Solve 1/a^4-4(b5/16a^2-1)=0

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

Solve for b_5

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

$$a\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 16a^{4}, the least common multiple of a^{4},16a^{2}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{16a^{2}}{16a^{2}}.

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

Since \frac{b_{5}}{16a^{2}} and \frac{16a^{2}}{16a^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Multiply 4 and 16 to get 64.

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

Express 64\times \frac{b_{5}-16a^{2}}{16a^{2}} as a single fraction.

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

Cancel out 16 in both numerator and denominator.

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

Express \frac{4\left(-16a^{2}+b_{5}\right)}{a^{2}}a^{4} as a single fraction.

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

Cancel out a^{2} in both numerator and denominator.

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

Use the distributive property to multiply -4a^{2} by -16a^{2}+b_{5}.

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

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

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

Subtract 64a^{4} from both sides.

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

The equation is in standard form.

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

Divide both sides by -4a^{2}.

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

Dividing by -4a^{2} undoes the multiplication by -4a^{2}.

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

Divide -16-64a^{4} by -4a^{2}.

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

Solve for a (complex solution)

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

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

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

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

Solve for a

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

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

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

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