Solve 1/sqrt{23-2sqrt{5}}-1/sqrt{23+2sqrt{5}}=asqrt{b}

$$\frac{ 1 }{ \sqrt{ 23 } -2 \sqrt{ 5 } } - \frac{ 1 }{ \sqrt{ 23 } +2 \sqrt{ 5 } } = a \sqrt{ b }$$

Solve for a (complex solution)

$$a=\frac{4\sqrt{5}b^{-\frac{1}{2}}}{3}$$

$$b\neq 0$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{1}{\sqrt{23}-2\sqrt{5}} by multiplying numerator and denominator by \sqrt{23}+2\sqrt{5}.

$$\frac{\sqrt{23}+2\sqrt{5}}{\left(\sqrt{23}-2\sqrt{5}\right)\left(\sqrt{23}+2\sqrt{5}\right)}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Consider \left(\sqrt{23}-2\sqrt{5}\right)\left(\sqrt{23}+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}.

$$\frac{\sqrt{23}+2\sqrt{5}}{\left(\sqrt{23}\right)^{2}-\left(-2\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

The square of \sqrt{23} is 23.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-\left(-2\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Expand \left(-2\sqrt{5}\right)^{2}.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-\left(-2\right)^{2}\left(\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-4\left(\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

The square of \sqrt{5} is 5.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-4\times 5}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Multiply 4 and 5 to get 20.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-20}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Subtract 20 from 23 to get 3.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Rationalize the denominator of \frac{1}{\sqrt{23}+2\sqrt{5}} by multiplying numerator and denominator by \sqrt{23}-2\sqrt{5}.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{\left(\sqrt{23}+2\sqrt{5}\right)\left(\sqrt{23}-2\sqrt{5}\right)}=a\sqrt{b}$$

Consider \left(\sqrt{23}+2\sqrt{5}\right)\left(\sqrt{23}-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}.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{\left(\sqrt{23}\right)^{2}-\left(2\sqrt{5}\right)^{2}}=a\sqrt{b}$$

The square of \sqrt{23} is 23.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-\left(2\sqrt{5}\right)^{2}}=a\sqrt{b}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-2^{2}\left(\sqrt{5}\right)^{2}}=a\sqrt{b}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-4\left(\sqrt{5}\right)^{2}}=a\sqrt{b}$$

The square of \sqrt{5} is 5.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-4\times 5}=a\sqrt{b}$$

Multiply 4 and 5 to get 20.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-20}=a\sqrt{b}$$

Subtract 20 from 23 to get 3.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{3}=a\sqrt{b}$$

Since \frac{\sqrt{23}+2\sqrt{5}}{3} and \frac{\sqrt{23}-2\sqrt{5}}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\sqrt{23}+2\sqrt{5}-\left(\sqrt{23}-2\sqrt{5}\right)}{3}=a\sqrt{b}$$

Do the multiplications in \sqrt{23}+2\sqrt{5}-\left(\sqrt{23}-2\sqrt{5}\right).

$$\frac{\sqrt{23}+2\sqrt{5}-\sqrt{23}+2\sqrt{5}}{3}=a\sqrt{b}$$

Do the calculations in \sqrt{23}+2\sqrt{5}-\sqrt{23}+2\sqrt{5}.

$$\frac{4\sqrt{5}}{3}=a\sqrt{b}$$

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

$$a\sqrt{b}=\frac{4\sqrt{5}}{3}$$

Multiply both sides of the equation by 3.

$$3a\sqrt{b}=4\sqrt{5}$$

The equation is in standard form.

$$3\sqrt{b}a=4\sqrt{5}$$

Divide both sides by 3\sqrt{b}.

$$\frac{3\sqrt{b}a}{3\sqrt{b}}=\frac{4\sqrt{5}}{3\sqrt{b}}$$

Dividing by 3\sqrt{b} undoes the multiplication by 3\sqrt{b}.

$$a=\frac{4\sqrt{5}}{3\sqrt{b}}$$

Divide 4\sqrt{5} by 3\sqrt{b}.

