Whakaoti i te x^2-x*2x+1-x^2=2x^2+4x-x-1

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

Whakaoti mō x

$$x=\frac{\sqrt{41}-3}{8}\approx 0.42539053$$

$$x=\frac{-\sqrt{41}-3}{8}\approx -1.17539053$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te x ki te x, ka x^{2}.

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

Pahekotia te x^{2} me -x^{2}\times 2, ka -x^{2}.

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

Pahekotia te -x^{2} me -x^{2}, ka -2x^{2}.

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

Pahekotia te 4x me -x, ka 3x.

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

Tangohia te 2x^{2} mai i ngā taha e rua.

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

Pahekotia te -2x^{2} me -2x^{2}, ka -4x^{2}.

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

Tangohia te 3x mai i ngā taha e rua.

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

Me tāpiri te 1 ki ngā taha e rua.

$$-4x^{2}+1-3x+1=0$$

Tāpirihia te 1 ki te 1, ka 2.

$$-4x^{2}+2-3x=0$$

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.

$$-4x^{2}-3x+2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -4 mō a, -3 mō b, me 2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-4\right)\times 2}}{2\left(-4\right)}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)\times 2}}{2\left(-4\right)}$$

Whakareatia -4 ki te -4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16\times 2}}{2\left(-4\right)}$$

Whakareatia 16 ki te 2.

$$x=\frac{-\left(-3\right)±\sqrt{9+32}}{2\left(-4\right)}$$

Tāpiri 9 ki te 32.

$$x=\frac{-\left(-3\right)±\sqrt{41}}{2\left(-4\right)}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±\sqrt{41}}{2\left(-4\right)}$$

Whakareatia 2 ki te -4.

$$x=\frac{3±\sqrt{41}}{-8}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{41}}{-8} ina he tāpiri te ±. Tāpiri 3 ki te \sqrt{41}.

$$x=\frac{\sqrt{41}+3}{-8}$$

Whakawehe 3+\sqrt{41} ki te -8.

$$x=\frac{-\sqrt{41}-3}{8}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{41}}{-8} ina he tango te ±. Tango \sqrt{41} mai i 3.

$$x=\frac{3-\sqrt{41}}{-8}$$

Whakawehe 3-\sqrt{41} ki te -8.

$$x=\frac{\sqrt{41}-3}{8}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{41}-3}{8} x=\frac{\sqrt{41}-3}{8}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te x ki te x, ka x^{2}.

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

Pahekotia te x^{2} me -x^{2}\times 2, ka -x^{2}.

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

Pahekotia te -x^{2} me -x^{2}, ka -2x^{2}.

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

Pahekotia te 4x me -x, ka 3x.

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

Tangohia te 2x^{2} mai i ngā taha e rua.

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

Pahekotia te -2x^{2} me -2x^{2}, ka -4x^{2}.

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

Tangohia te 3x mai i ngā taha e rua.

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

Tangohia te 1 mai i ngā taha e rua.

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

Tangohia te 1 i te -1, ka -2.

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

Whakawehea ngā taha e rua ki te -4.

$$\frac{-4x^{2}-3x}{-4}=-\frac{2}{-4}$$

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

$$x^{2}+\left(-\frac{3}{-4}\right)x=-\frac{2}{-4}$$

Whakawehe -3 ki te -4.

$$x^{2}+\frac{3}{4}x=-\frac{2}{-4}$$

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

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

Whakawehea te \frac{3}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{8}. Nā, tāpiria te pūrua o te \frac{3}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=\frac{1}{2}+\left(\frac{3}{8}\right)^{2}$$

Pūruatia \frac{3}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{1}{2}+\frac{9}{64}$$

Tāpiri \frac{1}{2} ki te \frac{9}{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.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{41}{64}$$

Tauwehea x^{2}+\frac{3}{4}x+\frac{9}{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}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{41}{64}$$

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

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{41}{64}}$$

Whakarūnātia.

$$x+\frac{3}{8}=\frac{\sqrt{41}}{8} x+\frac{3}{8}=-\frac{\sqrt{41}}{8}$$

Me tango \frac{3}{8} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{41}-3}{8} x=\frac{-\sqrt{41}-3}{8}$$