Whakaoti i te 1/a^4-4(b5/16a^2-1)=0

$$\frac { 1 } { a ^ { 4 } } - 4 ( \frac { b 5 } { 16 a ^ { 2 } } - 1 ) = 0$$

Whakaoti mō b_5

$$b_{5}=16a^{2}+\frac{4}{a^{2}}$$

$$a\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Me whakarea ngā taha e rua o te whārite ki te 16a^{4}, arā, te tauraro pātahi he tino iti rawa te kitea o a^{4},16a^{2}.

$$16-4\left(\frac{b_{5}}{16a^{2}}-1\right)\times 16a^{4}=0$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 1 ki te \frac{16a^{2}}{16a^{2}}.

$$16-4\left(\frac{b_{5}}{16a^{2}}-\frac{16a^{2}}{16a^{2}}\right)\times 16a^{4}=0$$

Tā te mea he rite te tauraro o \frac{b_{5}}{16a^{2}} me \frac{16a^{2}}{16a^{2}}, me tango rāua mā te tango i ō raua taurunga.

$$16-4\times \frac{b_{5}-16a^{2}}{16a^{2}}\times 16a^{4}=0$$

Whakareatia te 4 ki te 16, ka 64.

$$16-64\times \frac{b_{5}-16a^{2}}{16a^{2}}a^{4}=0$$

Tuhia te 64\times \frac{b_{5}-16a^{2}}{16a^{2}} hei hautanga kotahi.

$$16-\frac{64\left(b_{5}-16a^{2}\right)}{16a^{2}}a^{4}=0$$

Me whakakore tahi te 16 i te taurunga me te tauraro.

$$16-\frac{4\left(-16a^{2}+b_{5}\right)}{a^{2}}a^{4}=0$$

Tuhia te \frac{4\left(-16a^{2}+b_{5}\right)}{a^{2}}a^{4} hei hautanga kotahi.

$$16-\frac{4\left(-16a^{2}+b_{5}\right)a^{4}}{a^{2}}=0$$

Me whakakore tahi te a^{2} i te taurunga me te tauraro.

$$16-4a^{2}\left(-16a^{2}+b_{5}\right)=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -4a^{2} ki te -16a^{2}+b_{5}.

$$16+64a^{4}-4a^{2}b_{5}=0$$

Tangohia te 16 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$64a^{4}-4a^{2}b_{5}=-16$$

Tangohia te 64a^{4} mai i ngā taha e rua.

$$-4a^{2}b_{5}=-16-64a^{4}$$

He hanga arowhānui tō te whārite.

$$\left(-4a^{2}\right)b_{5}=-64a^{4}-16$$

Whakawehea ngā taha e rua ki te -4a^{2}.

$$\frac{\left(-4a^{2}\right)b_{5}}{-4a^{2}}=\frac{-64a^{4}-16}{-4a^{2}}$$

Mā te whakawehe ki te -4a^{2} ka wetekia te whakareanga ki te -4a^{2}.

$$b_{5}=\frac{-64a^{4}-16}{-4a^{2}}$$

Whakawehe -16-64a^{4} ki te -4a^{2}.

$$b_{5}=16a^{2}+\frac{4}{a^{2}}$$

Whakaoti mō a (complex solution)

$$a=\frac{\sqrt{2\left(\sqrt{b_{5}^{2}-256}+b_{5}\right)}}{8}$$

$$a=-\frac{\sqrt{2\left(\sqrt{b_{5}^{2}-256}+b_{5}\right)}}{8}$$

$$a=-\frac{\sqrt{2\left(-\sqrt{b_{5}^{2}-256}+b_{5}\right)}}{8}$$

$$a=\frac{\sqrt{2\left(-\sqrt{b_{5}^{2}-256}+b_{5}\right)}}{8}$$

Whakaoti mō a

$$a=-\frac{\sqrt{2\left(-\sqrt{b_{5}^{2}-256}+b_{5}\right)}}{8}$$

$$a=\frac{\sqrt{2\left(-\sqrt{b_{5}^{2}-256}+b_{5}\right)}}{8}$$

$$a=\frac{\sqrt{2\left(\sqrt{b_{5}^{2}-256}+b_{5}\right)}}{8}$$

$$a=-\frac{\sqrt{2\left(\sqrt{b_{5}^{2}-256}+b_{5}\right)}}{8}\text{, }b_{5}\geq 16$$