Whakaoti i te 4x(2x+1)=0

$$4 x ( 2 x + 1 ) = 0$$

Whakaoti mō x

$$x=-\frac{1}{2}=-0.5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te 2x+1.

$$8x^{2}+4x=0$$

Tauwehea te x.

$$x\left(8x+4\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 8x+4=0.

$$x=0 x=-\frac{1}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te 2x+1.

$$8x^{2}+4x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 8 mō a, 4 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}}}{2\times 8}$$

Tuhia te pūtakerua o te 4^{2}.

$$x=\frac{-4±4}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{-4±4}{16}$$

Nā, me whakaoti te whārite x=\frac{-4±4}{16} ina he tāpiri te ±. Tāpiri -4 ki te 4.

$$x=\frac{0}{16}$$

Whakawehe 0 ki te 16.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-4±4}{16} ina he tango te ±. Tango 4 mai i -4.

$$x=-\frac{8}{16}$$

Whakahekea te hautanga \frac{-8}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

$$x=-\frac{1}{2}$$

Kua oti te whārite te whakatau.

$$x=0 x=-\frac{1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 4x ki te 2x+1.

$$8x^{2}+4x=0$$

Whakawehea ngā taha e rua ki te 8.

$$\frac{8x^{2}+4x}{8}=\frac{0}{8}$$

Mā te whakawehe ki te 8 ka wetekia te whakareanga ki te 8.

$$x^{2}+\frac{4}{8}x=\frac{0}{8}$$

Whakahekea te hautanga \frac{4}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x^{2}+\frac{1}{2}x=\frac{0}{8}$$

Whakawehe 0 ki te 8.

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

Whakawehea te \frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{4}. Nā, tāpiria te pūrua o te \frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Tauwehea x^{2}+\frac{1}{2}x+\frac{1}{16}. 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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{1}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Whakarūnātia.

$$x+\frac{1}{4}=\frac{1}{4} x+\frac{1}{4}=-\frac{1}{4}$$

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

$$x=0 x=-\frac{1}{2}$$