Whakaoti i te 1/9x^2+1=4/9x

$$\frac { 1 } { 9 } x ^ { 2 } + 1 = \frac { 4 } { 9 } x$$

Whakaoti mō x (complex solution)

$$x=2+\sqrt{5}i\approx 2+2.236067977i$$

$$x=-\sqrt{5}i+2\approx 2-2.236067977i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te \frac{4}{9}x mai i ngā taha e rua.

$$\frac{1}{9}x^{2}+1-\frac{4}{9}x=0$$

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.

$$\frac{1}{9}x^{2}-\frac{4}{9}x+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi \frac{1}{9} mō a, -\frac{4}{9} mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūruatia -\frac{4}{9} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x=\frac{-\left(-\frac{4}{9}\right)±\sqrt{\frac{16}{81}-4\times \frac{1}{9}}}{2\times \frac{1}{9}}$$

Whakareatia -4 ki te \frac{1}{9}.

$$x=\frac{-\left(-\frac{4}{9}\right)±\sqrt{\frac{16}{81}-\frac{4}{9}}}{2\times \frac{1}{9}}$$

Tāpiri \frac{16}{81} ki te -\frac{4}{9} 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.

$$x=\frac{-\left(-\frac{4}{9}\right)±\sqrt{-\frac{20}{81}}}{2\times \frac{1}{9}}$$

Tuhia te pūtakerua o te -\frac{20}{81}.

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

Ko te tauaro o -\frac{4}{9} ko \frac{4}{9}.

$$x=\frac{\frac{4}{9}±\frac{2\sqrt{5}i}{9}}{2\times \frac{1}{9}}$$

Whakareatia 2 ki te \frac{1}{9}.

$$x=\frac{\frac{4}{9}±\frac{2\sqrt{5}i}{9}}{\frac{2}{9}}$$

Nā, me whakaoti te whārite x=\frac{\frac{4}{9}±\frac{2\sqrt{5}i}{9}}{\frac{2}{9}} ina he tāpiri te ±. Tāpiri \frac{4}{9} ki te \frac{2i\sqrt{5}}{9}.

$$x=\frac{4+2\sqrt{5}i}{\frac{2}{9}\times 9}$$

Whakawehe \frac{4+2i\sqrt{5}}{9} ki te \frac{2}{9} mā te whakarea \frac{4+2i\sqrt{5}}{9} ki te tau huripoki o \frac{2}{9}.

$$x=2+\sqrt{5}i$$

Nā, me whakaoti te whārite x=\frac{\frac{4}{9}±\frac{2\sqrt{5}i}{9}}{\frac{2}{9}} ina he tango te ±. Tango \frac{2i\sqrt{5}}{9} mai i \frac{4}{9}.

$$x=\frac{-2\sqrt{5}i+4}{\frac{2}{9}\times 9}$$

Whakawehe \frac{4-2i\sqrt{5}}{9} ki te \frac{2}{9} mā te whakarea \frac{4-2i\sqrt{5}}{9} ki te tau huripoki o \frac{2}{9}.

$$x=-\sqrt{5}i+2$$

Kua oti te whārite te whakatau.

$$x=2+\sqrt{5}i x=-\sqrt{5}i+2$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te \frac{4}{9}x mai i ngā taha e rua.

$$\frac{1}{9}x^{2}+1-\frac{4}{9}x=0$$

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$\frac{1}{9}x^{2}-\frac{4}{9}x=-1$$

Me whakarea ngā taha e rua ki te 9.

$$\frac{\frac{1}{9}x^{2}-\frac{4}{9}x}{\frac{1}{9}}=-\frac{1}{\frac{1}{9}}$$

Mā te whakawehe ki te \frac{1}{9} ka wetekia te whakareanga ki te \frac{1}{9}.

$$x^{2}+\left(-\frac{\frac{4}{9}}{\frac{1}{9}}\right)x=-\frac{1}{\frac{1}{9}}$$

Whakawehe -\frac{4}{9} ki te \frac{1}{9} mā te whakarea -\frac{4}{9} ki te tau huripoki o \frac{1}{9}.

$$x^{2}-4x=-\frac{1}{\frac{1}{9}}$$

Whakawehe -1 ki te \frac{1}{9} mā te whakarea -1 ki te tau huripoki o \frac{1}{9}.

$$x^{2}-4x=-9$$

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

$$x^{2}-4x+\left(-2\right)^{2}=-9+\left(-2\right)^{2}$$

Pūrua -2.

$$x^{2}-4x+4=-9+4$$

Tāpiri -9 ki te 4.

$$x^{2}-4x+4=-5$$

Tauwehea x^{2}-4x+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}.

$$\left(x-2\right)^{2}=-5$$

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

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{-5}$$

Whakarūnātia.

$$x-2=\sqrt{5}i x-2=-\sqrt{5}i$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$x=2+\sqrt{5}i x=-\sqrt{5}i+2$$