Solve frac{2+sqrt{5}}{2-sqrt{5}}+frac{2-sqrt{5}}{2+sqrt{5}}=a+sqrt{5b}

$$\frac{ 2+ \sqrt{ 5 } }{ 2- \sqrt{ 5 } } + \frac{ 2- \sqrt{ 5 } }{ 2+ \sqrt{ 5 } } = a+ \sqrt{ 5b }$$

Solve for b

$$b=\frac{\left(a+18\right)^{2}}{5}$$

$$a\leq -18$$

Steps for Solving Radical Equations

Rationalize the denominator of \frac{2+\sqrt{5}}{2-\sqrt{5}} by multiplying numerator and denominator by 2+\sqrt{5}.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Consider \left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 2. Square \sqrt{5}.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{4-5}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Subtract 5 from 4 to get -1.

$$\frac{\left(2+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{-1}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Multiply 2+\sqrt{5} and 2+\sqrt{5} to get \left(2+\sqrt{5}\right)^{2}.

$$\frac{\left(2+\sqrt{5}\right)^{2}}{-1}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{5}\right)^{2}.

$$\frac{4+4\sqrt{5}+\left(\sqrt{5}\right)^{2}}{-1}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

The square of \sqrt{5} is 5.

$$\frac{4+4\sqrt{5}+5}{-1}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Add 4 and 5 to get 9.

$$\frac{9+4\sqrt{5}}{-1}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Anything divided by -1 gives its opposite. To find the opposite of 9+4\sqrt{5}, find the opposite of each term.

$$-9-4\sqrt{5}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=a+\sqrt{5b}$$

Rationalize the denominator of \frac{2-\sqrt{5}}{2+\sqrt{5}} by multiplying numerator and denominator by 2-\sqrt{5}.

$$-9-4\sqrt{5}+\frac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}=a+\sqrt{5b}$$

Consider \left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$-9-4\sqrt{5}+\frac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{2^{2}-\left(\sqrt{5}\right)^{2}}=a+\sqrt{5b}$$

Square 2. Square \sqrt{5}.

$$-9-4\sqrt{5}+\frac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{4-5}=a+\sqrt{5b}$$

Subtract 5 from 4 to get -1.

$$-9-4\sqrt{5}+\frac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{-1}=a+\sqrt{5b}$$

Multiply 2-\sqrt{5} and 2-\sqrt{5} to get \left(2-\sqrt{5}\right)^{2}.

$$-9-4\sqrt{5}+\frac{\left(2-\sqrt{5}\right)^{2}}{-1}=a+\sqrt{5b}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-\sqrt{5}\right)^{2}.

$$-9-4\sqrt{5}+\frac{4-4\sqrt{5}+\left(\sqrt{5}\right)^{2}}{-1}=a+\sqrt{5b}$$

The square of \sqrt{5} is 5.

$$-9-4\sqrt{5}+\frac{4-4\sqrt{5}+5}{-1}=a+\sqrt{5b}$$

Add 4 and 5 to get 9.

$$-9-4\sqrt{5}+\frac{9-4\sqrt{5}}{-1}=a+\sqrt{5b}$$

Anything divided by -1 gives its opposite. To find the opposite of 9-4\sqrt{5}, find the opposite of each term.

$$-9-4\sqrt{5}-9+4\sqrt{5}=a+\sqrt{5b}$$

Subtract 9 from -9 to get -18.

$$-18-4\sqrt{5}+4\sqrt{5}=a+\sqrt{5b}$$

Combine -4\sqrt{5} and 4\sqrt{5} to get 0.

$$-18=a+\sqrt{5b}$$

Swap sides so that all variable terms are on the left hand side.

$$a+\sqrt{5b}=-18$$

Subtract a from both sides.

$$\sqrt{5b}=-18-a$$

Square both sides of the equation.

$$5b=\left(a+18\right)^{2}$$

Divide both sides by 5.

$$\frac{5b}{5}=\frac{\left(a+18\right)^{2}}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$b=\frac{\left(a+18\right)^{2}}{5}$$

Solve for a

$$a=-\left(\sqrt{5b}+18\right)$$

$$b\geq 0$$