Whakaoti i te x^2+0.5x-0.75=0

$${ x }^{ 2 } +0.5x-0.75=0$$

Whakaoti mō x

$$x=\frac{\sqrt{13}-1}{4}\approx 0.651387819$$

$$x=\frac{-\sqrt{13}-1}{4}\approx -1.151387819$$

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.

$$x^{2}+\frac{1}{2}x-0.75=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, \frac{1}{2} mō b, me -0.75 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\left(-0.75\right)}}{2}$$

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\left(-0.75\right)}}{2}$$

Whakareatia -4 ki te -0.75.

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

Tāpiri \frac{1}{4} ki te 3.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{13}{4}}}{2}$$

Tuhia te pūtakerua o te \frac{13}{4}.

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{13}}{2}}{2}$$

Nā, me whakaoti te whārite x=\frac{-\frac{1}{2}±\frac{\sqrt{13}}{2}}{2} ina he tāpiri te ±. Tāpiri -\frac{1}{2} ki te \frac{\sqrt{13}}{2}.

$$x=\frac{\sqrt{13}-1}{2\times 2}$$

Whakawehe \frac{-1+\sqrt{13}}{2} ki te 2.

$$x=\frac{\sqrt{13}-1}{4}$$

Nā, me whakaoti te whārite x=\frac{-\frac{1}{2}±\frac{\sqrt{13}}{2}}{2} ina he tango te ±. Tango \frac{\sqrt{13}}{2} mai i -\frac{1}{2}.

$$x=\frac{-\sqrt{13}-1}{2\times 2}$$

Whakawehe \frac{-1-\sqrt{13}}{2} ki te 2.

$$x=\frac{-\sqrt{13}-1}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{13}-1}{4} x=\frac{-\sqrt{13}-1}{4}$$

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.

$$x^{2}+\frac{1}{2}x-0.75=0$$

Me tāpiri 0.75 ki ngā taha e rua o te whārite.

$$x^{2}+\frac{1}{2}x-0.75-\left(-0.75\right)=-\left(-0.75\right)$$

Mā te tango i te -0.75 i a ia ake anō ka toe ko te 0.

$$x^{2}+\frac{1}{2}x=-\left(-0.75\right)$$

Tango -0.75 mai i 0.

$$x^{2}+\frac{1}{2}x=0.75$$

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.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=0.75+\left(\frac{1}{4}\right)^{2}$$

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=0.75+\frac{1}{16}$$

Tāpiri 0.75 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.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{13}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{13}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{13}{16}}$$

Whakarūnātia.

$$x+\frac{1}{4}=\frac{\sqrt{13}}{4} x+\frac{1}{4}=-\frac{\sqrt{13}}{4}$$

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

$$x=\frac{\sqrt{13}-1}{4} x=\frac{-\sqrt{13}-1}{4}$$