Whakaoti i te x^2+3x-17=0

$$x ^ { 2 } + 3 x - 17 = 0$$

Whakaoti mō x

$$x = \frac{\sqrt{77} - 3}{2} \approx 2.887482194$$

$$x=\frac{-\sqrt{77}-3}{2}\approx -5.887482194$$

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.

$$x^{2}+3x-17=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 3 mō b, me -17 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-17\right)}}{2}$$

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\left(-17\right)}}{2}$$

Whakareatia -4 ki te -17.

$$x=\frac{-3±\sqrt{9+68}}{2}$$

Tāpiri 9 ki te 68.

$$x=\frac{-3±\sqrt{77}}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{77}}{2} ina he tāpiri te ±. Tāpiri -3 ki te \sqrt{77}.

$$x=\frac{\sqrt{77}-3}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{77}}{2} ina he tango te ±. Tango \sqrt{77} mai i -3.

$$x=\frac{-\sqrt{77}-3}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{77}-3}{2} x=\frac{-\sqrt{77}-3}{2}$$

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.

$$x^{2}+3x-17=0$$

Me tāpiri 17 ki ngā taha e rua o te whārite.

$$x^{2}+3x-17-\left(-17\right)=-\left(-17\right)$$

Mā te tango i te -17 i a ia ake anō ka toe ko te 0.

$$x^{2}+3x=-\left(-17\right)$$

Tango -17 mai i 0.

$$x^{2}+3x=17$$

Whakawehea te 3, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{2}. Nā, tāpiria te pūrua o te \frac{3}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=17+\left(\frac{3}{2}\right)^{2}$$

Pūruatia \frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+3x+\frac{9}{4}=17+\frac{9}{4}$$

Tāpiri 17 ki te \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{77}{4}$$

Tauwehea x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{77}{4}$$

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

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{77}{4}}$$

Whakarūnātia.

$$x+\frac{3}{2}=\frac{\sqrt{77}}{2} x+\frac{3}{2}=-\frac{\sqrt{77}}{2}$$

Me tango \frac{3}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{77}-3}{2} x=\frac{-\sqrt{77}-3}{2}$$