Whakaoti i te 20x^2-x-1=0

$$20 { x }^{ 2 } -x-1=0$$

Whakaoti mō x

$$x=-\frac{1}{5}=-0.2$$

$$x=\frac{1}{4}=0.25$$

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 20x^{2}+ax+bx-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=20\left(-1\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-20=-19 2-10=-8 4-5=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -1.

$$a=-5 b=4$$

Tuhia anō te 20x^{2}-x-1 hei \left(20x^{2}-5x\right)+\left(4x-1\right).

$$\left(20x^{2}-5x\right)+\left(4x-1\right)$$

Whakatauwehea atu 5x i te 20x^{2}-5x.

$$5x\left(4x-1\right)+4x-1$$

Whakatauwehea atu te kīanga pātahi 4x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(4x-1\right)\left(5x+1\right)$$

Hei kimi otinga whārite, me whakaoti te 4x-1=0 me te 5x+1=0.

$$x=\frac{1}{4} x=-\frac{1}{5}$$

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.

$$20x^{2}-x-1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 20 mō a, -1 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 20\left(-1\right)}}{2\times 20}$$

Whakareatia -4 ki te 20.

$$x=\frac{-\left(-1\right)±\sqrt{1-80\left(-1\right)}}{2\times 20}$$

Whakareatia -80 ki te -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+80}}{2\times 20}$$

Tāpiri 1 ki te 80.

$$x=\frac{-\left(-1\right)±\sqrt{81}}{2\times 20}$$

Tuhia te pūtakerua o te 81.

$$x=\frac{-\left(-1\right)±9}{2\times 20}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±9}{2\times 20}$$

Whakareatia 2 ki te 20.

$$x=\frac{1±9}{40}$$

Nā, me whakaoti te whārite x=\frac{1±9}{40} ina he tāpiri te ±. Tāpiri 1 ki te 9.

$$x=\frac{10}{40}$$

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

$$x=\frac{1}{4}$$

Nā, me whakaoti te whārite x=\frac{1±9}{40} ina he tango te ±. Tango 9 mai i 1.

$$x=-\frac{8}{40}$$

Whakahekea te hautanga \frac{-8}{40} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

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

Kua oti te whārite te whakatau.

$$x=\frac{1}{4} x=-\frac{1}{5}$$

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.

$$20x^{2}-x-1=0$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$20x^{2}-x-1-\left(-1\right)=-\left(-1\right)$$

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

$$20x^{2}-x=-\left(-1\right)$$

Tango -1 mai i 0.

$$20x^{2}-x=1$$

Whakawehea ngā taha e rua ki te 20.

$$\frac{20x^{2}-x}{20}=\frac{1}{20}$$

Mā te whakawehe ki te 20 ka wetekia te whakareanga ki te 20.

$$x^{2}-\frac{1}{20}x=\frac{1}{20}$$

Whakawehea te -\frac{1}{20}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{40}. Nā, tāpiria te pūrua o te -\frac{1}{40} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{20}x+\left(-\frac{1}{40}\right)^{2}=\frac{1}{20}+\left(-\frac{1}{40}\right)^{2}$$

Pūruatia -\frac{1}{40} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{20}x+\frac{1}{1600}=\frac{1}{20}+\frac{1}{1600}$$

Tāpiri \frac{1}{20} ki te \frac{1}{1600} 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.

$$x^{2}-\frac{1}{20}x+\frac{1}{1600}=\frac{81}{1600}$$

Tauwehea x^{2}-\frac{1}{20}x+\frac{1}{1600}. 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}.

$$\left(x-\frac{1}{40}\right)^{2}=\frac{81}{1600}$$

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

$$\sqrt{\left(x-\frac{1}{40}\right)^{2}}=\sqrt{\frac{81}{1600}}$$

Whakarūnātia.

$$x-\frac{1}{40}=\frac{9}{40} x-\frac{1}{40}=-\frac{9}{40}$$

Me tāpiri \frac{1}{40} ki ngā taha e rua o te whārite.

$$x=\frac{1}{4} x=-\frac{1}{5}$$