Whakaoti i te (x)=2/3x(2x+9)-5x+1

$$( x ) = \frac { 2 } { 3 } x ( 2 x + 9 ) - 5 x + 1$$

Whakaoti mō x (complex solution)

$$x=-\frac{\sqrt{3}i}{2}\approx -0-0.866025404i$$

$$x=\frac{\sqrt{3}i}{2}\approx 0.866025404i$$

Ngā Upane mā te Kimi Pūtakerua

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{2}{3}x ki te 2x+9.

$$x=\frac{2}{3}x\times 2x+\frac{2}{3}x\times 9-5x+1$$

Whakareatia te x ki te x, ka x^{2}.

$$x=\frac{2}{3}x^{2}\times 2+\frac{2}{3}x\times 9-5x+1$$

Tuhia te \frac{2}{3}\times 2 hei hautanga kotahi.

$$x=\frac{2\times 2}{3}x^{2}+\frac{2}{3}x\times 9-5x+1$$

Whakareatia te 2 ki te 2, ka 4.

$$x=\frac{4}{3}x^{2}+\frac{2}{3}x\times 9-5x+1$$

Tuhia te \frac{2}{3}\times 9 hei hautanga kotahi.

$$x=\frac{4}{3}x^{2}+\frac{2\times 9}{3}x-5x+1$$

Whakareatia te 2 ki te 9, ka 18.

$$x=\frac{4}{3}x^{2}+\frac{18}{3}x-5x+1$$

Whakawehea te 18 ki te 3, kia riro ko 6.

$$x=\frac{4}{3}x^{2}+6x-5x+1$$

Pahekotia te 6x me -5x, ka x.

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

Tangohia te \frac{4}{3}x^{2} mai i ngā taha e rua.

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

Tangohia te x mai i ngā taha e rua.

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

Pahekotia te x me -x, ka 0.

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

Me whakarea ngā taha e rua ki te -\frac{3}{4}, te tau utu o -\frac{4}{3}.

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

Whakareatia te 1 ki te -\frac{3}{4}, ka -\frac{3}{4}.

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

Kua oti te whārite te whakatau.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{2}{3}x ki te 2x+9.

$$x=\frac{2}{3}x\times 2x+\frac{2}{3}x\times 9-5x+1$$

Whakareatia te x ki te x, ka x^{2}.

$$x=\frac{2}{3}x^{2}\times 2+\frac{2}{3}x\times 9-5x+1$$

Tuhia te \frac{2}{3}\times 2 hei hautanga kotahi.

$$x=\frac{2\times 2}{3}x^{2}+\frac{2}{3}x\times 9-5x+1$$

Whakareatia te 2 ki te 2, ka 4.

$$x=\frac{4}{3}x^{2}+\frac{2}{3}x\times 9-5x+1$$

Tuhia te \frac{2}{3}\times 9 hei hautanga kotahi.

$$x=\frac{4}{3}x^{2}+\frac{2\times 9}{3}x-5x+1$$

Whakareatia te 2 ki te 9, ka 18.

$$x=\frac{4}{3}x^{2}+\frac{18}{3}x-5x+1$$

Whakawehea te 18 ki te 3, kia riro ko 6.

$$x=\frac{4}{3}x^{2}+6x-5x+1$$

Pahekotia te 6x me -5x, ka x.

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

Tangohia te \frac{4}{3}x^{2} mai i ngā taha e rua.

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

Tangohia te x mai i ngā taha e rua.

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

Pahekotia te x me -x, ka 0.

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

Tangohia te 1 mai i ngā taha e rua.

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

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{4}{3}\right)\left(-1\right)}}{2\left(-\frac{4}{3}\right)}$$

Pūrua 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{4}{3}\right)\left(-1\right)}}{2\left(-\frac{4}{3}\right)}$$

Whakareatia -4 ki te -\frac{4}{3}.

$$x=\frac{0±\sqrt{\frac{16}{3}\left(-1\right)}}{2\left(-\frac{4}{3}\right)}$$

Whakareatia \frac{16}{3} ki te -1.

$$x=\frac{0±\sqrt{-\frac{16}{3}}}{2\left(-\frac{4}{3}\right)}$$

Tuhia te pūtakerua o te -\frac{16}{3}.

$$x=\frac{0±\frac{4\sqrt{3}i}{3}}{2\left(-\frac{4}{3}\right)}$$

Whakareatia 2 ki te -\frac{4}{3}.

$$x=\frac{0±\frac{4\sqrt{3}i}{3}}{-\frac{8}{3}}$$

Nā, me whakaoti te whārite x=\frac{0±\frac{4\sqrt{3}i}{3}}{-\frac{8}{3}} ina he tāpiri te ±.

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

Nā, me whakaoti te whārite x=\frac{0±\frac{4\sqrt{3}i}{3}}{-\frac{8}{3}} ina he tango te ±.

$$x=\frac{\sqrt{3}i}{2}$$

Kua oti te whārite te whakatau.

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