Whakaoti i te 5x^2+4x-5=0

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

Whakaoti mō x

$$x=\frac{\sqrt{29}-2}{5}\approx 0.677032961$$

$$x=\frac{-\sqrt{29}-2}{5}\approx -1.477032961$$

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.

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

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

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

Pūrua 4.

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

Whakareatia -4 ki te 5.

$$x=\frac{-4±\sqrt{16-20\left(-5\right)}}{2\times 5}$$

Whakareatia -20 ki te -5.

$$x=\frac{-4±\sqrt{16+100}}{2\times 5}$$

Tāpiri 16 ki te 100.

$$x=\frac{-4±\sqrt{116}}{2\times 5}$$

Tuhia te pūtakerua o te 116.

$$x=\frac{-4±2\sqrt{29}}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-4±2\sqrt{29}}{10}$$

Nā, me whakaoti te whārite x=\frac{-4±2\sqrt{29}}{10} ina he tāpiri te ±. Tāpiri -4 ki te 2\sqrt{29}.

$$x=\frac{2\sqrt{29}-4}{10}$$

Whakawehe -4+2\sqrt{29} ki te 10.

$$x=\frac{\sqrt{29}-2}{5}$$

Nā, me whakaoti te whārite x=\frac{-4±2\sqrt{29}}{10} ina he tango te ±. Tango 2\sqrt{29} mai i -4.

$$x=\frac{-2\sqrt{29}-4}{10}$$

Whakawehe -4-2\sqrt{29} ki te 10.

$$x=\frac{-\sqrt{29}-2}{5}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{29}-2}{5} x=\frac{-\sqrt{29}-2}{5}$$

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.

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

Me tāpiri 5 ki ngā taha e rua o te whārite.

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

Mā te tango i te -5 i a ia ake anō ka toe ko te 0.

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

Tango -5 mai i 0.

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

Whakawehea ngā taha e rua ki te 5.

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

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

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

Whakawehe 5 ki te 5.

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

Whakawehea te \frac{4}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{2}{5}. Nā, tāpiria te pūrua o te \frac{2}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{2}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri 1 ki te \frac{4}{25}.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{29}{25}$$

Tauwehea x^{2}+\frac{4}{5}x+\frac{4}{25}. 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}.

$$\left(x+\frac{2}{5}\right)^{2}=\frac{29}{25}$$

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

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{\frac{29}{25}}$$

Whakarūnātia.

$$x+\frac{2}{5}=\frac{\sqrt{29}}{5} x+\frac{2}{5}=-\frac{\sqrt{29}}{5}$$

Me tango \frac{2}{5} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{29}-2}{5} x=\frac{-\sqrt{29}-2}{5}$$