Whakaoti i te 3x^2-8x+4x/x-2=5x+8

$$\frac{ 3 { x }^{ 2 } -8x+4x }{ x-2 } = 5x+8$$

Whakaoti mō x

$$x = \frac{\sqrt{33} - 1}{2} \approx 2.372281323$$

$$x=\frac{-\sqrt{33}-1}{2}\approx -3.372281323$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 2 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x-2.

$$3x^{2}-8x+4x=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Pahekotia te -8x me 4x, ka -4x.

$$3x^{2}-4x=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Whakamahia te āhuatanga tohatoha hei whakarea te 5x ki te x-2.

$$3x^{2}-4x=5x^{2}-10x+\left(x-2\right)\times 8$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te 8.

$$3x^{2}-4x=5x^{2}-10x+8x-16$$

Pahekotia te -10x me 8x, ka -2x.

$$3x^{2}-4x=5x^{2}-2x-16$$

Tangohia te 5x^{2} mai i ngā taha e rua.

$$3x^{2}-4x-5x^{2}=-2x-16$$

Pahekotia te 3x^{2} me -5x^{2}, ka -2x^{2}.

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

Me tāpiri te 2x ki ngā taha e rua.

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

Pahekotia te -4x me 2x, ka -2x.

$$-2x^{2}-2x=-16$$

Me tāpiri te 16 ki ngā taha e rua.

$$-2x^{2}-2x+16=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -2 mō a, -2 mō b, me 16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-2\right)\times 16}}{2\left(-2\right)}$$

Pūrua -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-2\right)\times 16}}{2\left(-2\right)}$$

Whakareatia -4 ki te -2.

$$x=\frac{-\left(-2\right)±\sqrt{4+8\times 16}}{2\left(-2\right)}$$

Whakareatia 8 ki te 16.

$$x=\frac{-\left(-2\right)±\sqrt{4+128}}{2\left(-2\right)}$$

Tāpiri 4 ki te 128.

$$x=\frac{-\left(-2\right)±\sqrt{132}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 132.

$$x=\frac{-\left(-2\right)±2\sqrt{33}}{2\left(-2\right)}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{33}}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{2±2\sqrt{33}}{-4}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{33}}{-4} ina he tāpiri te ±. Tāpiri 2 ki te 2\sqrt{33}.

$$x=\frac{2\sqrt{33}+2}{-4}$$

Whakawehe 2+2\sqrt{33} ki te -4.

$$x=\frac{-\sqrt{33}-1}{2}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{33}}{-4} ina he tango te ±. Tango 2\sqrt{33} mai i 2.

$$x=\frac{2-2\sqrt{33}}{-4}$$

Whakawehe 2-2\sqrt{33} ki te -4.

$$x=\frac{\sqrt{33}-1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{33}-1}{2} x=\frac{\sqrt{33}-1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki 2 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x-2.

$$3x^{2}-8x+4x=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Pahekotia te -8x me 4x, ka -4x.

$$3x^{2}-4x=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Whakamahia te āhuatanga tohatoha hei whakarea te 5x ki te x-2.

$$3x^{2}-4x=5x^{2}-10x+\left(x-2\right)\times 8$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-2 ki te 8.

$$3x^{2}-4x=5x^{2}-10x+8x-16$$

Pahekotia te -10x me 8x, ka -2x.

$$3x^{2}-4x=5x^{2}-2x-16$$

Tangohia te 5x^{2} mai i ngā taha e rua.

$$3x^{2}-4x-5x^{2}=-2x-16$$

Pahekotia te 3x^{2} me -5x^{2}, ka -2x^{2}.

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

Me tāpiri te 2x ki ngā taha e rua.

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

Pahekotia te -4x me 2x, ka -2x.

$$-2x^{2}-2x=-16$$

Whakawehea ngā taha e rua ki te -2.

$$\frac{-2x^{2}-2x}{-2}=-\frac{16}{-2}$$

Mā te whakawehe ki te -2 ka wetekia te whakareanga ki te -2.

$$x^{2}+\left(-\frac{2}{-2}\right)x=-\frac{16}{-2}$$

Whakawehe -2 ki te -2.

$$x^{2}+x=-\frac{16}{-2}$$

Whakawehe -16 ki te -2.

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

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=8+\left(\frac{1}{2}\right)^{2}$$

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

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

Tāpiri 8 ki te \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{33}{4}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=\frac{33}{4}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{\sqrt{33}}{2} x+\frac{1}{2}=-\frac{\sqrt{33}}{2}$$

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

$$x=\frac{\sqrt{33}-1}{2} x=\frac{-\sqrt{33}-1}{2}$$