Whakaoti i te 2x^2-11x-40=0

$$2 { x }^{ 2 } -11x-40=0$$

Whakaoti mō x

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

$$x=8$$

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-40. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-11 ab=2\left(-40\right)=-80$$

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 -80.

$$1,-80 2,-40 4,-20 5,-16 8,-10$$

Tātaihia te tapeke mō ia takirua.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

Ko te otinga te takirua ka hoatu i te tapeke -11.

$$a=-16 b=5$$

Tuhia anō te 2x^{2}-11x-40 hei \left(2x^{2}-16x\right)+\left(5x-40\right).

$$\left(2x^{2}-16x\right)+\left(5x-40\right)$$

Tauwehea te 2x i te tuatahi me te 5 i te rōpū tuarua.

$$2x\left(x-8\right)+5\left(x-8\right)$$

Whakatauwehea atu te kīanga pātahi x-8 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-8\right)\left(2x+5\right)$$

Hei kimi otinga whārite, me whakaoti te x-8=0 me te 2x+5=0.

$$x=8 x=-\frac{5}{2}$$

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.

$$2x^{2}-11x-40=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -11 mō b, me -40 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\left(-40\right)}}{2\times 2}$$

Pūrua -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\left(-40\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\left(-40\right)}}{2\times 2}$$

Whakareatia -8 ki te -40.

$$x=\frac{-\left(-11\right)±\sqrt{121+320}}{2\times 2}$$

Tāpiri 121 ki te 320.

$$x=\frac{-\left(-11\right)±\sqrt{441}}{2\times 2}$$

Tuhia te pūtakerua o te 441.

$$x=\frac{-\left(-11\right)±21}{2\times 2}$$

Ko te tauaro o -11 ko 11.

$$x=\frac{11±21}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{11±21}{4}$$

Nā, me whakaoti te whārite x=\frac{11±21}{4} ina he tāpiri te ±. Tāpiri 11 ki te 21.

$$x=\frac{32}{4}$$

Whakawehe 32 ki te 4.

$$x=8$$

Nā, me whakaoti te whārite x=\frac{11±21}{4} ina he tango te ±. Tango 21 mai i 11.

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

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

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

Kua oti te whārite te whakatau.

$$x=8 x=-\frac{5}{2}$$

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.

$$2x^{2}-11x-40=0$$

Me tāpiri 40 ki ngā taha e rua o te whārite.

$$2x^{2}-11x-40-\left(-40\right)=-\left(-40\right)$$

Mā te tango i te -40 i a ia ake anō ka toe ko te 0.

$$2x^{2}-11x=-\left(-40\right)$$

Tango -40 mai i 0.

$$2x^{2}-11x=40$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-11x}{2}=\frac{40}{2}$$

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

$$x^{2}-\frac{11}{2}x=\frac{40}{2}$$

Whakawehe 40 ki te 2.

$$x^{2}-\frac{11}{2}x=20$$

Whakawehea te -\frac{11}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{11}{4}. Nā, tāpiria te pūrua o te -\frac{11}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=20+\left(-\frac{11}{4}\right)^{2}$$

Pūruatia -\frac{11}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=20+\frac{121}{16}$$

Tāpiri 20 ki te \frac{121}{16}.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{441}{16}$$

Tauwehea te x^{2}-\frac{11}{2}x+\frac{121}{16}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{441}{16}$$

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

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{441}{16}}$$

Whakarūnātia.

$$x-\frac{11}{4}=\frac{21}{4} x-\frac{11}{4}=-\frac{21}{4}$$

Me tāpiri \frac{11}{4} ki ngā taha e rua o te whārite.

$$x=8 x=-\frac{5}{2}$$