Solve 20/2sqrt{3-sqrt{2}}

$$\frac { 20 } { 2 \sqrt { 3 } - \sqrt { 2 } }$$

Evaluate

$$2\sqrt{2}+4\sqrt{3}\approx 9.756630355$$

Solution Steps

Rationalize the denominator of \frac{20}{2\sqrt{3}-\sqrt{2}} by multiplying numerator and denominator by 2\sqrt{3}+\sqrt{2}.

$$\frac{20\left(2\sqrt{3}+\sqrt{2}\right)}{\left(2\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+\sqrt{2}\right)}$$

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

$$\frac{20\left(2\sqrt{3}+\sqrt{2}\right)}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

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

$$\frac{20\left(2\sqrt{3}+\sqrt{2}\right)}{2^{2}\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{20\left(2\sqrt{3}+\sqrt{2}\right)}{4\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{20\left(2\sqrt{3}+\sqrt{2}\right)}{4\times 3-\left(\sqrt{2}\right)^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{20\left(2\sqrt{3}+\sqrt{2}\right)}{12-\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{20\left(2\sqrt{3}+\sqrt{2}\right)}{12-2}$$

Subtract 2 from 12 to get 10.

$$\frac{20\left(2\sqrt{3}+\sqrt{2}\right)}{10}$$

Divide 20\left(2\sqrt{3}+\sqrt{2}\right) by 10 to get 2\left(2\sqrt{3}+\sqrt{2}\right).

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

Use the distributive property to multiply 2 by 2\sqrt{3}+\sqrt{2}.

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