Whakaoti i te (2x+1)^2=3-x

$$( 2 x + 1 ) ^ { 2 } = 3 - x$$

Whakaoti mō x

$$x=\frac{\sqrt{57}-5}{8}\approx 0.318729304$$

$$x=\frac{-\sqrt{57}-5}{8}\approx -1.568729304$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$4x^{2}+4x+1=3-x$$

Tangohia te 3 mai i ngā taha e rua.

$$4x^{2}+4x+1-3=-x$$

Tangohia te 3 i te 1, ka -2.

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

Me tāpiri te x ki ngā taha e rua.

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

Pahekotia te 4x me x, ka 5x.

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

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

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

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 4\left(-2\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

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

Whakareatia -16 ki te -2.

$$x=\frac{-5±\sqrt{25+32}}{2\times 4}$$

Tāpiri 25 ki te 32.

$$x=\frac{-5±\sqrt{57}}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-5±\sqrt{57}}{8}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{57}}{8} ina he tāpiri te ±. Tāpiri -5 ki te \sqrt{57}.

$$x=\frac{\sqrt{57}-5}{8}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{57}}{8} ina he tango te ±. Tango \sqrt{57} mai i -5.

$$x=\frac{-\sqrt{57}-5}{8}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{57}-5}{8} x=\frac{-\sqrt{57}-5}{8}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$4x^{2}+4x+1=3-x$$

Me tāpiri te x ki ngā taha e rua.

$$4x^{2}+4x+1+x=3$$

Pahekotia te 4x me x, ka 5x.

$$4x^{2}+5x+1=3$$

Tangohia te 1 mai i ngā taha e rua.

$$4x^{2}+5x=3-1$$

Tangohia te 1 i te 3, ka 2.

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

Whakawehea ngā taha e rua ki te 4.

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

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

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

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

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

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.

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

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

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

Tāpiri \frac{1}{2} 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.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{57}{64}$$

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}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{57}{64}$$

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

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{57}{64}}$$

Whakarūnātia.

$$x+\frac{5}{8}=\frac{\sqrt{57}}{8} x+\frac{5}{8}=-\frac{\sqrt{57}}{8}$$

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

$$x=\frac{\sqrt{57}-5}{8} x=\frac{-\sqrt{57}-5}{8}$$