Whakaoti i te 8x^2+2x-3=0

$$8 { x }^{ 2 } +2x-3=0$$

Whakaoti mō x

$$x=-\frac{3}{4}=-0.75$$

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

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 8x^{2}+ax+bx-3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=2 ab=8\left(-3\right)=-24$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -24.

$$-1,24 -2,12 -3,8 -4,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

Ko te otinga te takirua ka hoatu i te tapeke 2.

$$a=-4 b=6$$

Tuhia anō te 8x^{2}+2x-3 hei \left(8x^{2}-4x\right)+\left(6x-3\right).

$$\left(8x^{2}-4x\right)+\left(6x-3\right)$$

Tauwehea te 4x i te tuatahi me te 3 i te rōpū tuarua.

$$4x\left(2x-1\right)+3\left(2x-1\right)$$

Whakatauwehea atu te kīanga pātahi 2x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-1\right)\left(4x+3\right)$$

Hei kimi otinga whārite, me whakaoti te 2x-1=0 me te 4x+3=0.

$$x=\frac{1}{2} x=-\frac{3}{4}$$

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.

$$8x^{2}+2x-3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 8 mō a, 2 mō b, me -3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Pūrua 2.

$$x=\frac{-2±\sqrt{4-4\times 8\left(-3\right)}}{2\times 8}$$

Whakareatia -4 ki te 8.

$$x=\frac{-2±\sqrt{4-32\left(-3\right)}}{2\times 8}$$

Whakareatia -32 ki te -3.

$$x=\frac{-2±\sqrt{4+96}}{2\times 8}$$

Tāpiri 4 ki te 96.

$$x=\frac{-2±\sqrt{100}}{2\times 8}$$

Tuhia te pūtakerua o te 100.

$$x=\frac{-2±10}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{-2±10}{16}$$

Nā, me whakaoti te whārite x=\frac{-2±10}{16} ina he tāpiri te ±. Tāpiri -2 ki te 10.

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

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

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

Nā, me whakaoti te whārite x=\frac{-2±10}{16} ina he tango te ±. Tango 10 mai i -2.

$$x=-\frac{12}{16}$$

Whakahekea te hautanga \frac{-12}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=-\frac{3}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{1}{2} x=-\frac{3}{4}$$

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.

$$8x^{2}+2x-3=0$$

Me tāpiri 3 ki ngā taha e rua o te whārite.

$$8x^{2}+2x-3-\left(-3\right)=-\left(-3\right)$$

Mā te tango i te -3 i a ia ake anō ka toe ko te 0.

$$8x^{2}+2x=-\left(-3\right)$$

Tango -3 mai i 0.

$$8x^{2}+2x=3$$

Whakawehea ngā taha e rua ki te 8.

$$\frac{8x^{2}+2x}{8}=\frac{3}{8}$$

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

$$x^{2}+\frac{2}{8}x=\frac{3}{8}$$

Whakahekea te hautanga \frac{2}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}+\frac{1}{4}x=\frac{3}{8}$$

Whakawehea te \frac{1}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{8}. Nā, tāpiria te pūrua o te \frac{1}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\frac{3}{8}+\left(\frac{1}{8}\right)^{2}$$

Pūruatia \frac{1}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{3}{8}+\frac{1}{64}$$

Tāpiri \frac{3}{8} ki te \frac{1}{64} 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.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{25}{64}$$

Tauwehea x^{2}+\frac{1}{4}x+\frac{1}{64}. 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}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{25}{64}$$

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

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Whakarūnātia.

$$x+\frac{1}{8}=\frac{5}{8} x+\frac{1}{8}=-\frac{5}{8}$$

Me tango \frac{1}{8} mai i ngā taha e rua o te whārite.

$$x=\frac{1}{2} x=-\frac{3}{4}$$