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

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

Whakaoti mō x

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

$$x=2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

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

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

$$a+b=-5 ab=2\times 2=4$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$-1,-4 -2,-2$$

Tātaihia te tapeke mō ia takirua.

$$-1-4=-5 -2-2=-4$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-4 b=-1$$

Tuhia anō te 2x^{2}-5x+2 hei \left(2x^{2}-4x\right)+\left(-x+2\right).

$$\left(2x^{2}-4x\right)+\left(-x+2\right)$$

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

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

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

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

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

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

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.

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

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

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

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\times 2}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\times 2}}{2\times 2}$$

Whakareatia -8 ki te 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-16}}{2\times 2}$$

Tāpiri 25 ki te -16.

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

Tuhia te pūtakerua o te 9.

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

Ko te tauaro o -5 ko 5.

$$x=\frac{5±3}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{5±3}{4}$$

Nā, me whakaoti te whārite x=\frac{5±3}{4} ina he tāpiri te ±. Tāpiri 5 ki te 3.

$$x=\frac{8}{4}$$

Whakawehe 8 ki te 4.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{5±3}{4} ina he tango te ±. Tango 3 mai i 5.

$$x=\frac{2}{4}$$

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

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

Kua oti te whārite te whakatau.

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

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.

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

Me tango 2 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe -2 ki te 2.

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

Whakawehea te -\frac{5}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{4}. Nā, tāpiria te pūrua o te -\frac{5}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-1+\left(-\frac{5}{4}\right)^{2}$$

Pūruatia -\frac{5}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-1+\frac{25}{16}$$

Tāpiri -1 ki te \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{9}{16}$$

Tauwehea x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{9}{16}$$

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

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Whakarūnātia.

$$x-\frac{5}{4}=\frac{3}{4} x-\frac{5}{4}=-\frac{3}{4}$$

Me tāpiri \frac{5}{4} ki ngā taha e rua o te whārite.

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