Whakaoti i te 16x^2+8x-3=0

$$16 x ^ { 2 } + 8 x - 3 = 0$$

Whakaoti mō x

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

$$x=\frac{1}{4}=0.25$$

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 16x^{2}+ax+bx-3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=8 ab=16\left(-3\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Tātaihia te tapeke mō ia takirua.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

Ko te otinga te takirua ka hoatu i te tapeke 8.

$$a=-4 b=12$$

Tuhia anō te 16x^{2}+8x-3 hei \left(16x^{2}-4x\right)+\left(12x-3\right).

$$\left(16x^{2}-4x\right)+\left(12x-3\right)$$

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

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

Whakatauwehea atu te kīanga pātahi 4x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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

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

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.

$$16x^{2}+8x-3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 16 mō a, 8 mō b, me -3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 16\left(-3\right)}}{2\times 16}$$

Pūrua 8.

$$x=\frac{-8±\sqrt{64-4\times 16\left(-3\right)}}{2\times 16}$$

Whakareatia -4 ki te 16.

$$x=\frac{-8±\sqrt{64-64\left(-3\right)}}{2\times 16}$$

Whakareatia -64 ki te -3.

$$x=\frac{-8±\sqrt{64+192}}{2\times 16}$$

Tāpiri 64 ki te 192.

$$x=\frac{-8±\sqrt{256}}{2\times 16}$$

Tuhia te pūtakerua o te 256.

$$x=\frac{-8±16}{2\times 16}$$

Whakareatia 2 ki te 16.

$$x=\frac{-8±16}{32}$$

Nā, me whakaoti te whārite x=\frac{-8±16}{32} ina he tāpiri te ±. Tāpiri -8 ki te 16.

$$x=\frac{8}{32}$$

Whakahekea te hautanga \frac{8}{32} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

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

Nā, me whakaoti te whārite x=\frac{-8±16}{32} ina he tango te ±. Tango 16 mai i -8.

$$x=-\frac{24}{32}$$

Whakahekea te hautanga \frac{-24}{32} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

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

Kua oti te whārite te whakatau.

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

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.

$$16x^{2}+8x-3=0$$

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

$$16x^{2}+8x-3-\left(-3\right)=-\left(-3\right)$$

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

$$16x^{2}+8x=-\left(-3\right)$$

Tango -3 mai i 0.

$$16x^{2}+8x=3$$

Whakawehea ngā taha e rua ki te 16.

$$\frac{16x^{2}+8x}{16}=\frac{3}{16}$$

Mā te whakawehe ki te 16 ka wetekia te whakareanga ki te 16.

$$x^{2}+\frac{8}{16}x=\frac{3}{16}$$

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

$$x^{2}+\frac{1}{2}x=\frac{3}{16}$$

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.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{3}{16}+\left(\frac{1}{4}\right)^{2}$$

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{3+1}{16}$$

Tāpiri \frac{3}{16} ki te \frac{1}{16} 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.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{4}$$

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}.

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

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

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

Whakarūnātia.

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

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

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