Solve frac{sqrt{2}}{frac{sqrt{6}+sqrt{2}}{4}}=b/frac{1{2}}

$$\frac { \sqrt { 2 } } { \frac { \sqrt { 6 } + \sqrt { 2 } } { 4 } } = \frac { b } { \frac { 1 } { 2 } }$$

Solve for b

$$b=\sqrt{3}-1\approx 0.732050808$$

Steps for Solving Linear Equation

Divide \sqrt{2} by \frac{\sqrt{6}+\sqrt{2}}{4} by multiplying \sqrt{2} by the reciprocal of \frac{\sqrt{6}+\sqrt{2}}{4}.

$$\frac{\sqrt{2}\times 4}{\sqrt{6}+\sqrt{2}}=\frac{b}{\frac{1}{2}}$$

Rationalize the denominator of \frac{\sqrt{2}\times 4}{\sqrt{6}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{6}-\sqrt{2}.

$$\frac{\sqrt{2}\times 4\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}=\frac{b}{\frac{1}{2}}$$

Consider \left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{2}\times 4\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{2}\right)^{2}}=\frac{b}{\frac{1}{2}}$$

Square \sqrt{6}. Square \sqrt{2}.

$$\frac{\sqrt{2}\times 4\left(\sqrt{6}-\sqrt{2}\right)}{6-2}=\frac{b}{\frac{1}{2}}$$

Subtract 2 from 6 to get 4.

$$\frac{\sqrt{2}\times 4\left(\sqrt{6}-\sqrt{2}\right)}{4}=\frac{b}{\frac{1}{2}}$$

Cancel out 4 and 4.

$$\sqrt{2}\left(\sqrt{6}-\sqrt{2}\right)=\frac{b}{\frac{1}{2}}$$

Use the distributive property to multiply \sqrt{2} by \sqrt{6}-\sqrt{2}.

$$\sqrt{2}\sqrt{6}-\left(\sqrt{2}\right)^{2}=\frac{b}{\frac{1}{2}}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\sqrt{2}\sqrt{2}\sqrt{3}-\left(\sqrt{2}\right)^{2}=\frac{b}{\frac{1}{2}}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$2\sqrt{3}-\left(\sqrt{2}\right)^{2}=\frac{b}{\frac{1}{2}}$$

The square of \sqrt{2} is 2.

$$2\sqrt{3}-2=\frac{b}{\frac{1}{2}}$$

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

$$\frac{b}{\frac{1}{2}}=2\sqrt{3}-2$$

The equation is in standard form.

$$2b=2\sqrt{3}-2$$

Divide both sides by 2.

$$\frac{2b}{2}=\frac{2\sqrt{3}-2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$b=\frac{2\sqrt{3}-2}{2}$$

Divide 2\sqrt{3}-2 by 2.

$$b=\sqrt{3}-1$$