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

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

Whakaoti mō x

$$x=\frac{\sqrt{21}-1}{5}\approx 0.716515139$$

$$x=\frac{-\sqrt{21}-1}{5}\approx -1.116515139$$

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.

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

Me tango 4 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 4 i a ia ake anō ka toe ko te 0.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te 5.

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

Whakareatia -20 ki te -4.

$$x=\frac{-2±\sqrt{4+80}}{2\times 5}$$

Tāpiri 4 ki te 80.

$$x=\frac{-2±\sqrt{84}}{2\times 5}$$

Tuhia te pūtakerua o te 84.

$$x=\frac{-2±2\sqrt{21}}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-2±2\sqrt{21}}{10}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{21}}{10} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{21}.

$$x=\frac{2\sqrt{21}-2}{10}$$

Whakawehe -2+2\sqrt{21} ki te 10.

$$x=\frac{\sqrt{21}-1}{5}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{21}}{10} ina he tango te ±. Tango 2\sqrt{21} mai i -2.

$$x=\frac{-2\sqrt{21}-2}{10}$$

Whakawehe -2-2\sqrt{21} ki te 10.

$$x=\frac{-\sqrt{21}-1}{5}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{21}-1}{5} x=\frac{-\sqrt{21}-1}{5}$$

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.

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

Whakawehea ngā taha e rua ki te 5.

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

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

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

Whakawehea te \frac{2}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{5}. Nā, tāpiria te pūrua o te \frac{1}{5} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{5} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri \frac{4}{5} ki te \frac{1}{25} 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.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{21}{25}$$

Tauwehea x^{2}+\frac{2}{5}x+\frac{1}{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}.

$$\left(x+\frac{1}{5}\right)^{2}=\frac{21}{25}$$

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

$$\sqrt{\left(x+\frac{1}{5}\right)^{2}}=\sqrt{\frac{21}{25}}$$

Whakarūnātia.

$$x+\frac{1}{5}=\frac{\sqrt{21}}{5} x+\frac{1}{5}=-\frac{\sqrt{21}}{5}$$

Me tango \frac{1}{5} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{21}-1}{5} x=\frac{-\sqrt{21}-1}{5}$$