)}$$

Whakareatia 36 ki te -1.

$$x=\frac{-6±\sqrt{36-36}}{2\left(-9\right)}$$

Tāpiri 36 ki te -36.

$$x=\frac{-6±\sqrt{0}}{2\left(-9\right)}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{6}{2\left(-9\right)}$$

Whakareatia 2 ki te -9.

$$x=-\frac{6}{-18}$$

Whakahekea te hautanga \frac{-6}{-18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Ngā Upane mō te Whakaoti i te Pūrua

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

$$6x-1-9x^{2}=0$$

Me tāpiri te 1 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$6x-9x^{2}=1$$

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$-9x^{2}+6x=1$$

Whakawehea ngā taha e rua ki te -9.

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

Mā te whakawehe ki te -9 ka wetekia te whakareanga ki te -9.

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

Whakahekea te hautanga \frac{6}{-9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$x^{2}-\frac{2}{3}x=\frac{1}{-9}$$

Whakawehe 1 ki te -9.

$$x^{2}-\frac{2}{3}x=-\frac{1}{9}$$

Whakawehea te -\frac{2}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{3}. Nā, tāpiria te pūrua o te -\frac{1}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{1}{9}+\left(-\frac{1}{3}\right)^{2}$$

Pūruatia -\frac{1}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri -\frac{1}{9} ki te \frac{1}{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.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=0$$

Tauwehea x^{2}-\frac{2}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{3}\right)^{2}=0$$

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

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x-\frac{1}{3}=0 x-\frac{1}{3}=0$$

Me tāpiri \frac{1}{3} ki ngā taha e rua o te whārite.

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

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

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