Solve z/a^2+frac{1{a^2}}=4

$$\frac { z } { a ^ { 2 } + \frac { 1 } { a ^ { 2 } } } = 4$$

Solve for z

$$z=\frac{4\left(a^{4}+1\right)}{a^{2}}$$

$$a\neq 0$$

Steps for Solving Linear Equation

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

$$\frac{z}{\frac{a^{2}a^{2}}{a^{2}}+\frac{1}{a^{2}}}=4$$

Since \frac{a^{2}a^{2}}{a^{2}} and \frac{1}{a^{2}} have the same denominator, add them by adding their numerators.

$$\frac{z}{\frac{a^{2}a^{2}+1}{a^{2}}}=4$$

Do the multiplications in a^{2}a^{2}+1.

$$\frac{z}{\frac{a^{4}+1}{a^{2}}}=4$$

Divide z by \frac{a^{4}+1}{a^{2}} by multiplying z by the reciprocal of \frac{a^{4}+1}{a^{2}}.

$$\frac{za^{2}}{a^{4}+1}=4$$

Multiply both sides of the equation by a^{4}+1.

$$za^{2}=4\left(a^{4}+1\right)$$

Use the distributive property to multiply 4 by a^{4}+1.

$$za^{2}=4a^{4}+4$$

The equation is in standard form.

$$a^{2}z=4a^{4}+4$$

Divide both sides by a^{2}.

$$\frac{a^{2}z}{a^{2}}=\frac{4a^{4}+4}{a^{2}}$$

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

$$z=\frac{4a^{4}+4}{a^{2}}$$

Divide 4a^{4}+4 by a^{2}.

$$z=\frac{4\left(a^{4}+1\right)}{a^{2}}$$

Solve for a

$$a=-\frac{\sqrt{2\left(\sqrt{z^{2}-64}+z\right)}}{4}$$

$$a=\frac{\sqrt{2\left(\sqrt{z^{2}-64}+z\right)}}{4}$$

$$a=\frac{\sqrt{2\left(-\sqrt{z^{2}-64}+z\right)}}{4}$$

$$a=-\frac{\sqrt{2\left(-\sqrt{z^{2}-64}+z\right)}}{4}\text{, }z\geq 8$$