Réitigh 1/a^4-4(b5/16a^2-1)=0

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

Réitigh do b_5.

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

$$a\neq 0$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Iolraigh an dá thaobh den chothromóid faoi 16a^{4}, an comhiolraí is lú de a^{4},16a^{2}.

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

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Méadaigh 1 faoi \frac{16a^{2}}{16a^{2}}.

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

Tá an t-ainmneoir céanna ag \frac{b_{5}}{16a^{2}} agus \frac{16a^{2}}{16a^{2}} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

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

Méadaigh 4 agus 16 chun 64 a fháil.

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

Scríobh 64\times \frac{b_{5}-16a^{2}}{16a^{2}} mar chodán aonair.

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

Cealaigh 16 mar uimhreoir agus ainmneoir.

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

Scríobh \frac{4\left(-16a^{2}+b_{5}\right)}{a^{2}}a^{4} mar chodán aonair.

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

Cealaigh a^{2} mar uimhreoir agus ainmneoir.

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

Úsáid an t-airí dáileach chun -4a^{2} a mhéadú faoi -16a^{2}+b_{5}.

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

Bain 16 ón dá thaobh. Is ionann rud ar bith a dhealaítear ó nialas agus a shéanadh.

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

Bain 64a^{4} ón dá thaobh.

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

Tá an chothromóid i bhfoirm chaighdeánach.

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

Roinn an dá thaobh faoi -4a^{2}.

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

Má roinntear é faoi -4a^{2} cuirtear an iolrúchán faoi -4a^{2} ar ceal.

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

Roinn -16-64a^{4} faoi -4a^{2}.

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

Réitigh do a. (complex solution)

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

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

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

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

Réitigh do a.

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

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

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

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