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

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

Whakaoti mō x

$$x=-1$$

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

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 5 mai i ngā taha e rua.

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

Tangohia te 5 i te 1, ka -4.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 5x^{2}+ax+bx-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=5\left(-4\right)=-20$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. 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.

$$a=-4 b=5$$

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

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

Whakatauwehea atu x i te 5x^{2}-4x.

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

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

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

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

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

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.

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

Me tango 5 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 5 i a ia ake anō ka toe ko te 0.

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

Tango 5 mai i 1.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, 1 mō b, me -4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 5\left(-4\right)}}{2\times 5}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\times 5\left(-4\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-1±\sqrt{1-20\left(-4\right)}}{2\times 5}$$

Whakareatia -20 ki te -4.

$$x=\frac{-1±\sqrt{1+80}}{2\times 5}$$

Tāpiri 1 ki te 80.

$$x=\frac{-1±\sqrt{81}}{2\times 5}$$

Tuhia te pūtakerua o te 81.

$$x=\frac{-1±9}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-1±9}{10}$$

Nā, me whakaoti te whārite x=\frac{-1±9}{10} ina he tāpiri te ±. Tāpiri -1 ki te 9.

$$x=\frac{8}{10}$$

Whakahekea te hautanga \frac{8}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Nā, me whakaoti te whārite x=\frac{-1±9}{10} ina he tango te ±. Tango 9 mai i -1.

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

Whakawehe -10 ki te 10.

$$x=-1$$

Kua oti te whārite te whakatau.

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

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.

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

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

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

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

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

Tango 1 mai i 5.

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

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}+x}{5}=\frac{4}{5}$$

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

$$x^{2}+\frac{1}{5}x=\frac{4}{5}$$

Whakawehea te \frac{1}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{10}. Nā, tāpiria te pūrua o te \frac{1}{10} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{5}x+\left(\frac{1}{10}\right)^{2}=\frac{4}{5}+\left(\frac{1}{10}\right)^{2}$$

Pūruatia \frac{1}{10} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{4}{5}+\frac{1}{100}$$

Tāpiri \frac{4}{5} ki te \frac{1}{100} 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.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{81}{100}$$

Tauwehea x^{2}+\frac{1}{5}x+\frac{1}{100}. 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}.

$$\left(x+\frac{1}{10}\right)^{2}=\frac{81}{100}$$

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

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Whakarūnātia.

$$x+\frac{1}{10}=\frac{9}{10} x+\frac{1}{10}=-\frac{9}{10}$$

Me tango \frac{1}{10} mai i ngā taha e rua o te whārite.

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