Solve frac{sqrt{2}-2sqrt{3}}{sqrt{2}+2sqrt{3}}*frac{3sqrt{2}-2sqrt{3}}{3sqrt{2}-2sqrt{3}}

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

Evaluate

$$\frac{2\sqrt{6}-7}{5}\approx -0.420204103$$

Solution Steps

Divide 3\sqrt{2}-2\sqrt{3} by 3\sqrt{2}-2\sqrt{3} to get 1.

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

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

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

Consider \left(\sqrt{2}+2\sqrt{3}\right)\left(\sqrt{2}-2\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

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

$$\frac{\left(\sqrt{2}\right)^{2}-4\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}\times 1$$

The square of \sqrt{2} is 2.

$$\frac{2-4\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}\times 1$$

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

$$\frac{2-4\sqrt{6}+4\left(\sqrt{3}\right)^{2}}{\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}\times 1$$

The square of \sqrt{3} is 3.

$$\frac{2-4\sqrt{6}+4\times 3}{\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}\times 1$$

Multiply 4 and 3 to get 12.

$$\frac{2-4\sqrt{6}+12}{\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}\times 1$$

Add 2 and 12 to get 14.

$$\frac{14-4\sqrt{6}}{\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}\times 1$$

The square of \sqrt{2} is 2.

$$\frac{14-4\sqrt{6}}{2-\left(2\sqrt{3}\right)^{2}}\times 1$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{14-4\sqrt{6}}{2-2^{2}\left(\sqrt{3}\right)^{2}}\times 1$$

Calculate 2 to the power of 2 and get 4.

$$\frac{14-4\sqrt{6}}{2-4\left(\sqrt{3}\right)^{2}}\times 1$$

The square of \sqrt{3} is 3.

$$\frac{14-4\sqrt{6}}{2-4\times 3}\times 1$$

Multiply 4 and 3 to get 12.

$$\frac{14-4\sqrt{6}}{2-12}\times 1$$

Subtract 12 from 2 to get -10.

$$\frac{14-4\sqrt{6}}{-10}\times 1$$

Express \frac{14-4\sqrt{6}}{-10}\times 1 as a single fraction.

$$\frac{14-4\sqrt{6}}{-10}$$

Factor

$$\frac{2 \sqrt{6} - 7}{5} = -0.4202041028867288$$