Solve frac{sqrt{5}-2sqrt{3}}{3sqrt{5}+sqrt{2}}*(frac{3sqrt{5}-sqrt{2}}{3sqrt{5}-sqrt{2}})

$$\frac{ \sqrt{ 5 } -2 \sqrt{ 3 } }{ 3 \sqrt{ 5 } + \sqrt{ 2 } } \times ( \frac{ 3 \sqrt{ 5 } - \sqrt{ 2 } }{ 3 \sqrt{ 5 } - \sqrt{ 2 } } )$$

Evaluate

$$\frac{2\sqrt{6}+15-\sqrt{10}-6\sqrt{15}}{43}\approx -0.151190657$$

Solution Steps

Divide 3\sqrt{5}-\sqrt{2} by 3\sqrt{5}-\sqrt{2} to get 1.

$$\frac{\sqrt{5}-2\sqrt{3}}{3\sqrt{5}+\sqrt{2}}\times 1$$

Rationalize the denominator of \frac{\sqrt{5}-2\sqrt{3}}{3\sqrt{5}+\sqrt{2}} by multiplying numerator and denominator by 3\sqrt{5}-\sqrt{2}.

$$\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}-\sqrt{2}\right)}{\left(3\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{5}-\sqrt{2}\right)}\times 1$$

Consider \left(3\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{5}-\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}.

$$\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}-\sqrt{2}\right)}{\left(3\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}}\times 1$$

Expand \left(3\sqrt{5}\right)^{2}.

$$\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}-\sqrt{2}\right)}{3^{2}\left(\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}}\times 1$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}-\sqrt{2}\right)}{9\left(\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}}\times 1$$

The square of \sqrt{5} is 5.

$$\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}-\sqrt{2}\right)}{9\times 5-\left(\sqrt{2}\right)^{2}}\times 1$$

Multiply 9 and 5 to get 45.

$$\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}-\sqrt{2}\right)}{45-\left(\sqrt{2}\right)^{2}}\times 1$$

The square of \sqrt{2} is 2.

$$\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}-\sqrt{2}\right)}{45-2}\times 1$$

Subtract 2 from 45 to get 43.

$$\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}-\sqrt{2}\right)}{43}\times 1$$

Express \frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}-\sqrt{2}\right)}{43}\times 1 as a single fraction.

$$\frac{\left(\sqrt{5}-2\sqrt{3}\right)\left(3\sqrt{5}-\sqrt{2}\right)}{43}$$

Apply the distributive property by multiplying each term of \sqrt{5}-2\sqrt{3} by each term of 3\sqrt{5}-\sqrt{2}.

$$\frac{3\left(\sqrt{5}\right)^{2}-\sqrt{5}\sqrt{2}-6\sqrt{3}\sqrt{5}+2\sqrt{3}\sqrt{2}}{43}$$

The square of \sqrt{5} is 5.

$$\frac{3\times 5-\sqrt{5}\sqrt{2}-6\sqrt{3}\sqrt{5}+2\sqrt{3}\sqrt{2}}{43}$$

Multiply 3 and 5 to get 15.

$$\frac{15-\sqrt{5}\sqrt{2}-6\sqrt{3}\sqrt{5}+2\sqrt{3}\sqrt{2}}{43}$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{15-\sqrt{10}-6\sqrt{3}\sqrt{5}+2\sqrt{3}\sqrt{2}}{43}$$

To multiply \sqrt{3} and \sqrt{5}, multiply the numbers under the square root.

$$\frac{15-\sqrt{10}-6\sqrt{15}+2\sqrt{3}\sqrt{2}}{43}$$

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

$$\frac{15-\sqrt{10}-6\sqrt{15}+2\sqrt{6}}{43}$$