Whakaoti i te -3x^2-x+10=0

$$- 3 x ^ { 2 } - x + 10 = 0$$

Whakaoti mō x

$$x=-2$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -3x^{2}+ax+bx+10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=-3\times 10=-30$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -30.

$$1,-30 2,-15 3,-10 5,-6$$

Tātaihia te tapeke mō ia takirua.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -1.

$$a=5 b=-6$$

Tuhia anō te -3x^{2}-x+10 hei \left(-3x^{2}+5x\right)+\left(-6x+10\right).

$$\left(-3x^{2}+5x\right)+\left(-6x+10\right)$$

Tauwehea te -x i te tuatahi me te -2 i te rōpū tuarua.

$$-x\left(3x-5\right)-2\left(3x-5\right)$$

Whakatauwehea atu te kīanga pātahi 3x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-5\right)\left(-x-2\right)$$

Hei kimi otinga whārite, me whakaoti te 3x-5=0 me te -x-2=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

$$-3x^{2}-x+10=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -3 mō a, -1 mō b, me 10 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-3\right)\times 10}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+12\times 10}}{2\left(-3\right)}$$

Whakareatia 12 ki te 10.

$$x=\frac{-\left(-1\right)±\sqrt{1+120}}{2\left(-3\right)}$$

Tāpiri 1 ki te 120.

$$x=\frac{-\left(-1\right)±\sqrt{121}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 121.

$$x=\frac{-\left(-1\right)±11}{2\left(-3\right)}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±11}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{1±11}{-6}$$

Nā, me whakaoti te whārite x=\frac{1±11}{-6} ina he tāpiri te ±. Tāpiri 1 ki te 11.

$$x=\frac{12}{-6}$$

Whakawehe 12 ki te -6.

$$x=-2$$

Nā, me whakaoti te whārite x=\frac{1±11}{-6} ina he tango te ±. Tango 11 mai i 1.

$$x=-\frac{10}{-6}$$

Whakahekea te hautanga \frac{-10}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

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.

$$-3x^{2}-x+10=0$$

Me tango 10 mai i ngā taha e rua o te whārite.

$$-3x^{2}-x+10-10=-10$$

Mā te tango i te 10 i a ia ake anō ka toe ko te 0.

$$-3x^{2}-x=-10$$

Whakawehea ngā taha e rua ki te -3.

$$\frac{-3x^{2}-x}{-3}=-\frac{10}{-3}$$

Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.

$$x^{2}+\left(-\frac{1}{-3}\right)x=-\frac{10}{-3}$$

Whakawehe -1 ki te -3.

$$x^{2}+\frac{1}{3}x=-\frac{10}{-3}$$

Whakawehe -10 ki te -3.

$$x^{2}+\frac{1}{3}x=\frac{10}{3}$$

Whakawehea te \frac{1}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{6}. Nā, tāpiria te pūrua o te \frac{1}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{10}{3}+\left(\frac{1}{6}\right)^{2}$$

Pūruatia \frac{1}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{10}{3}+\frac{1}{36}$$

Tāpiri \frac{10}{3} ki te \frac{1}{36} 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.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{121}{36}$$

Tauwehea x^{2}+\frac{1}{3}x+\frac{1}{36}. 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}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{121}{36}$$

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

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Whakarūnātia.

$$x+\frac{1}{6}=\frac{11}{6} x+\frac{1}{6}=-\frac{11}{6}$$

Me tango \frac{1}{6} mai i ngā taha e rua o te whārite.

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