Whakaoti i te x^2+2x-5=05*2x^2

$${ x }^{ 2 } +2x-5=05 \times 2 { x }^{ 2 }$$

Whakaoti mō x (complex solution)

$$x=\sqrt{6}-1\approx 1.449489743$$

$$x=-\left(\sqrt{6}+1\right)\approx -3.449489743$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te 0 ki te 5, ka 0.

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

Whakareatia te 0 ki te 2, ka 0.

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

Ko te tau i whakarea ki te kore ka hua ko te kore.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te -5.

$$x=\frac{-2±\sqrt{4+20}}{2}$$

Tāpiri 4 ki te 20.

$$x=\frac{-2±\sqrt{24}}{2}$$

Tuhia te pūtakerua o te 24.

$$x=\frac{-2±2\sqrt{6}}{2}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{6}}{2} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-2}{2}$$

Whakawehe -2+2\sqrt{6} ki te 2.

$$x=\sqrt{6}-1$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{6}}{2} ina he tango te ±. Tango 2\sqrt{6} mai i -2.

$$x=\frac{-2\sqrt{6}-2}{2}$$

Whakawehe -2-2\sqrt{6} ki te 2.

$$x=-\sqrt{6}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{6}-1 x=-\sqrt{6}-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te 0 ki te 5, ka 0.

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

Whakareatia te 0 ki te 2, ka 0.

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

Ko te tau i whakarea ki te kore ka hua ko te kore.

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

Me tāpiri te 5 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

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

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

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

Pūrua 1.

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

Tāpiri 5 ki te 1.

$$x^{2}+2x+1=6$$

Tauwehea x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=6$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{6}$$

Whakarūnātia.

$$x+1=\sqrt{6} x+1=-\sqrt{6}$$

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

$$x=\sqrt{6}-1 x=-\sqrt{6}-1$$

Whakaoti mō x

$$x=\sqrt{6}-1\approx 1.449489743$$

$$x=-\sqrt{6}-1\approx -3.449489743$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te 0 ki te 5, ka 0.

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

Whakareatia te 0 ki te 2, ka 0.

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

Ko te tau i whakarea ki te kore ka hua ko te kore.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te -5.

$$x=\frac{-2±\sqrt{4+20}}{2}$$

Tāpiri 4 ki te 20.

$$x=\frac{-2±\sqrt{24}}{2}$$

Tuhia te pūtakerua o te 24.

$$x=\frac{-2±2\sqrt{6}}{2}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{6}}{2} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-2}{2}$$

Whakawehe -2+2\sqrt{6} ki te 2.

$$x=\sqrt{6}-1$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{6}}{2} ina he tango te ±. Tango 2\sqrt{6} mai i -2.

$$x=\frac{-2\sqrt{6}-2}{2}$$

Whakawehe -2-2\sqrt{6} ki te 2.

$$x=-\sqrt{6}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{6}-1 x=-\sqrt{6}-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te 0 ki te 5, ka 0.

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

Whakareatia te 0 ki te 2, ka 0.

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

Ko te tau i whakarea ki te kore ka hua ko te kore.

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

Me tāpiri te 5 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

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

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

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

Pūrua 1.

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

Tāpiri 5 ki te 1.

$$x^{2}+2x+1=6$$

Tauwehea x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=6$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{6}$$

Whakarūnātia.

$$x+1=\sqrt{6} x+1=-\sqrt{6}$$

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

$$x=\sqrt{6}-1 x=-\sqrt{6}-1$$