2/sqrt{2/2}=b/frac{sqrt{2+sqrt{6}}{4}}-এর সমাধান করুন

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

Solve for b

$$b=\sqrt{3}+1\approx 2.732050808$$

Steps for Solving Linear Equation

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

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

Multiply 2 and 2 to get 4.

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

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

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

The square of \sqrt{2} is 2.

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

Divide 4\sqrt{2} by 2 to get 2\sqrt{2}.

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

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

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

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

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

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

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

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

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

Subtract 6 from 2 to get -4.

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

Cancel out -4 and -4.

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

Use the distributive property to multiply b\left(-1\right) by \sqrt{2}-\sqrt{6}.

$$2\sqrt{2}=-b\sqrt{2}+b\sqrt{6}$$

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

$$-b\sqrt{2}+b\sqrt{6}=2\sqrt{2}$$

Combine all terms containing b.

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

The equation is in standard form.

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

Divide both sides by -\sqrt{2}+\sqrt{6}.

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

Dividing by -\sqrt{2}+\sqrt{6} undoes the multiplication by -\sqrt{2}+\sqrt{6}.

$$b=\frac{2\sqrt{2}}{\sqrt{6}-\sqrt{2}}$$

Divide 2\sqrt{2} by -\sqrt{2}+\sqrt{6}.

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