Whakaoti i te (5x-1)(5x+1)=-1-5x

$$(5x-1)(5x+1)=-1-5x$$

Whakaoti mō x

$$x=-\frac{1}{5}=-0.2$$

$$x=0$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakaarohia te \left(5x-1\right)\left(5x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

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

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

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

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

$$25x^{2}-1=-1-5x$$

Tangohia te -1 mai i ngā taha e rua.

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

Ko te tauaro o -1 ko 1.

$$25x^{2}-1+1=-5x$$

Me tāpiri te 5x ki ngā taha e rua.

$$25x^{2}-1+1+5x=0$$

Tāpirihia te -1 ki te 1, ka 0.

$$25x^{2}+5x=0$$

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

$$x=\frac{-5±\sqrt{5^{2}}}{2\times 25}$$

Tuhia te pūtakerua o te 5^{2}.

$$x=\frac{-5±5}{2\times 25}$$

Whakareatia 2 ki te 25.

$$x=\frac{-5±5}{50}$$

Nā, me whakaoti te whārite x=\frac{-5±5}{50} ina he tāpiri te ±. Tāpiri -5 ki te 5.

$$x=\frac{0}{50}$$

Whakawehe 0 ki te 50.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-5±5}{50} ina he tango te ±. Tango 5 mai i -5.

$$x=-\frac{10}{50}$$

Whakahekea te hautanga \frac{-10}{50} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

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

Kua oti te whārite te whakatau.

$$x=0 x=-\frac{1}{5}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakaarohia te \left(5x-1\right)\left(5x+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 1.

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

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

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

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

$$25x^{2}-1=-1-5x$$

Me tāpiri te 5x ki ngā taha e rua.

$$25x^{2}-1+5x=-1$$

Me tāpiri te 1 ki ngā taha e rua.

$$25x^{2}+5x=-1+1$$

Tāpirihia te -1 ki te 1, ka 0.

$$25x^{2}+5x=0$$

Whakawehea ngā taha e rua ki te 25.

$$\frac{25x^{2}+5x}{25}=\frac{0}{25}$$

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

$$x^{2}+\frac{5}{25}x=\frac{0}{25}$$

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

$$x^{2}+\frac{1}{5}x=\frac{0}{25}$$

Whakawehe 0 ki te 25.

$$x^{2}+\frac{1}{5}x=0$$

Whakawehea te \frac{1}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{10}. Nā, tāpiria te pūrua o te \frac{1}{10} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{5}x+\left(\frac{1}{10}\right)^{2}=\left(\frac{1}{10}\right)^{2}$$

Pūruatia \frac{1}{10} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{1}{100}$$

Tauwehea x^{2}+\frac{1}{5}x+\frac{1}{100}. 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}.

$$\left(x+\frac{1}{10}\right)^{2}=\frac{1}{100}$$

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

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Whakarūnātia.

$$x+\frac{1}{10}=\frac{1}{10} x+\frac{1}{10}=-\frac{1}{10}$$

Me tango \frac{1}{10} mai i ngā taha e rua o te whārite.

$$x=0 x=-\frac{1}{5}$$