$$a=\frac{20\times \left(5b\right)^{-\frac{1}{2}}}{3}$$

Solve for a

$$a=\frac{4\sqrt{\frac{5}{b}}}{3}$$

$$b>0$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{1}{\sqrt{23}-2\sqrt{5}} by multiplying numerator and denominator by \sqrt{23}+2\sqrt{5}.

$$\frac{\sqrt{23}+2\sqrt{5}}{\left(\sqrt{23}-2\sqrt{5}\right)\left(\sqrt{23}+2\sqrt{5}\right)}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Consider \left(\sqrt{23}-2\sqrt{5}\right)\left(\sqrt{23}+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}.

$$\frac{\sqrt{23}+2\sqrt{5}}{\left(\sqrt{23}\right)^{2}-\left(-2\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

The square of \sqrt{23} is 23.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-\left(-2\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Expand \left(-2\sqrt{5}\right)^{2}.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-\left(-2\right)^{2}\left(\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-4\left(\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

The square of \sqrt{5} is 5.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-4\times 5}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Multiply 4 and 5 to get 20.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-20}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Subtract 20 from 23 to get 3.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Rationalize the denominator of \frac{1}{\sqrt{23}+2\sqrt{5}} by multiplying numerator and denominator by \sqrt{23}-2\sqrt{5}.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{\left(\sqrt{23}+2\sqrt{5}\right)\left(\sqrt{23}-2\sqrt{5}\right)}=a\sqrt{b}$$

Consider \left(\sqrt{23}+2\sqrt{5}\right)\left(\sqrt{23}-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}.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{\left(\sqrt{23}\right)^{2}-\left(2\sqrt{5}\right)^{2}}=a\sqrt{b}$$

The square of \sqrt{23} is 23.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-\left(2\sqrt{5}\right)^{2}}=a\sqrt{b}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-2^{2}\left(\sqrt{5}\right)^{2}}=a\sqrt{b}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-4\left(\sqrt{5}\right)^{2}}=a\sqrt{b}$$

The square of \sqrt{5} is 5.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-4\times 5}=a\sqrt{b}$$

Multiply 4 and 5 to get 20.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-20}=a\sqrt{b}$$

Subtract 20 from 23 to get 3.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{3}=a\sqrt{b}$$

Since \frac{\sqrt{23}+2\sqrt{5}}{3} and \frac{\sqrt{23}-2\sqrt{5}}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\sqrt{23}+2\sqrt{5}-\left(\sqrt{23}-2\sqrt{5}\right)}{3}=a\sqrt{b}$$

Do the multiplications in \sqrt{23}+2\sqrt{5}-\left(\sqrt{23}-2\sqrt{5}\right).

$$\frac{\sqrt{23}+2\sqrt{5}-\sqrt{23}+2\sqrt{5}}{3}=a\sqrt{b}$$

Do the calculations in \sqrt{23}+2\sqrt{5}-\sqrt{23}+2\sqrt{5}.

$$\frac{4\sqrt{5}}{3}=a\sqrt{b}$$

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

$$a\sqrt{b}=\frac{4\sqrt{5}}{3}$$

Multiply both sides of the equation by 3.

$$3a\sqrt{b}=4\sqrt{5}$$

The equation is in standard form.

$$3\sqrt{b}a=4\sqrt{5}$$

Divide both sides by 3\sqrt{b}.

$$\frac{3\sqrt{b}a}{3\sqrt{b}}=\frac{4\sqrt{5}}{3\sqrt{b}}$$

Dividing by 3\sqrt{b} undoes the multiplication by 3\sqrt{b}.

$$a=\frac{4\sqrt{5}}{3\sqrt{b}}$$

Divide 4\sqrt{5} by 3\sqrt{b}.

