Whakaoti i te left(5xright)^2-4x-5=0

$${ \left(5x \right) }^{ 2 } -4x-5 = 0$$

Whakaoti mō x

$$x=\frac{\sqrt{129}+2}{25}\approx 0.534312668$$

$$x=\frac{2-\sqrt{129}}{25}\approx -0.374312668$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakarohaina te \left(5x\right)^{2}.

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

Tātaihia te 5 mā te pū o 2, kia riro ko 25.

$$25x^{2}-4x-5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 25 mō a, -4 mō b, me -5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 25\left(-5\right)}}{2\times 25}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 25\left(-5\right)}}{2\times 25}$$

Whakareatia -4 ki te 25.

$$x=\frac{-\left(-4\right)±\sqrt{16-100\left(-5\right)}}{2\times 25}$$

Whakareatia -100 ki te -5.

$$x=\frac{-\left(-4\right)±\sqrt{16+500}}{2\times 25}$$

Tāpiri 16 ki te 500.

$$x=\frac{-\left(-4\right)±\sqrt{516}}{2\times 25}$$

Tuhia te pūtakerua o te 516.

$$x=\frac{-\left(-4\right)±2\sqrt{129}}{2\times 25}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2\sqrt{129}}{2\times 25}$$

Whakareatia 2 ki te 25.

$$x=\frac{4±2\sqrt{129}}{50}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{129}}{50} ina he tāpiri te ±. Tāpiri 4 ki te 2\sqrt{129}.

$$x=\frac{2\sqrt{129}+4}{50}$$

Whakawehe 4+2\sqrt{129} ki te 50.

$$x=\frac{\sqrt{129}+2}{25}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{129}}{50} ina he tango te ±. Tango 2\sqrt{129} mai i 4.

$$x=\frac{4-2\sqrt{129}}{50}$$

Whakawehe 4-2\sqrt{129} ki te 50.

$$x=\frac{2-\sqrt{129}}{25}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{129}+2}{25} x=\frac{2-\sqrt{129}}{25}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakarohaina te \left(5x\right)^{2}.

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

Tātaihia te 5 mā te pū o 2, kia riro ko 25.

$$25x^{2}-4x-5=0$$

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

$$25x^{2}-4x=5$$

Whakawehea ngā taha e rua ki te 25.

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

Mā te whakawehe ki te 25 ka wetekia te whakareanga ki te 25.

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

Whakahekea te hautanga \frac{5}{25} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 5.

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

Whakawehea te -\frac{4}{25}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{2}{25}. Nā, tāpiria te pūrua o te -\frac{2}{25} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{4}{25}x+\left(-\frac{2}{25}\right)^{2}=\frac{1}{5}+\left(-\frac{2}{25}\right)^{2}$$

Pūruatia -\frac{2}{25} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{4}{25}x+\frac{4}{625}=\frac{1}{5}+\frac{4}{625}$$

Tāpiri \frac{1}{5} ki te \frac{4}{625} 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.

$$x^{2}-\frac{4}{25}x+\frac{4}{625}=\frac{129}{625}$$

Tauwehea te x^{2}-\frac{4}{25}x+\frac{4}{625}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{25}\right)^{2}=\frac{129}{625}$$

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

$$\sqrt{\left(x-\frac{2}{25}\right)^{2}}=\sqrt{\frac{129}{625}}$$

Whakarūnātia.

$$x-\frac{2}{25}=\frac{\sqrt{129}}{25} x-\frac{2}{25}=-\frac{\sqrt{129}}{25}$$

Me tāpiri \frac{2}{25} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{129}+2}{25} x=\frac{2-\sqrt{129}}{25}$$