Whakaoti i te 1/10x^2-3/2x+5=0

$$\frac { 1 } { 10 } x ^ { 2 } - \frac { 3 } { 2 } x + 5 = 0$$

Whakaoti mō x

$$x=5$$

$$x=10$$

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.

$$\frac{1}{10}x^{2}-\frac{3}{2}x+5=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi \frac{1}{10} mō a, -\frac{3}{2} mō b, me 5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\left(-\frac{3}{2}\right)^{2}-4\times \frac{1}{10}\times 5}}{2\times \frac{1}{10}}$$

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

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-4\times \frac{1}{10}\times 5}}{2\times \frac{1}{10}}$$

Whakareatia -4 ki te \frac{1}{10}.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-\frac{2}{5}\times 5}}{2\times \frac{1}{10}}$$

Whakareatia -\frac{2}{5} ki te 5.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-2}}{2\times \frac{1}{10}}$$

Tāpiri \frac{9}{4} ki te -2.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{1}{4}}}{2\times \frac{1}{10}}$$

Tuhia te pūtakerua o te \frac{1}{4}.

$$x=\frac{-\left(-\frac{3}{2}\right)±\frac{1}{2}}{2\times \frac{1}{10}}$$

Ko te tauaro o -\frac{3}{2} ko \frac{3}{2}.

$$x=\frac{\frac{3}{2}±\frac{1}{2}}{2\times \frac{1}{10}}$$

Whakareatia 2 ki te \frac{1}{10}.

$$x=\frac{\frac{3}{2}±\frac{1}{2}}{\frac{1}{5}}$$

Nā, me whakaoti te whārite x=\frac{\frac{3}{2}±\frac{1}{2}}{\frac{1}{5}} ina he tāpiri te ±. Tāpiri \frac{3}{2} ki te \frac{1}{2} 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.

$$x=\frac{2}{\frac{1}{5}}$$

Whakawehe 2 ki te \frac{1}{5} mā te whakarea 2 ki te tau huripoki o \frac{1}{5}.

$$x=10$$

Nā, me whakaoti te whārite x=\frac{\frac{3}{2}±\frac{1}{2}}{\frac{1}{5}} ina he tango te ±. Tango \frac{1}{2} mai i \frac{3}{2} mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x=\frac{1}{\frac{1}{5}}$$

Whakawehe 1 ki te \frac{1}{5} mā te whakarea 1 ki te tau huripoki o \frac{1}{5}.

$$x=5$$

Kua oti te whārite te whakatau.

$$x=10 x=5$$

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.

$$\frac{1}{10}x^{2}-\frac{3}{2}x+5=0$$

Me tango 5 mai i ngā taha e rua o te whārite.

$$\frac{1}{10}x^{2}-\frac{3}{2}x+5-5=-5$$

Mā te tango i te 5 i a ia ake anō ka toe ko te 0.

$$\frac{1}{10}x^{2}-\frac{3}{2}x=-5$$

Me whakarea ngā taha e rua ki te 10.

$$\frac{\frac{1}{10}x^{2}-\frac{3}{2}x}{\frac{1}{10}}=-\frac{5}{\frac{1}{10}}$$

Mā te whakawehe ki te \frac{1}{10} ka wetekia te whakareanga ki te \frac{1}{10}.

$$x^{2}+\left(-\frac{\frac{3}{2}}{\frac{1}{10}}\right)x=-\frac{5}{\frac{1}{10}}$$

Whakawehe -\frac{3}{2} ki te \frac{1}{10} mā te whakarea -\frac{3}{2} ki te tau huripoki o \frac{1}{10}.

$$x^{2}-15x=-\frac{5}{\frac{1}{10}}$$

Whakawehe -5 ki te \frac{1}{10} mā te whakarea -5 ki te tau huripoki o \frac{1}{10}.

$$x^{2}-15x=-50$$

Whakawehea te -15, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{15}{2}. Nā, tāpiria te pūrua o te -\frac{15}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-15x+\left(-\frac{15}{2}\right)^{2}=-50+\left(-\frac{15}{2}\right)^{2}$$

Pūruatia -\frac{15}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-15x+\frac{225}{4}=-50+\frac{225}{4}$$

Tāpiri -50 ki te \frac{225}{4}.

$$x^{2}-15x+\frac{225}{4}=\frac{25}{4}$$

Tauwehea x^{2}-15x+\frac{225}{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}.

$$\left(x-\frac{15}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(x-\frac{15}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

$$x-\frac{15}{2}=\frac{5}{2} x-\frac{15}{2}=-\frac{5}{2}$$

Me tāpiri \frac{15}{2} ki ngā taha e rua o te whārite.

$$x=10 x=5$$