Whakaoti i te -1/4x^2+2x-3=0

$$- \frac { 1 } { 4 } x ^ { 2 } + 2 x - 3 = 0$$

Whakaoti mō x

$$x=2$$

$$x=6$$

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.

$$-\frac{1}{4}x^{2}+2x-3=0$$

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

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

Pūrua 2.

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

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

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

Tāpiri 4 ki te -3.

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

Tuhia te pūtakerua o te 1.

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

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

$$x=\frac{-2±1}{-\frac{1}{2}}$$

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

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

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

$$x=2$$

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

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

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

$$x=6$$

Kua oti te whārite te whakatau.

$$x=2 x=6$$

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.

$$-\frac{1}{4}x^{2}+2x-3=0$$

Me tāpiri 3 ki ngā taha e rua o te whārite.

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

Mā te tango i te -3 i a ia ake anō ka toe ko te 0.

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

Tango -3 mai i 0.

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

Me whakarea ngā taha e rua ki te -4.

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

Mā te whakawehe ki te -\frac{1}{4} ka wetekia te whakareanga ki te -\frac{1}{4}.

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

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

$$x^{2}-8x=\frac{3}{-\frac{1}{4}}$$

Whakawehe 3 ki te -\frac{1}{4} mā te whakarea 3 ki te tau huripoki o -\frac{1}{4}.

$$x^{2}-8x=-12$$

Whakawehea te -8, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -4. Nā, tāpiria te pūrua o te -4 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-8x+\left(-4\right)^{2}=-12+\left(-4\right)^{2}$$

Pūrua -4.

$$x^{2}-8x+16=-12+16$$

Tāpiri -12 ki te 16.

$$x^{2}-8x+16=4$$

Tauwehea x^{2}-8x+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}.

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

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

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

Whakarūnātia.

$$x-4=2 x-4=-2$$

Me tāpiri 4 ki ngā taha e rua o te whārite.

$$x=6 x=2$$