Solve 1/sqrt{2-1}+sqrt{3}(sqrt{3}-sqrt{6})+sqrt{8}

$$\frac { 1 } { \sqrt { 2 } - 1 } + \sqrt { 3 } ( \sqrt { 3 } - \sqrt { 6 } ) + \sqrt { 8 }$$

Evaluate

$$4$$

Solution Steps

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

$$\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

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}.

$$\frac{\sqrt{2}+1}{\left(\sqrt{2}\right)^{2}-1^{2}}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

Square \sqrt{2}. Square 1.

$$\frac{\sqrt{2}+1}{2-1}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

Subtract 1 from 2 to get 1.

$$\frac{\sqrt{2}+1}{1}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

Anything divided by one gives itself.

$$\sqrt{2}+1+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\sqrt{2}+1+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+2\sqrt{2}$$

Combine \sqrt{2} and 2\sqrt{2} to get 3\sqrt{2}.

$$3\sqrt{2}+1+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)$$

Use the distributive property to multiply \sqrt{3} by \sqrt{3}-\sqrt{6}.

$$3\sqrt{2}+1+\left(\sqrt{3}\right)^{2}-\sqrt{3}\sqrt{6}$$

The square of \sqrt{3} is 3.

$$3\sqrt{2}+1+3-\sqrt{3}\sqrt{6}$$

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

$$3\sqrt{2}+1+3-\sqrt{3}\sqrt{3}\sqrt{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$3\sqrt{2}+1+3-3\sqrt{2}$$

Add 1 and 3 to get 4.

$$3\sqrt{2}+4-3\sqrt{2}$$

Combine 3\sqrt{2} and -3\sqrt{2} to get 0.

$$4$$

Factor

$$2^{2}$$