Whakaoti i te q+q^2=3/4

$$q + q ^ { 2 } = \frac { 3 } { 4 }$$

Whakaoti mō q

$$q = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

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

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.

$$q^{2}+q=\frac{3}{4}$$

Me tango \frac{3}{4} mai i ngā taha e rua o te whārite.

$$q^{2}+q-\frac{3}{4}=\frac{3}{4}-\frac{3}{4}$$

Mā te tango i te \frac{3}{4} i a ia ake anō ka toe ko te 0.

$$q^{2}+q-\frac{3}{4}=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 1 mō b, me -\frac{3}{4} mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-1±\sqrt{1^{2}-4\left(-\frac{3}{4}\right)}}{2}$$

Pūrua 1.

$$q=\frac{-1±\sqrt{1-4\left(-\frac{3}{4}\right)}}{2}$$

Whakareatia -4 ki te -\frac{3}{4}.

$$q=\frac{-1±\sqrt{1+3}}{2}$$

Tāpiri 1 ki te 3.

$$q=\frac{-1±\sqrt{4}}{2}$$

Tuhia te pūtakerua o te 4.

$$q=\frac{-1±2}{2}$$

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

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

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

$$q=-\frac{3}{2}$$

Kua oti te whārite te whakatau.

$$q=\frac{1}{2} q=-\frac{3}{2}$$

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.

$$q^{2}+q=\frac{3}{4}$$

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

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

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

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

Tāpiri \frac{3}{4} ki te \frac{1}{4} 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.

$$q^{2}+q+\frac{1}{4}=1$$

Tauwehea q^{2}+q+\frac{1}{4}. 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}.

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

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

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

Whakarūnātia.

$$q+\frac{1}{2}=1 q+\frac{1}{2}=-1$$

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$q=\frac{1}{2} q=-\frac{3}{2}$$