Whakaoti i te -1/2x^2+4-x=0

$$- \frac { 1 } { 2 } x ^ { 2 } + 4 - x = 0$$

Whakaoti mō x

$$x=-4$$

$$x=2$$

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.

$$-\frac{1}{2}x^{2}-x+4=0$$

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

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

Whakareatia -4 ki te -\frac{1}{2}.

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

Whakareatia 2 ki te 4.

$$x=\frac{-\left(-1\right)±\sqrt{1+8}}{2\left(-\frac{1}{2}\right)}$$

Tāpiri 1 ki te 8.

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

Tuhia te pūtakerua o te 9.

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

Ko te tauaro o -1 ko 1.

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

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

$$x=\frac{1±3}{-1}$$

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

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

Whakawehe 4 ki te -1.

$$x=-4$$

Nā, me whakaoti te whārite x=\frac{1±3}{-1} ina he tango te ±. Tango 3 mai i 1.

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

Whakawehe -2 ki te -1.

$$x=2$$

Kua oti te whārite te whakatau.

$$x=-4 x=2$$

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.

$$-\frac{1}{2}x^{2}-x+4=0$$

Me tango 4 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 4 i a ia ake anō ka toe ko te 0.

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

Me whakarea ngā taha e rua ki te -2.

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

Mā te whakawehe ki te -\frac{1}{2} ka wetekia te whakareanga ki te -\frac{1}{2}.

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

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

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

Whakawehe -4 ki te -\frac{1}{2} mā te whakarea -4 ki te tau huripoki o -\frac{1}{2}.

$$x^{2}+2x=8$$

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

$$x^{2}+2x+1^{2}=8+1^{2}$$

Pūrua 1.

$$x^{2}+2x+1=8+1$$

Tāpiri 8 ki te 1.

$$x^{2}+2x+1=9$$

Tauwehea x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=9$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{9}$$

Whakarūnātia.

$$x+1=3 x+1=-3$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x=2 x=-4$$