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

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

Whakaoti mō x

$$x=-1$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

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 2x^{2}+ax+bx-5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=2\left(-5\right)=-10$$

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 -10.

$$1,-10 2,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-10=-9 2-5=-3$$

Ko te otinga te takirua ka hoatu i te tapeke -3.

$$a=-5 b=2$$

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

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

Whakatauwehea atu x i te 2x^{2}-5x.

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

Whakatauwehea atu te kīanga pātahi 2x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-5\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te 2x-5=0 me te x+1=0.

$$x=\frac{5}{2} x=-1$$

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.

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

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

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

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-5\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-5\right)}}{2\times 2}$$

Whakareatia -8 ki te -5.

$$x=\frac{-\left(-3\right)±\sqrt{9+40}}{2\times 2}$$

Tāpiri 9 ki te 40.

$$x=\frac{-\left(-3\right)±\sqrt{49}}{2\times 2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-\left(-3\right)±7}{2\times 2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±7}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{3±7}{4}$$

Nā, me whakaoti te whārite x=\frac{3±7}{4} ina he tāpiri te ±. Tāpiri 3 ki te 7.

$$x=\frac{10}{4}$$

Whakahekea te hautanga \frac{10}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Nā, me whakaoti te whārite x=\frac{3±7}{4} ina he tango te ±. Tango 7 mai i 3.

$$x=-\frac{4}{4}$$

Whakawehe -4 ki te 4.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=\frac{5}{2} x=-1$$

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.

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

Me tāpiri 5 ki ngā taha e rua o te whārite.

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

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

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

Tango -5 mai i 0.

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

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

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

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.

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

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

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{5}{2}+\frac{9}{16}$$

Tāpiri \frac{5}{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.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{49}{16}$$

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

$$\left(x-\frac{3}{4}\right)^{2}=\frac{49}{16}$$

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

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Whakarūnātia.

$$x-\frac{3}{4}=\frac{7}{4} x-\frac{3}{4}=-\frac{7}{4}$$

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

$$x=\frac{5}{2} x=-1$$