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

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

Evaluate

$$5\sqrt{3}-6\sqrt{2}\approx 0.174972664$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

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

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

The square of \sqrt{3} is 3.

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

Add 2 and 3 to get 5.

$$\sqrt{3}\left(5-2\sqrt{6}\right)$$

Use the distributive property to multiply \sqrt{3} by 5-2\sqrt{6}.

$$5\sqrt{3}-2\sqrt{3}\sqrt{6}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$5\sqrt{3}-2\sqrt{3}\sqrt{3}\sqrt{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$5\sqrt{3}-2\times 3\sqrt{2}$$

Multiply -2 and 3 to get -6.

$$5\sqrt{3}-6\sqrt{2}$$