$$a=\frac{20}{3\sqrt{5b}}$$

Solve for b

$$b=\frac{80}{9a^{2}}$$

$$a>0$$

Steps for Solving Radical Equations

Rationalize the denominator of \frac{1}{\sqrt{23}-2\sqrt{5}} by multiplying numerator and denominator by \sqrt{23}+2\sqrt{5}.

$$\frac{\sqrt{23}+2\sqrt{5}}{\left(\sqrt{23}-2\sqrt{5}\right)\left(\sqrt{23}+2\sqrt{5}\right)}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Consider \left(\sqrt{23}-2\sqrt{5}\right)\left(\sqrt{23}+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}.

$$\frac{\sqrt{23}+2\sqrt{5}}{\left(\sqrt{23}\right)^{2}-\left(-2\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

The square of \sqrt{23} is 23.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-\left(-2\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Expand \left(-2\sqrt{5}\right)^{2}.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-\left(-2\right)^{2}\left(\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-4\left(\sqrt{5}\right)^{2}}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

The square of \sqrt{5} is 5.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-4\times 5}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Multiply 4 and 5 to get 20.

$$\frac{\sqrt{23}+2\sqrt{5}}{23-20}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Subtract 20 from 23 to get 3.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{1}{\sqrt{23}+2\sqrt{5}}=a\sqrt{b}$$

Rationalize the denominator of \frac{1}{\sqrt{23}+2\sqrt{5}} by multiplying numerator and denominator by \sqrt{23}-2\sqrt{5}.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{\left(\sqrt{23}+2\sqrt{5}\right)\left(\sqrt{23}-2\sqrt{5}\right)}=a\sqrt{b}$$

Consider \left(\sqrt{23}+2\sqrt{5}\right)\left(\sqrt{23}-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}.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{\left(\sqrt{23}\right)^{2}-\left(2\sqrt{5}\right)^{2}}=a\sqrt{b}$$

The square of \sqrt{23} is 23.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-\left(2\sqrt{5}\right)^{2}}=a\sqrt{b}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-2^{2}\left(\sqrt{5}\right)^{2}}=a\sqrt{b}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-4\left(\sqrt{5}\right)^{2}}=a\sqrt{b}$$

The square of \sqrt{5} is 5.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-4\times 5}=a\sqrt{b}$$

Multiply 4 and 5 to get 20.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{23-20}=a\sqrt{b}$$

Subtract 20 from 23 to get 3.

$$\frac{\sqrt{23}+2\sqrt{5}}{3}-\frac{\sqrt{23}-2\sqrt{5}}{3}=a\sqrt{b}$$

Since \frac{\sqrt{23}+2\sqrt{5}}{3} and \frac{\sqrt{23}-2\sqrt{5}}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\sqrt{23}+2\sqrt{5}-\left(\sqrt{23}-2\sqrt{5}\right)}{3}=a\sqrt{b}$$

Do the multiplications in \sqrt{23}+2\sqrt{5}-\left(\sqrt{23}-2\sqrt{5}\right).

$$\frac{\sqrt{23}+2\sqrt{5}-\sqrt{23}+2\sqrt{5}}{3}=a\sqrt{b}$$

Do the calculations in \sqrt{23}+2\sqrt{5}-\sqrt{23}+2\sqrt{5}.

$$\frac{4\sqrt{5}}{3}=a\sqrt{b}$$

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

$$a\sqrt{b}=\frac{4\sqrt{5}}{3}$$

Multiply both sides of the equation by 3.

$$3a\sqrt{b}=4\sqrt{5}$$

Divide both sides by 3a.

$$\frac{3a\sqrt{b}}{3a}=\frac{4\sqrt{5}}{3a}$$

Dividing by 3a undoes the multiplication by 3a.

$$\sqrt{b}=\frac{4\sqrt{5}}{3a}$$

Square both sides of the equation.

$$b=\frac{80}{9a^{2}}$$

Solve for b (complex solution)

$$b=\frac{80}{9a^{2}}$$

$$arg(\sqrt{\frac{1}{a^{2}}}a)<\pi \text{ and }a\neq 0$$