Whakaoti i te 20x^2+2x-08=0

$$20 x ^ { 2 } + 2 x - 08 = 0$$

Whakaoti mō x

$$x=-\frac{1}{10}=-0.1$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakareatia te 0 ki te 8, ka 0.

$$20x^{2}+2x-0=0$$

Whakaraupapatia anō ngā kīanga tau.

$$20x^{2}+2x=0$$

Tauwehea te x.

$$x\left(20x+2\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 20x+2=0.

$$x=0 x=-\frac{1}{10}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te 0 ki te 8, ka 0.

$$20x^{2}+2x-0=0$$

Whakaraupapatia anō ngā kīanga tau.

$$20x^{2}+2x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 20 mō a, 2 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}}}{2\times 20}$$

Tuhia te pūtakerua o te 2^{2}.

$$x=\frac{-2±2}{2\times 20}$$

Whakareatia 2 ki te 20.

$$x=\frac{-2±2}{40}$$

Nā, me whakaoti te whārite x=\frac{-2±2}{40} ina he tāpiri te ±. Tāpiri -2 ki te 2.

$$x=\frac{0}{40}$$

Whakawehe 0 ki te 40.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-2±2}{40} ina he tango te ±. Tango 2 mai i -2.

$$x=-\frac{4}{40}$$

Whakahekea te hautanga \frac{-4}{40} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

Kua oti te whārite te whakatau.

$$x=0 x=-\frac{1}{10}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te 0 ki te 8, ka 0.

$$20x^{2}+2x-0=0$$

Me tāpiri te 0 ki ngā taha e rua.

$$20x^{2}+2x=0+0$$

Tāpirihia te 0 ki te 0, ka 0.

$$20x^{2}+2x=0$$

Whakawehea ngā taha e rua ki te 20.

$$\frac{20x^{2}+2x}{20}=\frac{0}{20}$$

Mā te whakawehe ki te 20 ka wetekia te whakareanga ki te 20.

$$x^{2}+\frac{2}{20}x=\frac{0}{20}$$

Whakahekea te hautanga \frac{2}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}+\frac{1}{10}x=\frac{0}{20}$$

Whakawehe 0 ki te 20.

$$x^{2}+\frac{1}{10}x=0$$

Whakawehea te \frac{1}{10}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{20}. Nā, tāpiria te pūrua o te \frac{1}{20} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{10}x+\left(\frac{1}{20}\right)^{2}=\left(\frac{1}{20}\right)^{2}$$

Pūruatia \frac{1}{20} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{10}x+\frac{1}{400}=\frac{1}{400}$$

Tauwehea x^{2}+\frac{1}{10}x+\frac{1}{400}. 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}.

$$\left(x+\frac{1}{20}\right)^{2}=\frac{1}{400}$$

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

$$\sqrt{\left(x+\frac{1}{20}\right)^{2}}=\sqrt{\frac{1}{400}}$$

Whakarūnātia.

$$x+\frac{1}{20}=\frac{1}{20} x+\frac{1}{20}=-\frac{1}{20}$$

Me tango \frac{1}{20} mai i ngā taha e rua o te whārite.

$$x=0 x=-\frac{1}{10}$$