Whakaoti i te 4x^2+2x+1=21

$$4 x ^ { 2 } + 2 x + 1 = 21$$

Whakaoti mō x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 21 mai i ngā taha e rua.

$$4x^{2}+2x+1-21=0$$

Tangohia te 21 i te 1, ka -20.

$$4x^{2}+2x-20=0$$

Whakawehea ngā taha e rua ki te 2.

$$2x^{2}+x-10=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 2x^{2}+ax+bx-10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=2\left(-10\right)=-20$$

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 2x^{2}+x-10 hei \left(2x^{2}-4x\right)+\left(5x-10\right).

$$\left(2x^{2}-4x\right)+\left(5x-10\right)$$

Tauwehea te 2x i te tuatahi me te 5 i te rōpū tuarua.

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

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

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

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

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

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.

$$4x^{2}+2x+1=21$$

Me tango 21 mai i ngā taha e rua o te whārite.

$$4x^{2}+2x+1-21=21-21$$

Mā te tango i te 21 i a ia ake anō ka toe ko te 0.

$$4x^{2}+2x+1-21=0$$

Tango 21 mai i 1.

$$4x^{2}+2x-20=0$$

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

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

Pūrua 2.

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

Whakareatia -4 ki te 4.

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

Whakareatia -16 ki te -20.

$$x=\frac{-2±\sqrt{4+320}}{2\times 4}$$

Tāpiri 4 ki te 320.

$$x=\frac{-2±\sqrt{324}}{2\times 4}$$

Tuhia te pūtakerua o te 324.

$$x=\frac{-2±18}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-2±18}{8}$$

Nā, me whakaoti te whārite x=\frac{-2±18}{8} ina he tāpiri te ±. Tāpiri -2 ki te 18.

$$x=\frac{16}{8}$$

Whakawehe 16 ki te 8.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{-2±18}{8} ina he tango te ±. Tango 18 mai i -2.

$$x=-\frac{20}{8}$$

Whakahekea te hautanga \frac{-20}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

Kua oti te whārite te whakatau.

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

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.

$$4x^{2}+2x+1=21$$

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

$$4x^{2}+2x+1-1=21-1$$

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

$$4x^{2}+2x=21-1$$

Tango 1 mai i 21.

$$4x^{2}+2x=20$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}+2x}{4}=\frac{20}{4}$$

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

$$x^{2}+\frac{2}{4}x=\frac{20}{4}$$

Whakahekea te hautanga \frac{2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}+\frac{1}{2}x=\frac{20}{4}$$

Whakawehe 20 ki te 4.

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

Whakawehea te \frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{4}. Nā, tāpiria te pūrua o te \frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri 5 ki te \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{81}{16}$$

Tauwehea x^{2}+\frac{1}{2}x+\frac{1}{16}. 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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{81}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Whakarūnātia.

$$x+\frac{1}{4}=\frac{9}{4} x+\frac{1}{4}=-\frac{9}{4}$$

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

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