Solve frac{sqrt{2}+1}{sqrt{2}-1}-frac{sqrt{2}-1}{sqrt{2}+1}

$$\frac { \sqrt { 2 } + 1 } { \sqrt { 2 } - 1 } - \frac { \sqrt { 2 } - 1 } { \sqrt { 2 } + 1 }$$

Evaluate

$$4\sqrt{2}\approx 5.656854249$$

Solution Steps

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

$$\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{\sqrt{2}-1}{\sqrt{2}+1}$$

Consider \left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}\right)^{2}-1^{2}}-\frac{\sqrt{2}-1}{\sqrt{2}+1}$$

Square \sqrt{2}. Square 1.

$$\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{2-1}-\frac{\sqrt{2}-1}{\sqrt{2}+1}$$

Subtract 1 from 2 to get 1.

$$\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{1}-\frac{\sqrt{2}-1}{\sqrt{2}+1}$$

Anything divided by one gives itself.

$$\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)-\frac{\sqrt{2}-1}{\sqrt{2}+1}$$

Multiply \sqrt{2}+1 and \sqrt{2}+1 to get \left(\sqrt{2}+1\right)^{2}.

$$\left(\sqrt{2}+1\right)^{2}-\frac{\sqrt{2}-1}{\sqrt{2}+1}$$

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

$$\left(\sqrt{2}+1\right)^{2}-\frac{\left(\sqrt{2}-1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}$$

Consider \left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\sqrt{2}+1\right)^{2}-\frac{\left(\sqrt{2}-1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}\right)^{2}-1^{2}}$$

Square \sqrt{2}. Square 1.

$$\left(\sqrt{2}+1\right)^{2}-\frac{\left(\sqrt{2}-1\right)\left(\sqrt{2}-1\right)}{2-1}$$

Subtract 1 from 2 to get 1.

$$\left(\sqrt{2}+1\right)^{2}-\frac{\left(\sqrt{2}-1\right)\left(\sqrt{2}-1\right)}{1}$$

Anything divided by one gives itself.

$$\left(\sqrt{2}+1\right)^{2}-\left(\sqrt{2}-1\right)\left(\sqrt{2}-1\right)$$

Multiply \sqrt{2}-1 and \sqrt{2}-1 to get \left(\sqrt{2}-1\right)^{2}.

$$\left(\sqrt{2}+1\right)^{2}-\left(\sqrt{2}-1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+1\right)^{2}.

$$\left(\sqrt{2}\right)^{2}+2\sqrt{2}+1-\left(\sqrt{2}-1\right)^{2}$$

The square of \sqrt{2} is 2.

$$2+2\sqrt{2}+1-\left(\sqrt{2}-1\right)^{2}$$

Add 2 and 1 to get 3.

$$3+2\sqrt{2}-\left(\sqrt{2}-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}-1\right)^{2}.

$$3+2\sqrt{2}-\left(\left(\sqrt{2}\right)^{2}-2\sqrt{2}+1\right)$$

The square of \sqrt{2} is 2.

$$3+2\sqrt{2}-\left(2-2\sqrt{2}+1\right)$$

Add 2 and 1 to get 3.

$$3+2\sqrt{2}-\left(3-2\sqrt{2}\right)$$

To find the opposite of 3-2\sqrt{2}, find the opposite of each term.

$$3+2\sqrt{2}-3-\left(-2\sqrt{2}\right)$$

The opposite of -2\sqrt{2} is 2\sqrt{2}.

$$3+2\sqrt{2}-3+2\sqrt{2}$$

Subtract 3 from 3 to get 0.

$$2\sqrt{2}+2\sqrt{2}$$

Combine 2\sqrt{2} and 2\sqrt{2} to get 4\sqrt{2}.

$$4\sqrt{2}$$