Whakaoti i te {l}{18/text{febre}}{8x^2=5x-2}

$$\left. \begin{array} { l } { 18 / \text { febre } } \\ { 8 x ^ { 2 } = 5 x - 2 } \end{array} \right.$$

Whakaoti mō x

$$x=\frac{5+\sqrt{39}i}{16}\approx 0.3125+0.390312375i$$

$$x=\frac{-\sqrt{39}i+5}{16}\approx 0.3125-0.390312375i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 5x mai i ngā taha e rua.

$$8x^{2}-5x=-2$$

Me tāpiri te 2 ki ngā taha e rua.

$$8x^{2}-5x+2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 8 mō a, -5 mō b, me 2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 8\times 2}}{2\times 8}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 8\times 2}}{2\times 8}$$

Whakareatia -4 ki te 8.

$$x=\frac{-\left(-5\right)±\sqrt{25-32\times 2}}{2\times 8}$$

Whakareatia -32 ki te 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-64}}{2\times 8}$$

Tāpiri 25 ki te -64.

$$x=\frac{-\left(-5\right)±\sqrt{-39}}{2\times 8}$$

Tuhia te pūtakerua o te -39.

$$x=\frac{-\left(-5\right)±\sqrt{39}i}{2\times 8}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{39}i}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{5±\sqrt{39}i}{16}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{39}i}{16} ina he tāpiri te ±. Tāpiri 5 ki te i\sqrt{39}.

$$x=\frac{5+\sqrt{39}i}{16}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{39}i}{16} ina he tango te ±. Tango i\sqrt{39} mai i 5.

$$x=\frac{-\sqrt{39}i+5}{16}$$

Kua oti te whārite te whakatau.

$$x=\frac{5+\sqrt{39}i}{16} x=\frac{-\sqrt{39}i+5}{16}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 5x mai i ngā taha e rua.

$$8x^{2}-5x=-2$$

Whakawehea ngā taha e rua ki te 8.

$$\frac{8x^{2}-5x}{8}=-\frac{2}{8}$$

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

$$x^{2}-\frac{5}{8}x=-\frac{2}{8}$$

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

$$x^{2}-\frac{5}{8}x=-\frac{1}{4}$$

Whakawehea te -\frac{5}{8}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{16}. Nā, tāpiria te pūrua o te -\frac{5}{16} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{5}{8}x+\left(-\frac{5}{16}\right)^{2}=-\frac{1}{4}+\left(-\frac{5}{16}\right)^{2}$$

Pūruatia -\frac{5}{16} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{5}{8}x+\frac{25}{256}=-\frac{1}{4}+\frac{25}{256}$$

Tāpiri -\frac{1}{4} ki te \frac{25}{256} 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.

$$x^{2}-\frac{5}{8}x+\frac{25}{256}=-\frac{39}{256}$$

Tauwehea x^{2}-\frac{5}{8}x+\frac{25}{256}. 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}.

$$\left(x-\frac{5}{16}\right)^{2}=-\frac{39}{256}$$

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

$$\sqrt{\left(x-\frac{5}{16}\right)^{2}}=\sqrt{-\frac{39}{256}}$$

Whakarūnātia.

$$x-\frac{5}{16}=\frac{\sqrt{39}i}{16} x-\frac{5}{16}=-\frac{\sqrt{39}i}{16}$$

Me tāpiri \frac{5}{16} ki ngā taha e rua o te whārite.

$$x=\frac{5+\sqrt{39}i}{16} x=\frac{-\sqrt{39}i+5}{16}$$