Whakaoti i te x^2+x-2x-(2x^2-5)=0

$${ x }^{ 2 } +x-2x-(2 { x }^{ 2 } -5)=0$$

Whakaoti mō x

$$x = \frac{\sqrt{21} - 1}{2} \approx 1.791287847$$

$$x=\frac{-\sqrt{21}-1}{2}\approx -2.791287847$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Pahekotia te x me -2x, ka -x.

$$x^{2}-x-\left(2x^{2}-5\right)=0$$

Hei kimi i te tauaro o 2x^{2}-5, kimihia te tauaro o ia taurangi.

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

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

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

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

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

Whakareatia -4 ki te -1.

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

Whakareatia 4 ki te 5.

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

Tāpiri 1 ki te 20.

$$x=\frac{-\left(-1\right)±\sqrt{21}}{2\left(-1\right)}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±\sqrt{21}}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{1±\sqrt{21}}{-2}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{21}}{-2} ina he tāpiri te ±. Tāpiri 1 ki te \sqrt{21}.

$$x=\frac{\sqrt{21}+1}{-2}$$

Whakawehe 1+\sqrt{21} ki te -2.

$$x=\frac{-\sqrt{21}-1}{2}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{21}}{-2} ina he tango te ±. Tango \sqrt{21} mai i 1.

$$x=\frac{1-\sqrt{21}}{-2}$$

Whakawehe 1-\sqrt{21} ki te -2.

$$x=\frac{\sqrt{21}-1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{21}-1}{2} x=\frac{\sqrt{21}-1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Pahekotia te x me -2x, ka -x.

$$x^{2}-x-\left(2x^{2}-5\right)=0$$

Hei kimi i te tauaro o 2x^{2}-5, kimihia te tauaro o ia taurangi.

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

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

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

Tangohia te 5 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Whakawehea ngā taha e rua ki te -1.

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

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

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

Whakawehe -1 ki te -1.

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

Whakawehe -5 ki te -1.

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

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

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

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri 5 ki te \frac{1}{4}.

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

Tauwehea x^{2}+x+\frac{1}{4}. 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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{21}{4}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{\sqrt{21}}{2} x+\frac{1}{2}=-\frac{\sqrt{21}}{2}$$

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{21}-1}{2} x=\frac{-\sqrt{21}-1}{2}$$