Whakaoti i te 2x^2-5x+1=4

$$2 { x }^{ 2 } -5x+1=4$$

Whakaoti mō x

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

$$x=3$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 4 mai i ngā taha e rua.

$$2x^{2}-5x+1-4=0$$

Tangohia te 4 i te 1, ka -3.

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

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

$$a+b=-5 ab=2\left(-3\right)=-6$$

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 -6.

$$1,-6 2,-3$$

Tātaihia te tapeke mō ia takirua.

$$1-6=-5 2-3=-1$$

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

$$a=-6 b=1$$

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

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

Whakatauwehea atu 2x i te 2x^{2}-6x.

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

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

$$\left(x-3\right)\left(2x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-3=0 me te 2x+1=0.

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

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.

$$2x^{2}-5x+1=4$$

Me tango 4 mai i ngā taha e rua o te whārite.

$$2x^{2}-5x+1-4=4-4$$

Mā te tango i te 4 i a ia ake anō ka toe ko te 0.

$$2x^{2}-5x+1-4=0$$

Tango 4 mai i 1.

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

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

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

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-3\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-3\right)}}{2\times 2}$$

Whakareatia -8 ki te -3.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 2}$$

Tāpiri 25 ki te 24.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2\times 2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-\left(-5\right)±7}{2\times 2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±7}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{5±7}{4}$$

Nā, me whakaoti te whārite x=\frac{5±7}{4} ina he tāpiri te ±. Tāpiri 5 ki te 7.

$$x=\frac{12}{4}$$

Whakawehe 12 ki te 4.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{5±7}{4} ina he tango te ±. Tango 7 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.

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

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.

$$2x^{2}-5x+1=4$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$2x^{2}-5x+1-1=4-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$2x^{2}-5x=4-1$$

Tango 1 mai i 4.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

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.

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

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

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{3}{2}+\frac{25}{16}$$

Tāpiri \frac{3}{2} ki te \frac{25}{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.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{49}{16}$$

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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{49}{16}$$

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

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Whakarūnātia.

$$x-\frac{5}{4}=\frac{7}{4} x-\frac{5}{4}=-\frac{7}{4}$$

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

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