Solve (frac{sqrt{3}-sqrt{2}}{sqrt{3}+sqrt{2}})^2

$$( \frac { \sqrt { 3 } - \sqrt { 2 } } { \sqrt { 3 } + \sqrt { 2 } } ) ^ { 2 }$$

Evaluate

$$49-20\sqrt{6}\approx 0.010205144$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}-\sqrt{2}.

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

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

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

Multiply \sqrt{3}-\sqrt{2} and \sqrt{3}-\sqrt{2} to get \left(\sqrt{3}-\sqrt{2}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-\sqrt{2}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

The square of \sqrt{2} is 2.

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

Add 3 and 2 to get 5.

$$\left(5-2\sqrt{6}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-2\sqrt{6}\right)^{2}.

$$25-20\sqrt{6}+4\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$25-20\sqrt{6}+4\times 6$$

Multiply 4 and 6 to get 24.

$$25-20\sqrt{6}+24$$

Add 25 and 24 to get 49.

$$49-20\sqrt{6}$$

Expand

$$49-20\sqrt{6}$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}-\sqrt{2}.

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

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

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

Multiply \sqrt{3}-\sqrt{2} and \sqrt{3}-\sqrt{2} to get \left(\sqrt{3}-\sqrt{2}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-\sqrt{2}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

The square of \sqrt{2} is 2.

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

Add 3 and 2 to get 5.

$$\left(5-2\sqrt{6}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-2\sqrt{6}\right)^{2}.

$$25-20\sqrt{6}+4\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$25-20\sqrt{6}+4\times 6$$

Multiply 4 and 6 to get 24.

$$25-20\sqrt{6}+24$$

Add 25 and 24 to get 49.

$$49-20\sqrt{6}$$