Whakaoti i te 5x^2-2x-16=0

$$5 x ^ { 2 } - 2 x - 16 = 0$$

Whakaoti mō x

$$x = -\frac{8}{5} = -1\frac{3}{5} = -1.6$$

$$x=2$$

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 5x^{2}+ax+bx-16. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-2 ab=5\left(-16\right)=-80$$

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 -2.

$$a=-10 b=8$$

Tuhia anō te 5x^{2}-2x-16 hei \left(5x^{2}-10x\right)+\left(8x-16\right).

$$\left(5x^{2}-10x\right)+\left(8x-16\right)$$

Tauwehea te 5x i te tuatahi me te 8 i te rōpū tuarua.

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

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

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

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

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

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.

$$5x^{2}-2x-16=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -2 mō b, me -16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -2.

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

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-2\right)±\sqrt{4-20\left(-16\right)}}{2\times 5}$$

Whakareatia -20 ki te -16.

$$x=\frac{-\left(-2\right)±\sqrt{4+320}}{2\times 5}$$

Tāpiri 4 ki te 320.

$$x=\frac{-\left(-2\right)±\sqrt{324}}{2\times 5}$$

Tuhia te pūtakerua o te 324.

$$x=\frac{-\left(-2\right)±18}{2\times 5}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±18}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{2±18}{10}$$

Nā, me whakaoti te whārite x=\frac{2±18}{10} ina he tāpiri te ±. Tāpiri 2 ki te 18.

$$x=\frac{20}{10}$$

Whakawehe 20 ki te 10.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{2±18}{10} ina he tango te ±. Tango 18 mai i 2.

$$x=-\frac{16}{10}$$

Whakahekea te hautanga \frac{-16}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

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.

$$5x^{2}-2x-16=0$$

Me tāpiri 16 ki ngā taha e rua o te whārite.

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

Mā te tango i te -16 i a ia ake anō ka toe ko te 0.

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

Tango -16 mai i 0.

$$5x^{2}-2x=16$$

Whakawehea ngā taha e rua ki te 5.

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

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

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

Whakawehea te -\frac{2}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{5}. Nā, tāpiria te pūrua o te -\frac{1}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=\frac{16}{5}+\left(-\frac{1}{5}\right)^{2}$$

Pūruatia -\frac{1}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri \frac{16}{5} ki te \frac{1}{25} 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.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{81}{25}$$

Tauwehea x^{2}-\frac{2}{5}x+\frac{1}{25}. 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}.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{81}{25}$$

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

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{81}{25}}$$

Whakarūnātia.

$$x-\frac{1}{5}=\frac{9}{5} x-\frac{1}{5}=-\frac{9}{5}$$

Me tāpiri \frac{1}{5} ki ngā taha e rua o te whārite.

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