Whakaoti i te -3x^2-2x+5=0

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

Whakaoti mō x

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

$$x=1$$

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+5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-2 ab=-3\times 5=-15$$

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 -15.

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

Tātaihia te tapeke mō ia takirua.

$$1-15=-14 3-5=-2$$

Ko te otinga te takirua ka hoatu i te tapeke -2.

$$a=3 b=-5$$

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

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

Tauwehea te 3x i te tuatahi me te 5 i te rōpū tuarua.

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

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

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

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

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

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.

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

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

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

Pūrua -2.

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

Whakareatia -4 ki te -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12\times 5}}{2\left(-3\right)}$$

Whakareatia 12 ki te 5.

$$x=\frac{-\left(-2\right)±\sqrt{4+60}}{2\left(-3\right)}$$

Tāpiri 4 ki te 60.

$$x=\frac{-\left(-2\right)±\sqrt{64}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-\left(-2\right)±8}{2\left(-3\right)}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±8}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{2±8}{-6}$$

Nā, me whakaoti te whārite x=\frac{2±8}{-6} ina he tāpiri te ±. Tāpiri 2 ki te 8.

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

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

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

Nā, me whakaoti te whārite x=\frac{2±8}{-6} ina he tango te ±. Tango 8 mai i 2.

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

Whakawehe -6 ki te -6.

$$x=1$$

Kua oti te whārite te whakatau.

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

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.

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

Me tango 5 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 5 i a ia ake anō ka toe ko te 0.

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

Whakawehea ngā taha e rua ki te -3.

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

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

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

Whakawehe -2 ki te -3.

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

Whakawehe -5 ki te -3.

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

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.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{5}{3}+\left(\frac{1}{3}\right)^{2}$$

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

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{5}{3}+\frac{1}{9}$$

Tāpiri \frac{5}{3} 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.

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

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}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{16}{9}$$

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

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Whakarūnātia.

$$x+\frac{1}{3}=\frac{4}{3} x+\frac{1}{3}=-\frac{4}{3}$$

Me tango \frac{1}{3} mai i ngā taha e rua o te whārite.

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