Solve A/4quad4+a/2=frac{2+sqrt{a^2+4}}{a}

$$\frac { A } { 4 } \quad \frac { 4 + a } { 2 } = \frac { 2 + \sqrt { a ^ { 2 } + 4 } } { a }$$

Solve for A

$$A=\frac{8\left(\sqrt{a^{2}+4}+2\right)}{a\left(a+4\right)}$$

$$a\neq -4\text{ and }a\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4a, the least common multiple of 4,2,a.

$$aA\times \frac{4+a}{2}=4\left(2+\sqrt{a^{2}+4}\right)$$

Express a\times \frac{4+a}{2} as a single fraction.

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

Use the distributive property to multiply 4 by 2+\sqrt{a^{2}+4}.

$$\frac{a\left(4+a\right)}{2}A=8+4\sqrt{a^{2}+4}$$

Use the distributive property to multiply a by 4+a.

$$\frac{4a+a^{2}}{2}A=8+4\sqrt{a^{2}+4}$$

Divide each term of 4a+a^{2} by 2 to get 2a+\frac{1}{2}a^{2}.

$$\left(2a+\frac{1}{2}a^{2}\right)A=8+4\sqrt{a^{2}+4}$$

The equation is in standard form.

$$\left(\frac{a^{2}}{2}+2a\right)A=4\sqrt{a^{2}+4}+8$$

Divide both sides by 2a+\frac{1}{2}a^{2}.

$$\frac{\left(\frac{a^{2}}{2}+2a\right)A}{\frac{a^{2}}{2}+2a}=\frac{4\sqrt{a^{2}+4}+8}{\frac{a^{2}}{2}+2a}$$

Dividing by 2a+\frac{1}{2}a^{2} undoes the multiplication by 2a+\frac{1}{2}a^{2}.

$$A=\frac{4\sqrt{a^{2}+4}+8}{\frac{a^{2}}{2}+2a}$$

Divide 8+4\sqrt{a^{2}+4} by 2a+\frac{1}{2}a^{2}.

$$A=\frac{8\left(\sqrt{a^{2}+4}+2\right)}{a\left(a+4\right)}$$