Whakaoti i te 25x^2-40x+16=0

$$25 x ^ { 2 } - 40 x + 16 = 0$$

Whakaoti mō x

$$x=\frac{4}{5}=0.8$$

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 25x^{2}+ax+bx+16. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-40 ab=25\times 16=400$$

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 400.

$$-1,-400 -2,-200 -4,-100 -5,-80 -8,-50 -10,-40 -16,-25 -20,-20$$

Tātaihia te tapeke mō ia takirua.

$$-1-400=-401 -2-200=-202 -4-100=-104 -5-80=-85 -8-50=-58 -10-40=-50 -16-25=-41 -20-20=-40$$

Ko te otinga te takirua ka hoatu i te tapeke -40.

$$a=-20 b=-20$$

Tuhia anō te 25x^{2}-40x+16 hei \left(25x^{2}-20x\right)+\left(-20x+16\right).

$$\left(25x^{2}-20x\right)+\left(-20x+16\right)$$

Tauwehea te 5x i te tuatahi me te -4 i te rōpū tuarua.

$$5x\left(5x-4\right)-4\left(5x-4\right)$$

Whakatauwehea atu te kīanga pātahi 5x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(5x-4\right)\left(5x-4\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(5x-4\right)^{2}$$

Hei kimi i te otinga whārite, whakaotia te 5x-4=0.

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

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.

$$25x^{2}-40x+16=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 25 mō a, -40 mō b, me 16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 25\times 16}}{2\times 25}$$

Pūrua -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 25\times 16}}{2\times 25}$$

Whakareatia -4 ki te 25.

$$x=\frac{-\left(-40\right)±\sqrt{1600-100\times 16}}{2\times 25}$$

Whakareatia -100 ki te 16.

$$x=\frac{-\left(-40\right)±\sqrt{1600-1600}}{2\times 25}$$

Tāpiri 1600 ki te -1600.

$$x=\frac{-\left(-40\right)±\sqrt{0}}{2\times 25}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{-40}{2\times 25}$$

Ko te tauaro o -40 ko 40.

$$x=\frac{40}{2\times 25}$$

Whakareatia 2 ki te 25.

$$x=\frac{40}{50}$$

Whakahekea te hautanga \frac{40}{50} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

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

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.

$$25x^{2}-40x+16=0$$

Me tango 16 mai i ngā taha e rua o te whārite.

$$25x^{2}-40x+16-16=-16$$

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

$$25x^{2}-40x=-16$$

Whakawehea ngā taha e rua ki te 25.

$$\frac{25x^{2}-40x}{25}=-\frac{16}{25}$$

Mā te whakawehe ki te 25 ka wetekia te whakareanga ki te 25.

$$x^{2}+\left(-\frac{40}{25}\right)x=-\frac{16}{25}$$

Whakahekea te hautanga \frac{-40}{25} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 5.

$$x^{2}-\frac{8}{5}x=-\frac{16}{25}$$

Whakawehea te -\frac{8}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{4}{5}. Nā, tāpiria te pūrua o te -\frac{4}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=-\frac{16}{25}+\left(-\frac{4}{5}\right)^{2}$$

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

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{-16+16}{25}$$

Tāpiri -\frac{16}{25} ki te \frac{16}{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.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=0$$

Tauwehea te x^{2}-\frac{8}{5}x+\frac{16}{25}. 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}.

$$\left(x-\frac{4}{5}\right)^{2}=0$$

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

$$\sqrt{\left(x-\frac{4}{5}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x-\frac{4}{5}=0 x-\frac{4}{5}=0$$

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

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

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

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