Whakaoti i te 4x^2+5x+1=0

$$4 { x }^{ 2 } +5x+1=0$$

Whakaoti mō x

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

$$x=-1$$

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 4x^{2}+ax+bx+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=5 ab=4\times 1=4$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$1,4 2,2$$

Tātaihia te tapeke mō ia takirua.

$$1+4=5 2+2=4$$

Ko te otinga te takirua ka hoatu i te tapeke 5.

$$a=1 b=4$$

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

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

Whakatauwehea atu x i te 4x^{2}+x.

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

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

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

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

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

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.

$$4x^{2}+5x+1=0$$

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

$$x=\frac{-5±\sqrt{5^{2}-4\times 4}}{2\times 4}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 4}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-5±\sqrt{25-16}}{2\times 4}$$

Tāpiri 25 ki te -16.

$$x=\frac{-5±\sqrt{9}}{2\times 4}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-5±3}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-5±3}{8}$$

Nā, me whakaoti te whārite x=\frac{-5±3}{8} ina he tāpiri te ±. Tāpiri -5 ki te 3.

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

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

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

Nā, me whakaoti te whārite x=\frac{-5±3}{8} ina he tango te ±. Tango 3 mai i -5.

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

Whakawehe -8 ki te 8.

$$x=-1$$

Kua oti te whārite te whakatau.

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

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.

$$4x^{2}+5x+1=0$$

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

$$4x^{2}+5x+1-1=-1$$

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

$$4x^{2}+5x=-1$$

Whakawehea ngā taha e rua ki te 4.

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

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

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

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

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=-\frac{1}{4}+\left(\frac{5}{8}\right)^{2}$$

Pūruatia \frac{5}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=-\frac{1}{4}+\frac{25}{64}$$

Tāpiri -\frac{1}{4} ki te \frac{25}{64} 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.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{9}{64}$$

Tauwehea x^{2}+\frac{5}{4}x+\frac{25}{64}. 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}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{9}{64}$$

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

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{9}{64}}$$

Whakarūnātia.

$$x+\frac{5}{8}=\frac{3}{8} x+\frac{5}{8}=-\frac{3}{8}$$

Me tango \frac{5}{8} mai i ngā taha e rua o te whārite.

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