Whakaoti i te 4x^2-5x-1=0

$$4 x ^ { 2 } - 5 x - 1 = 0$$

Whakaoti mō x

$$x = \frac{\sqrt{41} + 5}{8} \approx 1.42539053$$

$$x=\frac{5-\sqrt{41}}{8}\approx -0.17539053$$

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.

$$4x^{2}-5x-1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, -5 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 4\left(-1\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-\left(-5\right)±\sqrt{25-16\left(-1\right)}}{2\times 4}$$

Whakareatia -16 ki te -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+16}}{2\times 4}$$

Tāpiri 25 ki te 16.

$$x=\frac{-\left(-5\right)±\sqrt{41}}{2\times 4}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{41}}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{5±\sqrt{41}}{8}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{41}}{8} ina he tāpiri te ±. Tāpiri 5 ki te \sqrt{41}.

$$x=\frac{\sqrt{41}+5}{8}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{41}}{8} ina he tango te ±. Tango \sqrt{41} mai i 5.

$$x=\frac{5-\sqrt{41}}{8}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{41}+5}{8} x=\frac{5-\sqrt{41}}{8}$$

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.

$$4x^{2}-5x-1=0$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

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

Mā te tango i te -1 i a ia ake anō ka toe ko te 0.

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

Tango -1 mai i 0.

$$4x^{2}-5x=1$$

Whakawehea ngā taha e rua ki te 4.

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

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

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

Whakawehea te -\frac{5}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{8}. Nā, tāpiria te pūrua o te -\frac{5}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{5}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{1}{4}+\frac{25}{64}$$

Tāpiri \frac{1}{4} ki te \frac{25}{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.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{41}{64}$$

Tauwehea x^{2}-\frac{5}{4}x+\frac{25}{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}.

$$\left(x-\frac{5}{8}\right)^{2}=\frac{41}{64}$$

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

$$\sqrt{\left(x-\frac{5}{8}\right)^{2}}=\sqrt{\frac{41}{64}}$$

Whakarūnātia.

$$x-\frac{5}{8}=\frac{\sqrt{41}}{8} x-\frac{5}{8}=-\frac{\sqrt{41}}{8}$$

Me tāpiri \frac{5}{8} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{41}+5}{8} x=\frac{5-\sqrt{41}}{8}$$