Solve [(sqrt{3}+sqrt{2})(2-sqrt{2})+(1/2sqrt{2}+sqrt{3})^2-3/2](sqrt{3}-sqrt{2})

$$[ ( \sqrt { 3 } + \sqrt { 2 } ) ( 2 - \sqrt { 2 } ) + ( \frac { 1 } { 2 } \sqrt { 2 } + \sqrt { 3 } ) ^ { 2 } - \frac { 3 } { 2 } ] ( \sqrt { 3 } - \sqrt { 2 } )$$

Evaluate

$$2$$

Solution Steps

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

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

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

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

The square of \sqrt{2} is 2.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}\sqrt{2}+\sqrt{3}\right)^{2}.

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

The square of \sqrt{2} is 2.

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

Multiply \frac{1}{4} and 2 to get \frac{1}{2}.

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

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

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

The square of \sqrt{3} is 3.

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

Add \frac{1}{2} and 3 to get \frac{7}{2}.

$$\left(2\sqrt{3}-\sqrt{6}+2\sqrt{2}-2+\frac{7}{2}+\sqrt{6}-\frac{3}{2}\right)\left(\sqrt{3}-\sqrt{2}\right)$$

Add -2 and \frac{7}{2} to get \frac{3}{2}.

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

Combine -\sqrt{6} and \sqrt{6} to get 0.

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

Subtract \frac{3}{2} from \frac{3}{2} to get 0.

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

Use the distributive property to multiply 2\sqrt{3}+2\sqrt{2} by \sqrt{3}-\sqrt{2} and combine like terms.

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

The square of \sqrt{2} is 2.

$$6-2\times 2$$

Multiply -2 and 2 to get -4.

$$6-4$$

Subtract 4 from 6 to get 2.

$$2$$

Factor

$$2$$